Math 29 Problem Set Compilation [fixed]
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EXERCISE 9.1
1.
BASIC INTEGRATION FORMULAS
6π₯ 2 β 4π₯ + 5 ππ₯ =
6π₯ 3 3
β
4π₯ 2 2
7.
+ 5π₯ + π
= Factor, (x-c), c = 2 P(c) = 0 β the (x-c ) is the factor P(c) = 0 2 1 0 0 -8 2 4 8 12 4 0
= πππ β πππ + ππ + π
3.
=
π₯( π₯ β 1)ππ₯ =
= π₯ ππ₯ β π₯ππ₯
5.
π π ππ π
π π π π
β
(π 2 +2π+4)(πβ2) (πβ2)
= (π₯ 2 + 2π₯ + 4)ππ₯
π₯ π₯ β π₯ ππ₯ 3 2
=
π₯ 3 β8 ππ₯ π₯β2
+π
9.
2π₯ 2 +4π₯β3 ππ₯ π₯2
=
π₯3 3
=
ππ + π
+
2π₯ 2 2
+ 4π₯ + π
ππ + ππ + π
π₯ 4 β 2π₯ 3 + π₯ 2 ππ₯ 2
=
2+
4 π
= 2ππ₯ + = 2π₯ + 4
β
3 π2
4 ππ₯ π₯ ππ₯ π₯
ππ₯ β
3 ππ₯ π₯2
=
ππ π
π
β
ππ π
+
ππ π
+πͺ
3π₯ β1 ππ₯ β1
β
= ππ + ππππ +
= π₯ 2 ππ₯ β 2π₯ 3 ππ₯ + π₯ππ₯
π π
+π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
1
EXERCISE 9.2
1.
INTEGRATION BY SUBSTITUTION
2 β 3π₯ ππ₯
(2π₯+3)ππ₯ π₯ 2 +3π₯+4
5.
ππ’
Let u = 2 - 3xππ₯ = β3
Let u = π₯ 2 + 3π₯ + 4 ππ₯ = 2π₯ + 3
βππ’ = ππ₯ 3
ππ’ = (2π₯ + 3)ππ₯
= π’
1 2
ππ’ π’
=
ππ’ (β 3 )
1 =β 3
ππ’
= πππ’ + π
1 2
π’ ππ’
= π₯π§ ππ + ππ + π + π
3
1 2π₯ 2 3
= β3
+π π
=β
π πβππ π π
2
π₯ 2 ππ₯ (π₯ 3 β1)4
7. +π
3
ππ’
Let u = π₯ 3 β 1 ππ₯ = 3π₯ 2 4
3. π₯ (2π₯ β 1) ππ₯ Let u = 2π₯ 3 β 1
ππ’ = π₯ 2 ππ₯ 3 ππ’ 3 π₯4
=
ππ’ = 6π₯ 2 ππ₯
=
1 3
ππ’ = π₯ 2 ππ₯ 6
=
1 π’ β3 3 β3 π’ β3 β9
π’β4 +π
=
π₯ 2 (2π₯ 3 β 1)4 ππ₯
=
=
ππ’ (π’ )( ) 6
= β π(ππβπ)π + π
=
1 (π’4 )ππ’ 6
=
1 π’5 +π 6 5
4
+π π
π’5
= 30 + π =
(πππ β π)π +π ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
2
EXERCISE 9.2
INTEGRATION BY SUBSTITUTION 13. cos4 π₯ sin π₯ππ₯
ππ₯ π₯ππ 2 π₯
9.
ππ’
1
Let u = πππ₯ ππ₯ = π₯
= π’4 βππ’
=
1 ππ₯ ( ) ππ 2 π₯ π₯
= - π’4 ππ’
=
1 ππ’ π’2
=β
π’5 5
=
π’β2 ππ’
=β
ππ¨π¬ π π π
=
π’ β1 β1
= βsin π₯
ππ’ = βsin π₯ ππ₯
ππ₯ π₯
ππ’ =
ππ’ ππ₯
Let u = cos π₯
+π +π
+π
π
= β πππ + π 15. ππ₯ π π₯ β1
11.
Let u = 3π₯
Let u = π π₯ ππ’ = π π₯ ππ₯ 1 π’β1
= ( =
1 π’
β )ππ’
1 ππ’ π’β1
1 π’
β
ππ’
ππ’ =3 ππ₯ ππ’ = ππ₯ 3
=
ππ£ = ππ’ π’ 1 π’
ππ£ β
ππ’
=
Let v = π’ β 1
=
1 + 2 sin 3π₯ πππ 3π₯ππ₯
1 + 2 sin π’ πππ π’ ( 3 ) 1 3
1 + 2 sin π’ πππ π’ππ’
Let v =1 + 2 sin π’ 1 ππ£ π’
= πππ’ β πππ’ + π π’ =π’β1
ππ£ ππ’
=
; π’ = ππ₯
= lnβ‘|π’ β 1| β lnβ‘|π π₯ | + π = lnβ‘ (1 β π π₯ )
ππ£ 2
= 2πππ π’ ;
= πππ π’ππ’
1 + 2 sin 3π₯ πππ 3π₯ππ₯ 1
=
1 [ 3
ππ£
=
1 2π£ 2 6 3
=
(π+πππππ)π π
π£ 2 ( 2 )] 3
= π₯π§ π β ππ β π + π
+π π
+π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
3
EXERCISE 9.2
INTEGRATION BY SUBSTITUTION
π ππ 2 π₯ππ₯ ` π+π π‘πππ₯
17.
3π₯ 2 +14π₯+14 ππ₯ π₯+4
21.
π (π₯) π (π₯)
Let u = π + π π‘πππ₯
=
ππ’ ππ₯
* using synthetic division
=π
;
ππ’ π
= π ππ 2 π₯ππ₯
ππ’ π
=
π π₯ π π₯
π(π₯)
-4 3 14 13
π’
1 =π
=
sin π₯ πππ π₯
= π π₯ π π₯ +
-12 -8
ππ’ π’ π π₯π§ π
3 2 5 - R(x) π + πππππ + π
π π₯ = 3π₯ + 2 π₯ + 4 = πππππππππ‘ππ π(π₯) 5 ππ₯ π₯+4
= (3π₯ + 2) ππ₯ +
For the second integral : 2
19.
π‘ππ3π₯ π ππ 3π₯ππ₯ πππ‘ π’ = π₯ + 4
Let u = π‘ππ3π₯ ππ’ ππ₯
= 3π ππ 2 3π₯ ;
= π’
1 2
ππ’ 3
=
1 [ 3 3
3π₯ 2 2
=
1 3
=
π(πππ§ ππ)π π
=
]+π
ππ¦ = 1 ; ππ’ = ππ₯ ππ₯
= (3π₯ + 2)ππ₯ + 5 =[
ππ’ (3)
3 2π’ 2
= π ππ 2 3π₯ππ₯
;
πππ π
ππ’ π’
+ 2π₯ + 5πππ’ + π]
+ ππ + ππ₯π§β‘ (π + π) + π
3
2tan 3π₯ 2 3
+π
π
+π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
4
EXERCISE 9.2
INTEGRATION BY SUBSTITUTION
π₯ 5 β2π₯ 3 β2π₯ π₯ 2 +1
23.
ππ₯
π₯ 3 β 3π₯ π₯ 2 + 1 π₯ 5 β 2π₯ 3 β 2π₯ π₯5 + π₯3 β3π₯ 3 β 2π₯ β3π₯ 3 β 3π₯ π₯ π(π₯) dx = π(π₯)
π π₯ ππ₯ +
= π₯ 3 β 3π₯ ππ₯ + π₯4
=4 β
3π₯ 2 + 2
π (π₯) ππ₯ π(π₯) π₯ ππ₯ π₯ 2 +1
π₯ ππ₯ π₯ 2 +1
For the 2nd term Let u = x2+1 ππ’ = 2π₯ ππ₯ ππ’ = π₯ππ₯ 2 π₯4 =4
=
3π₯ 2 β 2
ππ π
β
+
πππ π
ππ’ 2
π’ π
+ π π₯π§ ππ + π + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
5
EXERCISE 9.3 1.
INTEGRATION OF TRIGONOMETRIC FUNCTIONS
π ππ5π₯π‘ππ5π₯ππ₯ πΏππ‘ π’ = 5π₯ ππ’ ππ₯
ππ’ 5
=5
=
cos 3 π₯ππ₯ 1+π πππ₯ . 1βπ πππ₯ 1+π πππ₯
7. =
(cos 3 π₯) 1+π πππ₯ ππ₯ (1βπ πππ₯ )(1+π πππ₯ )
=
(co s 3 π₯+cos 3 π₯π πππ₯ )ππ₯ 1βsin 2 π₯
=
cos 3 π₯ 1+π πππ₯ ππ₯ cos 3 π₯
= ππ₯
π πππ’π‘πππ’
ππ’ 5
1
= 5 π πππ’π‘πππ’ππ’ 1 =5
π πππ’ + π
π =π
πππππ + π
= πππ π₯ 1 + π πππ₯ ππ₯ = π πππ₯ + πππ π₯π πππ₯ππ₯ πΏππ‘ π’ = π πππ₯ ππ’ ππ₯
π πππ₯ +πππ π₯ π ππ 2 π₯
3.
ππ₯
=
π πππ₯ π ππ 2 π₯
=
1 π πππ₯
=
ππ πππ₯ +
= πππ π₯ ; ππ’ = πππ π₯ππ₯
= π πππ₯ + π’ππ’ πππ π₯ π ππ 2 π₯
ππ₯ +
ππ₯
ππ₯ + πππ‘π₯ππ ππ₯ππ₯
= π πππ₯ +
π’2 2
= ππππ +
π¬π’π§π π π
+π +π
πππ‘π₯ππ ππ₯ππ₯
= β ππ ππππ + ππππ β ππππ + π
9.
1 + π‘πππ₯ 2 ππ₯ = (1 + 2π‘πππ₯ + tan2 π₯)ππ₯
5.
ππ₯ 1 2
1
; Let u= 2 π₯
1 2
sin π₯ cot π₯ ππ’ ππ₯
=
= =2 =2 =2
1 2
2ππ’ = ππ₯
= [2π‘πππ₯ + (1 + tan2 π₯)]ππ₯ = 2 π‘πππ₯ππ₯ +
sec 2 π₯ππ₯
= 2 βππ|πππ π₯| + π‘πππ₯ + π
2ππ’ π πππ’πππ‘π’
= βπ ππ |ππππ| + ππππ + π
ππ’ π πππ’ πππ‘π’ ππ’ π πππ’ ( 1 πππ π’
πππ π’ π πππ’
)
(ππ’)
= 2 π πππ’ππ’ = πππ ππππ + ππππ + π DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
6
EXERCISE 9.3
INTEGRATION OF TRIGONOMETRIC FUNCTIONS
πππ 6π₯ππ₯ cos 2 3π₯
11.
Let u = 3x ; 2u = 6x ππ’ ππ₯
=3 ;
ππ’ 3
πππ 2π’
ππ’ 3
= =
= ππ₯
cos 2 π’ 2 3
4 sin 2 π₯ππ π 2 π₯ π ππ 2π₯πππ 2π₯
15.
1 πππ π’
πππ π’ πππ π’
ππ’
=
(4π πππ₯πππ π₯ )(π πππ₯πππ π₯ ) ππ₯ 2π πππ₯πππ π₯ πππ 2π₯
=
2π πππ₯πππ π₯ πππ 2π₯
=
π ππ 2π₯ πππ 2π₯
ππ’ ππ₯
2
ππ₯ πππ π₯
=
1 πππ π₯
= ππ₯
ππ’ 2
π‘πππ’ππ’
π = β π₯π§ πππππ + π π
π ππ 2π₯ππ₯ 2π πππ₯ππ π 2 π₯
=
.
1 2
=
2π πππ₯πππ π₯ππ₯ (2π πππ₯πππ π₯ )πππ π₯
ππ₯
ππ’ 2
=2
π πππ’ πππ π’
π
= π π₯π§ πππππ + πππππ + π
=
ππ₯
Let u = 2x
= 3 π πππ’ππ’
13.
ππ₯
ππ₯ π ππ 3π₯π‘ππ 3π₯
17.
Let u = 3x ππ₯
= π πππ₯ππ₯ = ππ ππππ + ππππ + π
ππ’ ππ₯
ππ’ 3
=3
= ππ₯
ππ’ 3
=
π πππ’π‘πππ’ 1
=3 ππ ππ’ + π π
= β π πππππ + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
7
EXERCISE 9.4
1.
ππ₯ π 2π₯
=
π β2π₯ dx
INTEGRATION OF EXPONENTIAL FUNCTIONS
ππ’ ππ’ πππ‘ π’ = 2π₯ ; = β2 ; β = ππ₯ ππ₯ 2 = π π’ (β =β
1 2
2ππ’ ) 3
2
π π’ ππ’
= 3 π π’ ππ’
1
1
π
=β π (πβππ ) + π
π π ππ 4π₯ πππ 4π₯ππ₯
2 3
=
π πππ π
ππ’ 4
1 4
=
π
π πππ’
1 4
)
1 =4
=
ππ£ + π
ππππππ π
53β2π₯ ππ₯
= 5π’ (β
πππ π’ππ’
π π£ ππ£
+π
πππ‘ π’ = 3 β 2π₯ ;
ππ£ πππ‘ π£ = π πππ’ ; = cos π’ ; ππ£ = πππ π’ππ’ ππ’ =
π2 +π
7.
ππ’ ππ’ πππ‘ π’ = 4π₯ ; =4; = ππ₯ ππ₯ 4 π π πππ’ πππ π’(
3π₯
=
=β 2π π’ + π
=
3π₯ 2ππ’ ; = ππ₯ 2 3
πππ‘ π’ = = ππ’ (
ππ’ ) 2
=β 2 π π’ + π
3.
3π₯
π 3π₯ ππ₯ = π 2 ππ₯
5.
ππ’ ππ’ = β2 ; β = ππ₯ ππ₯ 2
ππ’ ) 2
1
= β 2 5π’ ππ’ 1 53β2π₯ ππ 5
= β2 =β
ππβππ ππππ
+π
+π
+π 3π₯ 2π₯ ππ₯
9.
π π₯ π π₯ = (ππ)π₯ = 6π₯ ππ₯ ππ
= πππ + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
8
EXERCISE 9.5
1.
INTEGRATION OF HYPERBOLIC FUNCTIONS
π πππ 3π₯ β 1 ππ₯
π ππ π 2 πππ₯ ππ₯ π₯
5.
Let u = 3π₯ β 1 ππ’ ππ₯
πππ‘ π’ = πππ₯ ; ππ’ 3
= 3 ; ππ₯ =
=
ππ’
= 1 3
=
1 πππ π π’ππ’ 3
=
π ππππ π
π πππ2 π’ππ’
= π‘ππππ’ + π
π πππ π’ ( 3 )
=
ππ’ 1 ππ₯ = ; ππ’ = ππ₯ π₯ π₯
= ππππβ‘ (πππ) + π
π πππ π’ππ’ +c
ππ β π + π
7.
1
1
ππ ππ 2 π₯ πππ‘π 2 π₯ππ₯ 1
Let u = 2 π₯ ;
ππ’ ππ₯
=
1 2
; 2ππ’ = ππ₯
= 2 ππ πππ’ πππ‘ππ’ππ’ 3.
ππ ππ2 1 β π₯ 2 π₯ππ₯
= 2(βππ πππ’ + π)
Let u=1 β π₯ 2
= βπππππ π + π
ππ’ ππ₯
β
π π
= -2π₯
ππ’ = π₯ππ₯ 2 = ππ ππ2 π’(β =β
1 2
ππ’ ) 2
ππ ππ2 π’ππ’
1
=β 2 (βπππ‘ππ’ + π) =
π ππππ π
π β ππ + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
9
EXERCISE 9.6
APPLICATION OF INDEFINITE INTEGRATION
1. Given slope 3π₯ 2 + 4 ππ¦ = 3π₯ 2 + 4 ππ₯ ππ¦ = 3π₯ 2 + 4 ππ₯
π¦=
π = πππ + ππ + π
π₯+1 π¦β1
ππ¦ π₯ + 1 = ππ₯ π¦ β 1
ππ¦ = π¦2
π¦ 2 β 2π¦ =
ππ₯ π₯
β ln π₯ β
1 =π 4
β ln 1 β
1 =π 4
π=β
1 4
β ln π₯ β
π¦ β 1 ππ¦ = π₯2 2
through 1,4
1 β = πππ₯ + π 4
+ 4π₯ + π
3. Given slope
π¦2 , π₯
ππ¦ π¦ 2 = ππ₯ π₯
3π₯ 2 + 4 ππ₯
ππ¦ = 3π₯ 3 3
7. Given slope
π₯ + 1 ππ₯ +π₯+π 2
1 1 + = 0 4π¦ π¦ 4
β4π¦ ln π₯ β 4 + π¦ = 0 ππ π₯π§ π β π + π = π
π¦ 2 β 2π¦ = π₯ 2 + 2π₯ + 2π ππ β ππ + ππ + ππ + ππ = π
1 5. Given slope π₯π¦
ππ¦ 1 = ππ₯ π₯π¦
π
=
ππ¦ = ππ₯
π¦
1
π¦ β2 ππ¦ =
ππ₯
1
π¦2 ππ₯ π₯
π¦ππ¦ = π¦2 2
9. Given slope π¦, through 1,1
ln π₯ 2 2 π
+π 2
π = πππ + ππ
1 2
2π¦
=π₯+π 1
2
=π₯+π
When π₯ = 1 , π¦ = 1 2 1 =1+π ; π =1 2π¦
1
2
=π₯+π
ππ = π + π
2
π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
10
EXERCISE 9.6
APPLICATION OF INDEFINITE INTEGRATION
11. Given slope π₯ β2 , through 1,2
v = -32t + vo
ππ¦ 1 = 2 ππ₯ π₯
when t = 1 sec, s=h=48ft ππ₯ π₯2
ππ¦ =
1 π¦ =β +π π₯
h=-16t2+ vot + c1 48 = -16(1)2 + vo(1) + c2 64 - vo = c2 When t = 0, s = 0, c2 = 0
1 2=β +π 1
s = -16t2 + vot
2 = β1 + π
when t = 1 sec, s = 48
π=3
s = -16t2 + c1t 1 π₯
π¦ = β +3 x π₯π¦ = β1 + 3π₯ ππ β ππ + π = π
c1=64 s=-16t2 + 64t v = -32t + 64 @ max, v = 0
13. a=-32 ft/sec2 a=-2
0 = -32t + 64 32t=64 t = 2 sec
ππ¦ = β32 ππ‘ ππ£ =
48 = -16(1)2 + c1(1)
s = -16t2 + 64t β32ππ‘
s = -16(2)2 + 64(2) s = 64ft
v=-32t+c ππ = β32π‘ + π1 ππ‘ ππ =
(β32π‘ + π1 )ππ‘
s=16t2 + c1t + c2 when t = 0, v = vo v=-32t + c1 vo= -32(0) + c1 vo =c1 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
11
EXERCISE 9.6
APPLICATION OF INDEFINITE INTEGRATION
15. a = 32ft/sec2 a = 32 ππ£ = 32 ππ‘ ππ£ =
32ππ‘
v = 32t + c1 ππ = 32π‘ + π1 ππ‘ ππ =
32π‘ + π1 ππ‘
S = 16t2 + c1 + c2 when t = 0, v = 0 c1 = 0 v = 32t when t = 0 , s = 0 c2 = 0 s = 16t2 400 16
π‘=
π‘=
20 4
t = 5 sec v = vt *since it is a free falling body, its velocity is ( - ) vt = -32t vt = -32(5) vt = -160 ft/sec
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
12
EXERCISE 10.1
PRODUCT OF SINES AND COSINES
1. Κ sin 5π₯ sin π₯ ππ₯ =
5. Κ cos 3π₯ β 2π cos π₯ + π ππ₯ 1 = Κ[cos π’ + π£ + cosβ‘ (π’ β π£)]ππ₯ 2
2 sin π’ sin π£ ππ₯
πππ‘ π’ = 3π₯ β 2π
= [cos π’ β π₯ β cos(π’ + π£)]ππ₯ π’ = 5π₯
π£ =π₯+π
π£=π₯
1 = Κ[cos 5π₯ β π₯ β cosβ‘ (5π₯ + π₯)]ππ₯ 2 1 = Κ[cos 4π₯ β cos 6π₯]ππ₯ 2
= =
πππ ππ πππππ β +πͺ π ππ
β
1 sin 6π₯ 6
π’ + π£ = 3π₯ β 2π + π₯ + π = 4π₯ β π
ο·
1 = [Κ cos 4π₯ππ₯ β Κ cos 6π₯ππ₯ 2 1 1 [ sin 4π₯ 2 4
ο·
π’ β π£ = 3π₯ β 2π β π₯ + π = 2π₯ β 3π 1 = Κ[cos 4π₯ β π + cos(2π₯ β 3π)]ππ₯ 2
]+πΆ
πππ cos 4π₯ β π = cos 4π₯πππ π + π ππ4π₯π πππ = βπππ 4π₯ πππ cos 2π₯ β 3π = cos 2π₯πππ 3π + sin 2π₯π ππ 3π = β cos 2π₯
3. Κ sin 9π₯ β 3 cos π₯ + 5 ππ₯ 1 = Κ [sin 9x β 3 + x + 5 + sin 9π₯ β 3 β π₯ β 5 ππ₯ 2
1 = Κ[sin 5π₯ + 2 + sin(3π₯ β 8)]ππ₯ 2 πππ‘ π§ = 5π₯ + 2 ππ§ =5 ππ₯ ππ§ = ππ₯ 5 1
= 2 [β cos π§ =β
1 5
1 2
= Κ(cos 4π₯ β cos 2π₯)ππ₯ 1
1
1
= 2 [β 4 sin 4π₯ β 2 sin 2π₯] + πΆ π π = β π¬π’π§ ππ β π¬π’π§ ππ + πͺ π π
; πππ‘ π€ = 3π₯ β 8 ππ€ ; =3 ππ₯ ππ€ ; = ππ₯ 3
1
β 3 πππ π€] + πΆ
π π πππ ππ + π β πππ ππ β π + πͺ ππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
13
EXERCISE 10.1
7.
PRODUCT OF SINES AND COSINES
π
4 π ππ 8π₯ πππ 3π₯ππ₯
= 2Κ[sin 8π₯ + 3π₯ + π πππ₯ 8π₯ β 3π₯ ππ₯
5 = Κ[cos π’ β π£ β cosβ‘ (π’ + π£)]ππ₯ 2
= 2Κ[π ππ11π₯ + sin 5π₯]ππ₯
πππ‘ π’ = 4π₯ +
πππ‘ π’ = 11π₯ ; πππ‘ π£ = 5π₯
1
ππ’ = 11 ; ππ₯
ππ£ =5 ππ₯
ππ’ = ππ₯ ; 11
ππ£ = ππ₯ 5
1
= 2[β 11 cos 11π₯ β 5 cos 5π₯ ] + πΆ =β
π
9. 5 π ππ 4π₯ + 3 π ππ 2π₯ β 6 ππ₯
π 3
;
π
π
π
π
ο·
π’ β π£ = 4π₯ + 3 β 2π₯ β 6 = 2π₯ + π/2 π π£ = 2π₯ β 6
ο·
π’ + π£ = 4π₯ + 3 + 2π₯ β 6 = 6π₯ + π/6
π π ππ¨π¬ πππ β πππππ + πͺ ππ π
5 π π = Κ[πππ 2π₯ + β πππ 6π₯ + ]ππ₯ 2 2 6 ο·
π
πππ cos 2π₯ + 2 = cos 2π₯πππ = βπ ππ2π₯
ο·
πππ cos 6π₯ +
π π β sin 2π₯ sin 2 2
π 6
π π β π ππ 6π₯ π ππ 6 6 3 1 = πππ 6π₯ β π ππ 6π₯ 2 2 = πππ 6π₯πππ
5 3 1 = Κ[β π ππ 2π₯ β πππ 6π₯ + π ππ 6π₯ ]ππ₯ 2 2 2 5 1
3
2 2
12
= [ πππ 2π₯ β
=
π πππ ππ π
β
π ππ 6π₯ β
π π πππ ππ ππ
1 12
π ππ₯ 6π₯ + πΆ π
β ππ πππ ππ + πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
14
EXERCISE 10.2
POWER OF SINES AND COSINES
1. π ππ3 π₯πππ 4 π₯ππ₯; ππ¦ πΆππ π πΌ = =
=
1 π’5 π’7 β +πΆ 3 5 7
=
1 5 π’ 15
=
π π ππππ ππ β ππππ ππ + πͺ ππ ππ
π ππ4 π₯πππ 4 π₯π πππ₯ππ₯ (1 β πππ 2 π₯)2πππ 4 π₯π πππ₯ππ₯
=
(1 β 2πππ 2 π₯ + πππ 4 π₯)πππ 4 π₯π πππ₯ππ₯
=
(πππ 4 π₯ β 2πππ 6 π₯ + πππ 8 π₯)π πππ₯ππ₯
1
β 21 π’7 + πΆ
5.
π ππ4 π₯πππ 2 π₯ππ₯
Let u = cosx
=
(π ππ2 π₯)2πππ 2 π₯ππ₯
ππ’ = βπ πππ₯ ππ₯
=
1 β πππ 2π₯ 2 1 + πππ 2π₯ ( ) ππ₯ 2 2
-ππ’ = π πππ₯ππ₯ =
= - (π’4 β 2π’6 + π’8 )ππ’ = =
2π’ 6 7
π’5 5
π’9 9
+πΆ
=
π π π ππππ π β ππππ π β ππππ π + πͺ π π π
=
β
β
=
3. π ππ4 3π₯πππ 3 3π₯ππ₯ ; ππ¦ πΆππ π πΌπΌ =
4
2
π ππ 3π₯πππ 3π₯πππ 3π₯ππ₯
1 1 1 1 πππ 2π₯ β ( πππ 2π₯ + πππ 2 2π₯) + ππ₯ 4 2 4 2 2
1 8
1 β 2πππ 2π₯ + πππ 2 2π₯ 1 + πππ 2π₯ ππ₯
1 (1 β 2πππ 2π₯ + πππ 2 π₯ + πππ 2π₯ β 2πππ 2 2π₯ + πππ 3 2π₯)ππ₯ 8
=
1 8
π ππ4 3π₯ 1 β π ππ2 3π₯ πππ 3π₯ππ₯ (π ππ4 3π₯ β π ππ6 3π₯)πππ 3π₯ππ₯
4
(πππ 2 2π₯ 1 + πππ 2π₯ ππ₯ 2
1 (1 β πππ 2π₯ β πππ 2 2π₯ + πππ 3 2π₯)ππ₯ 8
= =
1 4
=
=
=
1 β 2πππ 2π₯ +
1 8
ππ₯ β
πππ 2 2π₯ππ₯ +
πππ 2π₯ππ₯ β
πππ 3 2π₯ππ₯
1
1
1
1
1
2
2
8
2
6
[π₯ β π ππ2π₯ β ( π₯ + π ππ4π₯ + π ππ2π₯ β π ππ3 2π₯]
π πππππ ππππ ππ β β +πͺ ππ ππ ππ
Let u = sin3x ππ’ ππ₯
=
= 3πππ 3π₯ ; ( π’4 β π’6 )
ππ’ 3
= πππ 3π₯ππ₯
ππ’ 3 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
15
EXERCISE 10.2
POWER OF SINES AND COSINES
7. ( π πππ₯ + πππ π₯)2dx =
(π πππ₯ + 2 π πππ₯ πππ π₯ + πππ 2 π₯)ππ₯
=
π πππ₯ππ₯ + 2
=
π πππ₯ππ₯ + 2 π ππ2 πππ π₯ππ₯ + (
1
πππ 2 π₯ππ₯
π ππ2 πππ π₯ππ₯ + 1
1+πππ 2π₯ )ππ₯ 2
Let u = sinx
=
π ππ2 2π₯ π ππ2π₯ ππ₯
=
1 β πππ 2 2π₯ π ππ2π₯ ππ₯
πππ‘ π’ = πππ 2π₯
= -πππ π₯ + 2
π’ ππ’ +
2 3 π’2 3 π
π
1 ππ₯ + 2
1 2
π₯
+2+ π
= -ππππ + π ππππ + π +
π ππ 2π₯ 4
πππππ π
πππ 2π₯ ππ₯ 2
1
βπ π πππππ + ππππ ππ + π π π
= πβ
π₯
ππππ
+π
π ππ7 π₯ πππ 2 π₯ ππ₯
=
π ππ7 π₯ πππ 2 π₯ πππ π₯ ππ₯
=
π ππ7 π₯ 1 β π ππ2 π₯ πππ π₯ππ₯
=
π ππ7 π₯ β π ππ9 π₯ πππ π₯ ππ₯ u=sinx du=cosxdx
=
π’7 β π’9 ππ’
πππ 2 2π₯ ππ₯
1
= 2 β 2 π ππ6π₯ β 5 πππ 5π₯ β πππ π₯ + 2 + 8 π ππ4π₯ + π πππππ πππππ β π β ππ
π’β
+πͺ
π ππ3π₯πππ 2π₯ ππ₯ +
1
ππ’ 2
=
ππ.
= π ππ2 3π₯ + 2π ππ3π₯πππ 2π₯ + πππ 2 2π₯ ππ₯
π₯
β
β1 2
+πΆ
9. (π ππ3π₯ + πππ 2π₯)2 ππ₯
= π ππ2 3π₯ ππ₯ + 2
π’3 3
; π·π’ = β2π ππ2π₯ ππ₯
=
ππ’ = πππ π₯ππ₯ π πππ₯ππ₯ + 2
1 β π’2
=
ππ’ = πππ π₯ ππ₯
=
π ππ3 2π₯ ππ₯
ππ.
πππππ + π +
π
= =
π’8 π’10 β +π 8 10 π π ππππ π β ππ πππππ π + π
π
11. πππ 2 4π₯ ππ₯ 1 + πππ 8π₯ ππ₯ 2
= 1
=2 =
1 + πππ 8π₯ ππ₯
π πππππ + +π π ππ DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
16
EXERCISE 10.3
POWER OF TANGENTS AND SECANTS
1. π‘ππ2 2π₯π ππ 4 2π₯ππ₯
5. ____ 2 π₯ππ₯ β πππ . π¦ = 3 π‘ππ
= π‘ππ2 2π₯π ππ 2 2π₯π ππ 2 2π₯ππ₯
πΉπππ π‘ππ πππ π πππ π‘πππ:
= π‘ππ2 2π₯(1 + π‘ππ2 2π₯)π ππ 2 2π₯ππ₯
ππ¦ π₯ π₯ π₯ = π‘ππ2 π ππ 2 β π ππ 2 + 1 ππ₯ 2 2 2
1
= (π‘ππ2 2π₯ + π‘ππ4 2π₯)π ππ 2 2π₯ππ₯
ππ’
=
1 π’3 2 3
=
+
ππππ ππ π
+
2
π₯ π₯ = π‘ππ2 (π‘ππ2 ) 2 2
(π’2 + π’4 )ππ’ π’5 5
π₯
β 2π‘ππ + π₯ + π
π₯ π₯ = π‘ππ2 (π ππ 2 β 1) 2 2
= (π’2 + π’4 )( 2 ) =
2
π₯ π₯ π₯ = π‘ππ2 π ππ 2 β π‘ππ2 2 2 2
ππ’ = π ππ 2 2π₯ππ₯ 2
1 2
3π₯
π₯ π₯ π₯ = π‘ππ2 π ππ 2 β (π ππ 2 β 1) 2 2 2
ππ’ = 2π ππ 2 2π₯ ππ₯
πππ‘ π’ = π‘ππ2π₯ ;
2
π₯ 2
ππ¦ = π‘ππ4 dx
+π
= πππππππππ, πππ πππππππ ππππ ππ "ππππ "
ππππ ππ
+π
ππ
7. (π πππ₯ + π‘ππ π₯)2 ππ₯ 3.
π‘πππ₯ π ππ 6 π₯ππ₯ ; πΆπ΄ππΈ πΌ
= (π ππ 2 π₯ + 2π πππ₯ π‘ππ π₯ + π‘ππ2 π₯) ππ₯
1
= π‘πππ₯ + 2π πππ₯ + π‘ππ2 π₯ ππ₯
= π‘ππ2π₯ π ππ 4 π₯π ππ 2 π₯ππ₯ 1
= π‘πππ₯ + 2π πππ₯ + (π ππ 2 π₯ β 1) ππ₯
1
= π‘πππ₯ + 2π πππ₯ + π‘πππ₯ β π₯ + π
= π‘ππ2 π₯(1 + π‘ππ2 π₯)2 π ππ 2 π₯ππ₯ = π‘ππ2 π₯(1 + 2π‘ππ2 π₯ + π‘ππ4 π₯)π ππ 2 π₯ππ₯ 1 2
5 2
= πππππ + πππππ β π + π
9 2
2
= (π‘ππ π₯ + 2π‘ππ π₯ + π‘ππ π₯)π ππ π₯ππ₯ πππ‘ π’ = π‘πππ₯ ; 1
ππ’ = π ππ 2 π₯ ; ππ’ = π ππ 2 π₯ ππ₯ ππ₯
5
9
= (π’2 π₯ + 2π’2 π₯ + π’2 π₯)ππ’ 3
7
4π’ 2 + 7
11
=
2π’ 2 3
=
πππππ π πππππ π + π π
π
+
2π’2 11 π
+π ππ
π πππ π π + ππ
+π DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
17
EXERCISE 10.3
POWER OF TANGENTS AND SECANTS
π ππ 3π₯
9. (π‘ππ 3π₯ )4 ππ₯ π ππ 4 3π₯ π‘ππ 4 3π₯
=
1
ππ₯
= π‘ππ3 π₯π ππ β2 π₯ππ₯
= π ππ 4 3π₯π‘ππβ4 3π₯ππ₯
3
= π‘ππ2 π₯π‘πππ₯π ππ β2 π₯π πππ₯ππ₯
= π ππ 2 3π₯π ππ 2 3π₯π‘ππβ4 3π₯ππ₯
3
= (π ππ 2 π₯ β 1)π‘πππ₯π ππ β2 π₯π πππ₯ππ₯
= π ππ 2 3π₯(1 + π‘ππ2 3π₯)π‘ππβ4 3π₯ππ₯ = (π‘ππβ4 3π₯ + π‘ππβ2 3π₯)π ππ 2 3π₯ππ₯ ππ’ πππ‘ π’ = π‘ππ 3π₯ ; = 3π ππ 2 3π₯ ππ₯ ππ’ = π ππ 2 3π₯ππ₯ 3
1
=
β
π’ β2 3
1 π‘ππ β3 3π₯ 3 β3
= β
β
ππ’ = π πππ₯π‘πππ₯ππ₯ 1
3
3
=
+π
2π’ 2 3
1
β 2π’β2 + π π
π‘ππ β1 3π₯ 3
ππππ ππ πππππ β π + π
ππ’ = π πππ₯π‘πππ₯ ππ₯
πππ‘ π’ = π ππ π₯ ;
1
1 π’ β3 β3
3
= (π ππ 2 π₯ β π ππ β2 π₯)π‘πππ₯π πππ₯ππ₯
= (π’2 β π’β2 )ππ’
=3 (π’β4 + π’β2 ) ππ’ =3
π‘ππ 3 π₯ ππ₯ π πππ₯
11.
+π
=
πππππ π β π
π π
+π
ππππ π
π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
18
EXERCISE 10.4
POWER OF COTANGENTS AND COSECANTS
1. πππ‘ 4 π₯ππ π 4 π₯ππ₯
= πππ‘ 4 π₯(1 + πππ‘ 2 π₯)ππ π 2 π₯ππ₯
1
= πππ‘ 2 3π₯ ππ π 2 3π₯ ππ π 2 3π₯ ππ₯
= (πππ‘ 4 π₯ + πππ‘ 6 π₯)ππ π 2 π₯ππ₯ πππ‘ π’ = πππ‘ π₯ ;
1
= πππ‘ 2 3π₯ 1 + πππ‘ 2 3π₯ ππ π 2 3π₯ ππ₯
ππ’ = βππ π 2 π₯ ; ππ’ = βππ π 2 π₯ππ₯ ππ₯
= β (π’4 + π’6 )ππ’ =-
π’5 5
= -
3.
+
π’7 7
5
ππ’ π π₯ = β3 ππ π 2 3π₯ ππ₯ ππ₯
c β
πππ‘ 5 4π₯ππ₯
= πππ‘ 3 4π₯πππ‘ 2 4π₯ππ₯ 3
1
= πππ‘ 2 3π₯ + πππ‘ 2 3π₯ ππ π 2 3π₯ ππ₯ πππ‘ π’ = πππ‘ 3π₯
+c
ππππ π ππππ π + + π π
πππ 3π₯ ππ π 4 3π₯ ππ₯
5.
2
= πππ‘ 4π₯(ππ π 4π₯ β 1)ππ₯
ππ’ = ππ π 2 3π₯ ππ₯ 3
=β
1 3
=β
1 3
1
5
π’2 + π’2 ππ’ 3
π’2 3 2
7
+
π’2 7 2
+π
π π π π = β ππππ ππ β ππππ ππ + π π ππ
= (πππ‘ 3 4π₯ππ π 2 4π₯ β πππ‘ 3 4π₯)ππ₯ = [πππ‘ 3 4π₯ππ π 2 4π₯ β (ππ π 2 4π₯ β 1)πππ‘4π₯]ππ₯ =
πππ‘ 3 4π₯ππ π2 4π₯ππ₯ β πππ‘4π₯ππ π2 4π₯ππ₯ β πππ‘4π₯ππ₯
πππ‘ π’ = πππ‘ 4π₯ ; 1
ππ’ = ππ π 2 4π₯ππ₯ β4
1
1
=β 4 π’3 ππ’ β 4 π’ππ’ + 4 ππβ‘(πππ 4π₯) 1 π’4 ππ₯ 4
=β 4
= β
β
π’2 2
1 4
+ ππ πππ 4π₯ + π
ππππ ππ ππππ ππ + ππ π
π
+ π ππ πππππ + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
19
EXERCISE 10.4
POWER OF COTANGENTS AND COSECANTS
7.
πππ 5 2π₯ππ₯ π ππ 8 2π₯
πππ 5 2π₯ππ₯ π ππ 5 2π₯
=
πππ‘ 5 2π₯ ππ π 3 2π₯ ππ₯ \
=
πππ‘ 4 2π₯ ππ π 2 2π₯ ππ π 2π₯ πππ‘ 2π₯ ππ₯
=
ππ π 2 2π₯ β 1
=
2
1 π ππ 3 2π₯
ππ₯
ππ π 2 2π₯ ππ π 2π₯ πππ‘ 2π₯ ππ₯
=
ππ π 4 2π₯ β 2 ππ π2 2π₯ + 1 ππ π 2 2π₯ ππ π 2π₯ πππ‘ 2π₯ ππ₯
=
ππ π 6 2π₯ β 2 ππ π 4 2π₯ + ππ π 2 2π₯ ππ π 2π₯ πππ‘ 2π₯ ππ₯
πππ‘ π’ = ππ π 2π₯
=β
1 2
π’6 β 2π’4 + π’2 ππ’
=β
1 π’7 2 7
β
= β
ππ₯
=
πππ‘ β6 π₯ ππ π 2 π₯ ππ π 2 π₯ ππ₯
=
πππ‘ β6 π₯ 1 + πππ‘ 2 π₯ ππ π 2 π₯ ππ₯
=
πππ‘ β6 π₯ + πππ‘ β4 π₯ ππ π 2 π₯ ππ₯
πππ‘: π’ = πππ‘ π₯ ππ’ = β ππ π 2 π₯ ππ₯ ππ₯
= β1
ππ’ = ππ π 2π₯ πππ‘ 2π₯ ππ₯ 2
2π’ 5 5
ππ π 4 π₯ πππ‘ 6 π₯
βππ’ = ππ π 2 π₯ππ₯
ππ’ π(π₯) = β2 ππ π 2π₯ πππ‘ 2π₯ ππ₯ ππ₯ β
9.
+
π’3 3
π’β6 + π’β4 ππ₯
= β1 β =
+π
ππππ ππ ππππ ππ ππππ ππ + β +π ππ π π
πππ‘ β5 π₯ 5
π’β5 π’β3 β +π 5 3
+
πππ‘ β3 π₯ 3
+π
ππππ π ππππ π = + +π π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
20
EXERCISE 10.5
1.
TRIGONOMETRIC SUBSTITUTIONS
π₯ 2 ππ₯
3.
4βπ₯ 2
ππ₯
π₯ = π’ ; π = 2 ; 22 β π₯ 2
π’ = 3π₯
π’ = π π ππ π π₯ π ππ π = 2 π₯ = 2π ππ π ππ₯ = 2πππ πππ
π=2
= =
π’ = ππ‘πππ 3π₯ = 2π‘πππ 2 3π₯ π₯ = π‘πππ ; π‘πππ = 3 2
π₯ 2 ππ₯ 4 β π₯2
2 ππ₯ = π ππ 2 πππ 3
4 π ππ π 2πππ πππ 2πππ π π ππ2 πππ
=4
1 β πππ 2πππ 2 ππ β
π₯ 2
= πππππππ
1 2
ππ₯
=
π₯ 9π₯ 2 + 4 2 π ππ 2 πππ 3
=
2π ππππππ π ]2 + C
π π βπ π π
9π₯ 2 + 4
2π πππ =
1 π₯ π ππ2π + πΆ ; π = π ππβ( ) 2 2
= 2[ππππ ππ β
9π₯ 2 + 4 2
π πππ =
=4
=2
;
π₯ 9π₯ 2 +4
π β ππ +πͺ π
2 π‘πππ2π πππ 3
π πππππ 2π‘πππ
= 1 = 2
1 πππ π π πππ πππ π
ππ
=
1 2
1 ππ π πππ
=
1 2
ππ ππππ
=
1 [-ππβ‘|ππ ππ 2
= β
π ππ π
+ πππ‘π|] + πΆ
πππ + π π β +πͺ ππ ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
21
EXERCISE 10.5
5.
π₯ 2 ππ₯ 3 9βπ₯ 2 2
7.
π₯ 2 ππ₯
=
=
TRIGONOMETRIC SUBSTITUTIONS
9β
9 β π₯2
3 ππ₯ = πππ πππ 2
π₯ = 3π πππ ππ₯ = 3πππ π
=
= =
2π₯
π = π ππβ1 ( )
(3π πππ)2 3 πππ π 9 β (3π πππ)2 3 πππ π 9π ππ2 π 1 β π ππ2 π
3
= =
π ππ2 π (1 β π ππ2 π)
=
π ππ2 π ππ πππ 2 π π‘ππ2 πππ β
3 π₯ = π πππ 2 2π₯ = π πππ 3
π’ = ππ πππ
=
ππ₯
π’ = ππ πππ ; 2π₯ = 3π πππ
π₯ 2 ππ₯
π’=π₯ ; π=3
=
π₯2
π = 3 ; π’ = 2π₯
π₯2 3
9 β π₯2
9β4π₯ 2
3 2
3πππ π ( πππ πππ ) 3 2
( π πππ )2 3πππ π (3πππ πππ ) 3 2
2( π πππ )2 9πππ 2 πππ 2(
9 ) 4π ππ 2 π
π ππ 2 πππ
= π‘πππ β π =
π
π β π¨πππππ + π π π β ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
22
EXERCISE 10.5
ππ₯ π₯ 2 +4 2
9.
TRIGONOMETRIC SUBSTITUTIONS
π₯ = 2 π‘πππ ;
ππ₯
11.
; π€ππππ: π’ = π₯ , π = 2
π₯ π₯ 2 β9
π = 3 ;π’ = π₯
π₯ π‘πππ = 2
π’ = ππ πππ π₯ = 3π πππ ; ππ₯ = 3π ππππ‘πππππ π₯ π₯ π πππ = ; π = π΄πππ ππ 3 3
π₯
π·π₯ = 2 π ππ 2 πππ ; π = ππππ‘ππ 2
x
π₯2 + 4
x π₯2 β 9
2 π₯2 + 4 2
π πππ =
π₯2 + 4
2π πππ =
3
2
4 π ππ 2 π = π₯ 2 + 4 π‘πππ =
2
2 π ππ πππ 4 π ππ 2 π 2
=
2 π ππ 2 πππ = 16 π ππ 4 π
= 1 8
πππ 2 πππ
=
1 8
1 + πππ 2π ππ 2
=
1 8
=
1 ππ + 2
1 1 π 2
=8 =
ππ 1 = 2 8 π ππ π 8
πππ 2π ππ 2
ππ π ππ 2 π
= =
π₯2 β 9 ; 3π‘πππ = 3 ππ₯
π₯ π₯2 β 9
=
π₯2 β 9
3π ππππ‘πππππ 3π πππ(3π‘πππ)
ππ 3
=
1 π 3
=
π π π¨πππππ + π π π
1
+ 4 (2)π ππππππ π + π
π π π½+ ππππ½ππππ½ + π ππ ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
23
EXERCISE 10.5
TRIGONOMETRIC SUBSTITUTIONS
ππ₯
3
π₯ 2 β 16 2 ππ. ( ) π₯3
ππ.
π’ = π₯; π = 4
5 β 12π₯ + 4π₯ 2 = 2π₯ β 9 β 4
π’ = π ππβ ; π₯ = 4π ππβ ; π ππβ =
π₯ 4
π’ = ππ ππβ ; 2π₯ β 3 = 2π ππβ 2π₯ β 3 = 2π ππβ ; 2π₯ = 2π ππβ + 3
π₯ 3 = 64 sec 3 β ; ππ₯ = 4π ππβ π‘ππβ πβ π₯ 2 β 16 ; 4π‘ππβ = 4
π₯ 2 β 16
=4 =4
2ππ₯ = 2π ππβ π‘ππβ ππ₯ = π ππβ π‘ππβ π ππβ =
3
=
5 β 12π₯ + 4π₯ 2
π = 2 ; π’ = 2π₯ β 3
π₯ β = ππππ ππβ 4
π‘ππβ =
2π₯ β 3
( 4π‘ππβ (4π ππβ π‘ππβ πβ )) (64sec3 )
2π₯ β 3 2
β = ππππ ππ
tan4 β πβ secβ‘^2β
π‘ππβ =
(sec 2 β1)^2πβ sec 2 β
2π‘ππβ =
2π₯ β 3 2
2π₯ β 3 2
2π₯ β 3
=4
sec 4 β2 sec 2 β + 1 sec 2 β
=
(π ππβ π‘ππβ ) 2π ππβ 2π‘ππβ
=4
sec 4 β β 2 sec 2 β + 1 sec 2 β
=
1 4
π ππβ π‘ππβ π ππβ π‘ππβ
=4
sec 2 β β 2 + 1/ sec 2 β πβ
=
1 4
πβ
1
= 4(π‘ππβ β 2β + 2 β + π ππβ πππ β =
1 4
= β ; β = ππππ ππ
π π ππ β ππ ππ β ππ β π ππ«ππ¬ππ + +π π ππ
=
2
β4 2
β4
2π₯β3 2
π ππ β π ππππππ + π π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
24
EXERCISE 10.6
π.
π₯2
ADDITIONAL STANDARD FORMULAS
ππ₯ + 25
36 β 9π₯ 2 ππ₯
π.
Let: π’ = π₯
Let: π = 6
π=5
π’ = 3π₯
ππ’ = ππ₯
ππ’ = ππ₯ 3
=
π π π¨πππππ + π π π
π₯ππ₯
π.
1 β π₯4
=
1 3π₯ 1 3π₯ 36 β 9π₯ 2 + 8π΄πππ ππ +π 3 2 3 6 1 3π₯ 2
1
=
Let: π’ = π₯ 2
π₯
36 β 9π₯ 2 + 3 8π΄πππ ππ 2 + π
=3
π π ππ β πππ + ππ¨πππππ + π π π
π=1 ππ’ = ππ₯ 2
π.
1 π₯2 = π΄πππ ππ + π 2 1
Let: π = 5
π π
= π¨πππππππ + π
16π₯ 2 + 25ππ₯
π’ = 4π₯ ππ’ = ππ₯ 4
π.
ππ₯ 49 β 25π₯ 2
Let: π = 7
=
=
1 4π₯ 4
2
16π₯ 2 + 25 +
1 52 4
2
ππ 4π₯ + 16π₯ 2 + 25 + π
π ππ ππππ + ππ + ππ ππ + ππππ + ππ + π ππ π
π’ = 5π₯ ππ’ 2
= ππ₯
=
π ππ β π ππ +π ππ ππ + π
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25
EXERCISE 10.7
INTEGRANDS INVOLVING QUADRATIC EQUATIONS
π.
ππ₯ π₯ 2 β3π₯+2
1.
πππππππ‘πππ π‘ππ π ππ’πππ
2π₯ 2
ππ₯
=
1 2
π₯β2
2
π₯ β 3π₯ = β2 π₯ 2 β 3π₯ = 3 2
2
3 π₯β 2
2
π₯β
9 9 = β2 + 4 4
=
(π₯ β
=
π’2 1
β
2
1 2
= ππ
ππ’ 1 π’βπ = ππ +π 2 βπ 2π π’+π
ππ
1 2
1 4
1 2
=
ππ’ + π2
= ππππ‘ππ
3 π’=π₯β 2 π=
π’2
1 1 π= , π’=π₯β 2 2
1 β 4 3 2 ) 2
=
1
+4
1 π’ = ππππ‘ππ + π 2 π
1 4
ππ₯
=
ππ₯ β 2π₯ + 1
3 1 2 2 3 1 π₯β + 2 2
π₯β β
+π
πβπ +π πβπ
1 2
π₯β 1 2
+π
=
π ππππππππ β π + π π
π.
3 β 2π₯ β π₯ 2
=
4β π₯+1
2
π’ = π₯ + 1, π = 2 π2
= =
π₯+1 2
=
β
π’2
π’ π2 π’ 2 2 = π β π’ + ππππ ππ + π 2 π 4
3 β 2π₯ β π₯ 2 + 2 ππππ ππ
π₯+1 2
+π
π+π π+π π β ππ β ππ + πππππππ +π π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
26
EXERCISE 10.7 π.
π₯2
INTEGRANDS INVOLVING QUADRATIC EQUATIONS
ππ₯ β 8π₯ + 7
2π₯β3ππ₯ 4π₯ 2 β1
11.
2π₯ππ₯ 4π₯ 2 β1
Completing the square
=
π₯ 2 β 8π₯ = β7
=2
π₯ 2 β 8π₯ + 16 = β7 + 16 π₯β4
2
=9
π₯β4
2
β9=0
π₯ππ₯ 4π₯ 2 β1
β3
ππ₯ 4π₯ 2 β1
πππ‘ π’ = 4π₯ 2 β 1 ; ππ’ 8
=2
ππ₯ (π₯ β 4)2 + 9
=
3ππ₯ 4π₯ 2 β1
β
=
π’
1
β 3[2 ππ
π ππ|πππ π
ππ’ = π₯ππ₯ 8 4π₯ 2 β1 4π₯ 2 +1 π
β π| β π ππ
+ π] ππβπ ππ+π
+π
π = 3 ;π’ = π₯ β4 = =
π’2
1 π’βπ ππ +π 2π π’+π 1
= 6 ππ =
ππ’ β π2
π₯β4β3 π₯β4+3
+π
=
π πβπ ππ +π π πβπ
(2π₯+7)ππ₯ π₯ 2 +2π₯+5
13.
=
2π₯+2 +5ππ₯ π₯ 2 +2π₯+5 2π₯+2 π₯ 2 +2π₯+5
+5
ππ₯ (π₯+1)2 +4
πππ‘ π’ = π₯ 2 + 2π₯ + 5 ; ππ’ = (2π₯π‘2)ππ₯ =
3+2π₯ ππ₯ π₯ 2 +9
9. = =3
2π₯ππ₯ π₯ 2 +9
ππ₯ + π₯ 2 +9
1
+ 2 π΄πππ‘ππ
π₯+1 2
+π π
=
3ππ₯ + π₯ 2 +9
ππ’ π’
2
β‘ππ|ππ + ππ + π| + π π¨πππππ
π+π + π
π
π₯ππ₯ π₯ 2 +9
πππ‘ π’ = π₯ 2 + 9 ; ππ’ = 2π₯ππ₯ 1
π₯
= 33 π΄πππ‘ππ 3 + 2
ππ’ 2
π’
π
= π¨πππππ π + ππ ππ + π + π
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EXERCISE 10.7 (π₯β3)ππ₯
15. = =
INTEGRANDS INVOLVING QUADRATIC EQUATIONS 19.
4π₯βπ₯ 2 π₯β2 β1ππ₯
=2
4π₯βπ₯ 2 π₯β2 ππ₯ 4π₯βπ₯ 2
πππ‘ π’ = ππ’ =
4π₯βπ₯ 2
4π₯ β
π₯2
β2(πβ2)ππ₯ 2 4π₯βπ₯ 2
+
17 2
ππ₯ π₯ 2 β4π₯+20
π₯ 2 β 4π₯ + 20 = π₯ β 2
4 β 2π₯
; ππ’ =
2 4π₯ β π₯ 2
ππ₯ = 2[
; 4π₯ β π₯ 2 = 4 β (2 β π₯)2
ππ’ π’
+
17 2
2
πβπ + π
+ 16
ππ₯ ] π₯β2 2 +16
= 2[ππ π₯ 2 β 4π₯ + 20 +
17 1 π₯β2 ( )Arctan 4 2 4
= π ππ ππ β ππ + ππ +
4β(2βπ₯)2
= β ππ β ππ β π¨πππππ
ππ πβπ Arctan + π π
+ π] π
π
π₯+3 ππ₯
17.
8π₯βπ₯ 2 π₯β4 +7ππ₯
=
8π₯βπ₯ 2 π₯β4 ππ₯ 8π₯βπ₯ 2
πππ‘ π’ = ππ’ = =
2π₯+4ππ₯ π₯ 2 β4π₯+20
2(2π₯+4+17)ππ₯ π₯ 2 β4π₯+20
ππ’
= ππ’ β
=
=
πππ‘ π’ = π₯ 2 β 4π₯ + 20 ; ππ’ = (2π₯ β 4)ππ₯
ππ₯
β
(4π₯+9)ππ₯ π₯ 2 β4π₯+20
+7
ππ₯ 8π₯βπ₯ 2
8π₯ β π₯ 2 ; ππ’ =
β2(π₯ β 4)ππ₯ 2 8π₯
β π₯2
8 β 2π₯ 2 8π₯ β π₯ 2
ππ₯
; 8π₯ β π₯ 2
16 β (4 β π₯)2
=β ππ’ + 7
ππ’ 16β(4βπ₯)2
= - ππ β ππ + ππ¨πππππ
πβπ + π
π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
28
EXERCISE 10.8
ALGEBRAIC SUBSTITUTION
ππ₯
1.
5
π§=
3
π₯
5
=
π§ 2 ππ§ π§3 β π§2
=3
3
9π₯ 6 β10π₯ 4 30 π
+π
π
π
ππ (πππ β ππππ ) = +π ππ
ππ§ π§β1
=3
7
3π₯ 6 π₯ 4 = β +π 10 3
2 π₯βπ₯ 3
π’ = π§β1 ππ’ = ππ§ 5.
ππ’ π’
=3
ππ₯ π₯+2
= 3 ππ |π§ β 1 | + π
π§ = π₯+2 π§4 = π₯ + 2 π₯ = 2 β π§4 ππ₯ = β4π§ 3 ππ§
π
= π ππ | π β π | + π
1
1
π₯+2 2
4
= 3 ππ π’ + π
= β4
π§ 3 ππ§ π§3 β π§2
= β4
π§ππ§ π§β1
1
(π₯ 3 βπ₯ 4 ππ₯
3.
3 4β
π’ = π§β1
1
4π₯ 2
π§=
12
ππ’ = ππ§
π₯
π§ = π’+1
ππ₯ = 12π§11 ππ§ 5
=3
(π§ 4 βπ§ 3 ) π§ 11 ππ§ π§8
= 3 (π§ 9 β π§ 8 ) ππ₯ = 3[ π§ 9 ππ§ β
π§ 8 ππ§]
= β4
π’ + 1 ππ’ π’
= β4[
π’ ππ’ + π’
ππ’ ] π’
= β4[π’ + ππ π’ + π = βπ[π β π + ππ π β π + π]
10
9
π§ π§ = 3[ β + π] 10 9 =
3π§10 π§ 9 β +π 10 3 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
29
EXERCISE 10.8
7.
ALGEBRAIC SUBSTITUTION
1
4 + π₯ππ₯ ;
9. π₯ π₯ + 4 3 ππ₯ π§ = π₯+4
π§ =(4+ π₯)1/2
12
π₯ (π§ 2 β 4)2= π₯ππ₯ = 4π§ 3 β 16π§ ππ§
=
π§ 4π§ 3 β 16π§ ππ§
=
4π§ 4 β 16π§ 2 ππ§
=4
π§5 π§3 β 16 +πΆ 5 3
=
4 (4 + 5
= 4+ π₯
= 4+ π₯
= 4+ π₯
π₯)5/2β
4 4+ π₯ 5
3 2
3 2
3 2
=
4 15
=
π π+ π π
4+ π₯
16 (4 + 3
; π§3 = π₯ + 4
π₯ = π§ 3 β 4 ; ππ₯ = 3π§ 3 ππ§
π§ 2 β 4 = π₯π§ 4 β 8π§ 2 + 16 = π₯ π§=
1 3
π§ 3 β 4 π§ 3π§ 2 ππ§
=
π§ 6 ππ§ β 4
=3
π§ 3 ππ§
7
4 3π§ 3 = β 3π§ 3 + π 7
3 π₯+4 = 7
π₯)3/2+C
7 3
β3 π₯+4
4 3
+π
4
16 β +πΆ 3
12 4 + π₯ β 80 +πΆ 15
=
3 π₯+4 3 7
π₯+4β7 +π 1 3
π₯ π₯ + 4 ππ₯ =
π π+π
π π
π
πβπ
+π
48 + 12 π₯ β 80 +πΆ 15 3 2
π π
12 + 3 π₯ β 20 + πΆ π πβπ +πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
30
EXERCISE 10.8
11.
4β 2π₯+1 1β2π₯
ALGEBRAIC SUBSTITUTION
ππ₯
π§ = 2π₯ + 1 ; π§ 2 = 2π₯ + 1 π§2 β 1 2π₯ = π§ 2 β 1 ; π₯ = ; ππ₯ = π§ππ§ 2 =
= = = =
ππ.
x 5 4 + x 3 dx
π§=
4 + π₯3π§2 = 4 + π₯3 ; π₯ =
ππ₯ =
1 4 β π§2 3
=β
4 β π§ π§ππ§ 1β2
π§ 2 β1 2
ππ§ β 2
=π§β2
2π§ππ§ β π§2 β 2
= π§ β 2 ππ π§ 2 β 2 +
ππ§ 2 π§ β2 1 ππ 2
= ππ + π β π ππ ππ β π +
π π
2π₯π‘1β 2 2π₯+1+ 2
ππ
+π
ππππ β π ππ + π + π
+π
2 3
4 β π§2
5
(π§)
=
1 3
=
1 2π§ 5 8π§ 3 β +π 3 5 3
=
2π§ 5 8π§ 3 β +π 15 9
=
β2π§ππ§ 3 4 β π§2
2 3
β8π§ 2 + 2π§ 4 ππ§ 3
=
2π§ β 1 ππ§ π§2 β 2
3 4 β π§2
4 β π§ 2 π§ β2π§ππ§ 3
=
4π§ β 2ππ§ π§2 β 2
β2π§ππ§
2π§ππ§
3
=
4π§ β 2ππ§ 1+ β π§2 β 2 1β
6
2π§ 4 ππ§ β 8π§ 2 ππ§
4 + π₯ 3 β 40 4 β π₯ 3 +π 45
=
4 + π₯ 3 6 4 + π₯ 3 β 40 +π 45
=
4 + π₯ 3 24 + 6π₯ 3 β 40 +π 45
= =
4 β π§2
π₯ 5 4 + π₯ 3 ππ₯
=
4π§ β π§ 2 ππ§ 2 β π§2
2 3
3
4+π₯ 3 β16+6π₯ 3 45
+π
π π + ππ πππ β π +π ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
31
EXERCISE 10.8
ALGEBRAIC SUBSTITUTION
3
1
15. π₯ 3 (4 + π₯ 2 )2 ππ₯
17. Κ
3+1 =2 2
π₯ = π‘ππ π’ ; ππ₯ = π ππ 2 π’ ππ’
π§2 = 4 + π₯2
= Κ πππ‘ 3 π’ ππ π π’ ππ’
X = 4 β π§2
= Κ πππ‘ π’ ππ π π’ ππ π 2 ( π’ β 1)ππ’ 1
1
dx = 2 4 β π§ 2 β2 (-2zdz)
= π π’ππ . π = ππ π π’ πππ ππ = β πππ‘ π’ ππ π π’ ππ’
= βΚ π 2 β 1 ππ
π§ππ§
1 (4βπ§ 2 )2
= Κ1ππ β Κπ 2 ππ
= (4 β π§ 2 )(π§ 3 )(βπ§ππ§) = Κπ β = =
ππ₯
π‘πππ π₯ 2 + 1 = π‘ππ2 π’ + 1 = π ππ π’ & π’ = π‘ππβ1 π₯
Z = 4 + π₯2
=-
π₯ 4 π₯ 2 +1
β4π§ 4 + π§ 6 ππ§
π 3 3
= ππ π π’ β
4π§ 5 π§ 7 β + +πΆ 5 7
=
=
28π§ 5 + 5π§ 7 +πΆ 35
=
β28( 4+π₯ 2 )5 +5( 4+π₯ 2 )7 +C 35
ππ π 3 π’ 3
+πΆ
ππ + π πππ β π +πͺ πππ
5
= =
=
4 + π₯ 2 (β28 + 5(4 + π₯ 2 ) +πΆ 35 4+π₯ 2
5
(β28+20+5π₯ 2 ) 35
π + ππ
π
+πΆ
(πππ β π)
ππ
+πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
32
EXERCISE 10.8
ππ. ( π₯=
ALGEBRAIC SUBSTITUTION
ππ₯ π₯ 2 (81 +
π₯4
)
1 1 ; ππ₯ = β 2 ππ§ π§ π§
1 π§2
βππ§ π§2
81 +
πππ‘ π₯ = 1 π§
=
1
3 4
1 π₯
ππ§
β π₯3
β π§2
1 π§4
ππ§
β π§2
1/π§ 4
1
(π§ 2 β1) 3 π§ π§4
=
πππ‘ π’ = 81π§ 4 + 1 ; ππ’ = 324π§ 3 ππ§
β1 81
π§=
1
81π§ 4 + 1
1 324
1 , π§2
π§ 2 β1 3 π§3
π§3
=
=
(π₯βπ₯ 3 )1/3 ππ₯ π₯4
=
1 π4
βππ§ π§2 81π§ 4 +1 π§6
=
21.
(β
ππ§ ) π§2
1
π§ 2 β 1 3 π§ππ§
=β
ππ’ πππ‘ π’ = π§ 2 β 1
3
π’4 81π§ 4 + 1
1 4
+π
=β
1 2
1
π’3 ππ’ 4
π ππ + ππ =β +π ππ ππ
=
1 π’3 β 2 ( 4 )+c 3
=β
3 2 π§ β1 8 3
= β8
1 π₯2
β1
4 3
4 3
+π +π
4
=
3 1-x2 3 β 8 x2 +c
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
33
EXERCISE 10.9
INTEGRATION OF RATIONAL FUNCTIONS OF SINES AND COSINES
π·π₯ 1+πππ π₯
1.
ππ₯ = π πππ₯ + πππ π₯ + 3
π.
2ππ§ 1+π§ 2 2π§ 1βπ§ 2 + +3 1+π§ 2 1+π§ 2
2 π·π§
=
1+π§ 2 1βπ§ 2 1+π§ 2
1+
2ππ§
=
1+π§ 2 1+π§ 2 + 1βπ§ 2 1+π§ 2
=
2ππ§ 1 2
π§+2
π§+2 2 π’ 2 = ππππ‘ππ + π = ππππ‘ππ + π 7 7 π π 2
= ππ₯ 4+2 π πππ₯
2 ππ§
=
=
1+ π§ 2 2π§ 1+π§ 2
1 2π§+1 ππππ‘ππ 7 7
π π
4+2
2ππ§
=
1+π§ 2
ππ§
=
=
1 2 2
+
3
= 2
π’2
3 2
ππππ‘ππ
π π
π§+2 3 2
ππ’ 2 π’ = ππππ‘ππ + π 2 +π π π
4
1
2
π
ππππππ
ππππ π + π π
+ π
1+π§ 2
π.
π πππ₯ππ₯ = 2ππ§ =2 1 β π§2
= π§+
+ πΆ
4+4π§ 2 + 4π§ 1+π§ 2
2 ππ§ 1 3 ; π€ππππ: π’ = π§ + ; π = 4π§ 2 + 4π§ + 4 2 2
=2
2
2ππ§
=
4π§ 4+ 1+π§ 2
=
7 4
1
π = πππ + π π
3.
+
ππ’ 7 1 βΆ π€ππππ π = , π’ = π§ + π’2 + π2 2 2
=2
ππ§ = π§ + π
1+2π§β2π§ 2 +3+3π§ 2 1+π§ 2
2ππ§ = 4 + 2π§ + 2π§ 2
=
2ππ§ 2
=
1+π§ 2
1+2π§β2π§ 2 +3 1+π§ 2
2 ππ§
=
2ππ§
=
1+π§ 2
+ π=
1 3
ππππ‘ππ
π
ππππππ
π πππ π + π π
2π§+1 3
+ π
2
= 2π ππ
π+π’ πβπ’
1 + π§ 2 2ππ§ . 1 β π§2 1 + π§2 π2
ππ’ π€ππππ π = 1, π’ = π§ β π’2
+ π = ππ
1+π§ 1βπ§
+πΆ π
π + πππ π π π+π = ππ + π = ππ π + π ππ πβπ π β πππ π
+π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
34
EXERCISE 10.10
INTEGRATION BY PARTS
1. π₯πππ π₯ππ₯ =
π₯πππ π₯ππ₯
; πππ‘ ππ£ = πππ₯ππ₯ π£ = π πππ₯
= π₯π πππ₯ β
π’=π₯ ππ’ = ππ₯
π πππ₯ππ₯
= πππππ + ππππ + πͺ
3. π βπ₯ πππ 2π₯ππ₯ ; π’ = πππ 2π₯
ππ£ = π βπ₯ ππ₯ ; π’ = π ππ2π₯
;
π£ = -π βπ₯
ππ’ = π ππ2π₯ππ₯ ;
; ππ£ = π βπ₯ ππ₯
; ππ’ = 2πππ 2π₯ππ₯ ; π£ = -π βπ₯
= -π π₯ πππ 2π₯ β 2π βπ₯ π ππ2π₯ππ₯ = -π π₯ πππ 2π₯ β 2 π βπ₯ π ππ2π₯ππ₯ = -π βπ₯ πππ 2π₯ β 2[-π βπ₯ π ππ2π₯ β
-π βπ₯ 2πππ 2π₯ππ₯
= -π βπ₯ πππ 2π₯ + 2π βπ₯ π ππ2π₯ β 4 π βπ₯ πππ 2π₯ππ₯ πππ 4 π βπ₯ πππ 2π₯ππ₯ π‘π πππ‘π π ππππ =
2π βπ₯ π ππ 2π₯βπ βπ₯ πππ 2π₯ 5
=
πβπ π
+πΆ
πππππ β πππππ + πͺ
5. ππππ‘ππ2π₯ππ₯ ; ππ£ = ππ₯ ; π’ = ππππ‘ππ2π₯ 2ππ₯
π£ = π₯ ; ππ’ = 1+π₯ 2 2π₯ππ₯ 1+4π₯ 2
= π₯ππππ‘ππ2π₯ β = π₯ππππ‘ππ2π₯ β 2 1
ππ₯ 1+4π₯ 2 ππ’ π’
= π₯ππππ‘ππ2π₯ β 4 π π
= πππππππππ β ππ π + πππ + πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
35
EXERCISE 10.10
INTEGRATION BY PARTS
7. π ππ 3 π₯ππ₯ ; ππ£ = π ππ 2 π₯ππ₯ ; π’ = π πππ₯ π£ = π‘πππ₯
ππ’ = π πππ₯π‘πππ₯
π ππ 2 π₯π πππ₯ππ₯
=
= π πππ₯π‘πππ₯ β
π ππ 2 π₯ β 1 π πππ₯ππ₯
= π πππ₯π‘πππ₯ + ππ π πππ₯π‘πππ₯ β π ππ 3 π₯ππ₯ ; πππ
π ππ 3 π₯ππ₯ ππ πππ‘π π ππππ
2 π ππ 3 π₯ππ₯ = π πππ₯π‘πππ₯ + ππ π πππ₯ + π‘πππ₯ π π
π ππ 3 π₯ππ₯ =
ππππππππ + ππ ππππ + ππππ + πͺ
9. π₯πππ 2 2π₯ππ₯ ; ππ£ = π2π₯ππ₯ π£= 1 π₯ 2
=π₯ =
π₯2 2
1 π₯ 2
;
1
+ 8 π ππ4π₯
1
1
ππ’ = π₯ 1
+ 8 π ππ4π₯ β (2 π₯ + 8 π ππ4π₯)ππ₯
1 8
1 4
+ π₯π ππ4π₯ β π₯ 2 +
π
π’ =π₯
π
1 πππ 4π₯ 32
+πΆ
π
= π ππ + π πππππ + ππ πππππ + πͺ
11.
π₯ππππ πππ₯ππ₯ 1βπ₯ 2
;
ππ£ =
π₯ππ₯
;
1βπ₯ 2
π£ = - 1 β π₯2
;
π’ = ππππ πππ₯ ππ’ =
= - 1 β π₯ 2 ππππ πππ₯ β (- 1 β π₯ 2 )(
ππ₯ 1βπ₯ 2
ππ₯ 1βπ₯ 2
)
= - 1 β π₯ 2 ππππ πππ₯ + ππ₯ = π β π β ππ πππππππ + πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
36
EXERCISE 10.10
INTEGRATION BY PARTS
13. π πππ₯ππ 1 + π πππ₯ ππ₯ ;
π’ = ππ 1 + π πππ₯ ππ’ =
1 1+π πππ₯
;
ππ£ = π πππ₯ππ₯
πππ π₯ππ₯ ;
π£ = πππ π₯
πππ 2 π₯ππ₯ 1+π πππ₯
= -πππ ππ 1 + π πππ₯ +
1βπ ππ 2 π₯ππ₯ 1+π πππ₯
= -πππ ππβ‘|1 + π πππ₯ | + = -πππ ππ 1 + π πππ₯ +
1 β π πππ₯ ππ₯
= -πππ ππ 1 + π πππ₯ + π₯ + πππ π₯ + πΆ = β πππ ππ π + ππππ + π + ππππ + πͺ
15.
π π₯ π₯ππ₯ (π₯+1)2
;
π’ = ππ₯ π₯ ππ’ = π π₯ π₯ + 1 ππ₯
=-
ππ₯π₯ π₯+1
+ π π₯ ππ₯ =
ππ£ = (π₯+1)2 ππ₯
;
π£ =-
π₯3
=
1 ππππ ππ 3
=
1 3 π₯ ππππ πππ₯ 3
+ (3 πππ π β
=
1 3 π₯ ππππ πππ₯ 3
+
=
1 3 π₯ ππππ ππ 3
ππππ ππππ; 1
=3
1
1 3
1βπ₯ 2
3β 1βπ₯ 2 9 1βπ₯ 2 2+π₯ 2
π₯3 1βπ₯ 2
ππ₯ 1βπ₯ 2
;
ππ£ = π₯ 2 ππ₯ ; π£ =
π₯3 3
ππ₯
1
+
1 π₯+1
ππ +πͺ π+π
17. π₯ 2 ππππ πππ₯ππ₯; π’ = ππππ πππ₯ ; ππ’ = β3
1
;
9
πππ 3 π )+ 9
πΆ
+πΆ +πΆ
ππ₯ ; π = 1 ; π£ = π₯ ; π₯ = π ππ π ; ππ₯ = πππ π ππ ;
1 β π₯ 2 = πππ π
π ππ 3 π (πππ πππ) πππ π
1
= 3 π ππ2 π π πππππ π π
= (βππππ +
ππππ π )+ π
πͺ DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
37
EXERCISE 10.11
INTEGRATION OF RATIONAL FUNCTIONS
12π₯+18 π₯+2 π₯+4 (π₯β1)
1.
12π₯ + 18 π΄ π΅ πΆ = + + π₯ + 2 π₯ + 4 (π₯ β 1) (π₯ + 2) (π₯ + 4) (π₯ β 1) 12π₯ + 18 = π΄ π₯ + 4 π₯ β 1 + π΅ π₯ + 2 π₯ β 1 + πΆ π₯ + 2 (π₯ + 4) 12π₯ + 18 = π΄(π₯ 2 + 3π₯ β 4) + π΅ π₯ 2 + π₯ β 2 + πΆ(π₯ 2 + 6π₯ + 8) 12π₯ + 18 = π΄π₯ 2 + 3π΄π₯ β 4π΄ + π΅π₯ 2 + π΅ β 2π΅ + πΆπ₯ 2 + 6πΆπ₯ + 8πΆ π΄π₯ 2 + π΅π₯ 2 + πΆπ₯ 2 = 0 3π΄π₯ + π΅π₯ + 6πΆπ₯ = 12π₯ 4π΄ + π΅ + 8πΆ = 18 π΄=1 π΅ = β3 πΆ=2
=
ππ₯ + (π₯+2)
β3ππ₯ + (π₯+4)
2ππ₯ (π₯β1)
= ππ π + π β π ππ π + π + πππ π β π β‘
3.
1=
ππ₯ π₯β1 (π₯β4) π΄ π΅ + (π₯ β 1) (π₯ β 4)
1 = π΄ π₯ β 4 + π΅(π₯ β 1) 1 = π΄π₯ β 4π΄ + π΅π₯ β π΅
=
β1 3 (π₯ β1)
1
ππ₯ +
1 3 (π₯ β4)
1
π΄+π΅ =0 β4π΄ β π΅ = 1 π΄ = βπ΅ 1 π΅= 3
ππ₯
= β 3 ππ π₯ β 1 + 3 ππ π₯ β 4 + πΆ
=
π ππ πβπ π ππ πβπ
+C
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
38
EXERCISE 10.11
INTEGRATION OF RATIONAL FUNCTIONS
6π₯ 2 +23π₯β9 ππ₯ (π₯ 3 +2π₯ 2 β3π₯)
5.
6π₯ 2 + 23π₯ β 9 ππ₯ π₯ π₯ + 3 (π₯ β 2) 6π₯ 2 + 23π₯ β 9 =
π΄ π΅ πΆ + + π₯ (π₯ + 3) (π₯ β 1)
6π₯ 2 + 23π₯ β 9 = π΄ π₯ + 3 π₯ β 1 + π΅ π₯ π₯ β 1 + πΆ π₯ (π₯ + 3) 6π₯ 2 + 23 β 9 = π΄ π₯ 2 + 2π β 3 + π΅ π₯ 2 β π₯ + πΆ(π₯ 2 + 3π₯) π΄+π΅+πΆ =6 2π΄ β π΅ + 3πΆ = 23 β3π΄ + 0π΅ + 0πΆ = β9 π΄=3 π΅ = β2 πΆ=5 =3
ππ₯ π₯
β2
ππ₯ + (π₯+3)
5
ππ₯ (π₯β1)
= πππ π β πππ π + π + πππ π β π + πͺ π₯ 3 +5π₯ 2 +9π₯+7 ππ₯ π₯ 2 +5π₯+4
7.
π₯ 3 + 5π₯ 2 + 9π₯ + 7 ππ₯ π₯ + 4 (π₯ + 1)
By division of polynomials, 5π₯ + 7 π΄ π΅ = + π₯ + 4 (π₯ + 1) (π₯ + 4) (π₯ + 1) 5π₯ + 7 = π΄ π₯ + 1 + π΅(π₯ + 4)
= =
π₯ππ₯ + ππ π
+
ππ ππ π
13 ππ₯ 3
(π₯ + 4) π+π +
+ π ππ π
ππ π₯ = 4, 13 π΄= 3 ππ π₯ = β1 2 π΅= 3
2 ππ₯ 3
(π₯ + 1) π+π +πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
39
EXERCISE 10.11
9.
INTEGRATION OF RATIONAL FUNCTIONS
2π₯+1 π₯β2 (π₯β3)2
2π₯ + 1 =
π΄ π΅ πΆ + + (π₯ β 2) (π₯ β 3) (π₯ β 3)2
2π₯ + 1 = π΄ π₯ β 3
2
+π΅ π₯β3 π₯β2 +πΆ π₯β2
2π₯ + 1 = π΄ π₯ 2 β 6π₯ + 9 + π΅ π₯ 2 β 5π₯ + 6 + πΆ π₯ β 2 π΄+π΅ =0 β6π΄ β 5π΅ + πΆ = 2 9π΄ + 6π΅ β 2πΆ = 1 π΄=5 π΅ = β5 πΆ=7 =
5ππ₯ + π₯β2
β5ππ₯ + π₯β3
7ππ₯ π₯β3
2
7
= 5ππ π₯ β 2 β 5ππ π₯ β 3 + π₯β3 = πππ
πβπ π + πβπ (π β π)
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
40
EXERCISE 10.11
11.
INTEGRATION OF RATIONAL FUNCTIONS
2π₯β5 ππ₯ π₯(π₯β1)
2π₯ β 5 π΄ π΅ πΆ π· = + + + π₯(π₯ β 1) π₯ (π₯ β 1) (π₯ β 1)2 (π₯ β 13 2π₯ β 5 = π΄ π₯ β 1
3
+ ππ₯ π₯ β 1
2
+ ππ₯ π₯ β 1 + π·π₯
2π₯ β 5 = π΄π₯ 3 β 3π΄π₯ 2 + 3π΄π₯ β π΄ + π΅π₯ 3 β 2π΅π₯ 2 + π΅π₯ +C π₯ 2 β πΆπ₯ + π·π₯ 2π₯ β 5 = π΄π₯ 3 = 3π΄π₯ 2 β 2π΅π₯ 2 + π΅π₯ + πΆπ₯ 2 β πΆπ₯ + π·π₯ 2π₯ β 5 = π΄π₯ 3 + π΅π₯ 2 β 3π΄π₯ 2 β 2π΅π₯ 2 + πΆπ₯ 2 + 3π΄π₯ + π΅π₯ β πΆπ₯ + π·π₯ β π΄ 2π₯ β 5 = π΄ + π΅ π₯ 3 + β3π΄ β 2π΅ + π π₯ 2 + 3π΄ + π΅ β πΆ + π· π₯ β π΄ π΄+π΅ =0 β3π΄ β 2π΅ + πΆ = 0 3π΄ + π΅ β πΆ + π· = 2 βπ΄ = β5 π΄=5 π΅ = β5 πΆ=5 π· = β3 =
5ππ₯ + π₯
=5
ππ₯ β5 π₯
β5ππ₯ + (π₯ β 1)
5ππ₯ + (π₯ β 1)2
ππ₯ +5 (π₯ β 1) 5
β3ππ₯ (π₯ β 1)3
ππ₯ β3 (π₯ β 1)2
ππ₯ (π₯ β 1)3
3
= 5ππ π₯ β 5ππ π₯ β 1 β (π₯β1) + 2(π₯β1)2 + πΆ = πππ
π π π β + +πͺ π β π (π β π) π(π β π)π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
41
EXERCISE 10.11
INTEGRATION OF RATIONAL FUNCTIONS
3π₯ 2 +17π₯+32 π₯ 3 +8π₯ 2 +16π₯
13.
3π₯ 2 + 17π₯ + 32 π₯(π₯ + 4)2 3π₯ 2 + 17π₯ + 32 π΄ π΅ πΆ = + + 2 π₯(π₯ + 4) π₯ (π₯ + 4) (π₯ + 4)2 π΄+π΅ =3 8π΄ + 4π΅ + πΆ = 17 16π΄ = 32 π΄=2 π΅=1 πΆ=3 2ππ₯ + π₯
=
ππ₯ + (π₯ + 4)
β3ππ₯ (π₯ + 4)2 π
= πππ π + ππ π + π + π+π 2π₯+1 3π₯β1 (π₯ 2 +2π₯+2)
15.
2π₯ + 1 π΄ π΅ 2π₯ + 2 + πΆ = + 2 2 3π₯ β 1 (π₯ + 2π₯ + 2) (3π₯ β 1) π₯ + 2π₯ + 2 2π₯ + 1 = π΄ π₯ 2 + 2π₯ + 2 + π΅ 2π₯ + 2 3π₯ β 1 + πΆ 3π₯ β 1 2π₯ + 1 = π΄ π₯ 2 + 2π₯ + 2 + π΅(6π₯ 2 + 4π₯ + 2) + πΆ 3π₯ β 1 π΄+π΅ =0 2π΄ + 4π΅ + 3πΆ = 2 2π΄ + 2π΅ β πΆ = 1 5 π΄=β 2 5 π΅= 2 πΆ = β1 =β
5 2 5
ππ₯ 5 + (3π₯ β 1) 2
(2π₯ + 2) ππ₯ β π₯ 2 + 2π₯ + 2
π₯2
ππ₯ + 2π₯ + 2
5
= β 2 ππ 3π₯ β 1 + 2 ππ π₯ 2 + 2π₯ + 2 β ππ π₯ 2 + 2π₯ + 2 =
π ππ + ππ + π ππ β ππ ππ + ππ + π π ππ β π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
42
EXERCISE 10.11
17.
INTEGRATION OF RATIONAL FUNCTIONS
5π₯ 2 βπ₯+17 ππ₯ π₯+2 (π₯ 2 +9)
5π₯ 2 β π₯ + 17 π΄ π΅ 2π₯ + πΆ = + 2 π₯ + 2 (π₯ + 9) π₯ + 2 π₯2 + 9 5π₯ 2 β π₯ + 17 = π΄ π₯ 2 + 9 + 2π΅π₯ + πΆ (π₯ + 2) 5π₯ 2 β π₯ + 17 = π΄π₯ 2 + 9π΄ + 2π΅π₯ 2 + 4π΅π₯ + πΆπ₯ + 2πΆ 5π₯ 2 β π₯ + 17 = π΄π₯ 2 + 2π΅π₯ 2 + 4π΅π₯ + πΆπ₯ + 9π΄ + 2πΆ 5π₯ 2 β π₯ + 17 = π΄ + 2π΅ π₯ 2 + 4π΅ + πΆ π₯ + 9π΄ + 2πΆ π₯ 2 = π΄ + 2π΅ = 5 π₯ = 4π΅ + πΆ = β1 π = 9π΄ + 2πΆ = 17 π΄ + 2π΅ = 5 β 2 4π΅ + πΆ = β1
= β2π΄ β 4π΅ = β10 =
4π΅ + πΆ = β1 β2π΄ + πΆ = β11
β2π΄ + πΆ = β1 β 2 9π΄ + 2πΆ = 17
= 4π΄ β 2πΆ = 22 =
9π΄ + 2πΆ = 17 13π΄ = 39
A=3 9(3)+2C=17
4B-5=-1
27+2C=17
4B=-1+5
2C=17-27
4B=4
2C=-10
B=1
C=-5 =
=3
3 1 2π₯ β 5 + ππ₯ π₯+2 π₯2 + 9 ππ₯ π₯+2
+2
π₯ππ₯ β π₯ 2 +9
5
ππ₯ π₯ 2 +9 π
= πππ π + π + ππ ππ + π β π π¨πππππ
π + π
πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
43
EXERCISE 10.11
19.
INTEGRATION OF RATIONAL FUNCTIONS
4π₯ 2 +21π₯+54 π₯ 2 +6π₯+13
4 β 3π₯ β 2 π₯ 2 + 6π₯ + 13 π΄ 2π₯ + 6 + π΅ π₯ 2 + 6π₯ + 13 π΄ 2π₯ + 6 + π΅ = 3π₯ β 2 2π΄ + π΅ = 3 π΅ = β11 π΄=
3 2
3 2π₯ + 6 ππ₯ ππ₯ =Κ4β[ Κ 2 + (β11Κ 2 + 6π₯ + 13)] 2 π₯ + 6π₯ + 13 π₯ = β11Κ
π₯2
ππ₯ + 6π₯ + 9 + 13 β 9
ππ₯ π₯ + 3 2 + 13 β 9
= β11Κ 1 2
= β11( ππππ‘ππ = 4π₯ β
π₯+3 ) 2
3 ππβ‘| π₯ 2 2 π
2
+ 6π₯ + 13| β
= ππ β π ππβ‘| ππ + ππ + ππ| +
11 π₯+3 ππππ‘ππ 2 2 ππ π+π ππππππ π + π
πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
44
EXERCISE 10.11
21.
INTEGRATION OF RATIONAL FUNCTIONS
π₯ 3 +7π₯ 2 +25π₯+35 π₯ 2 +5π₯+6
π₯+2+
9π₯ + 23 ππ₯ + 5π₯ + 6
π₯2
9π₯ + 23 π΄ π΅ = + π₯ + 3 (π₯ + 2) π₯ + 3 π₯ + 2 9π₯ + 23 = π΄ π₯ + 2 + π΅(π₯ + 3)
x=-3 9(-3)+23= A(-3+2)+B(-3+3) -27+23=A(-1)+B(0) -4=-A A=4 If x=-2 9(-2)+23= A(-2+2)+B(-2+3) -18+23=A(0)+B 5=B B=5 β2 5 + ππ₯ π₯+3 π₯+2
=
π₯+2+
=
π₯ππ₯ + 2 ππ₯ β 4
ππ₯ π₯+3
+5
ππ₯ π₯+2
ππ = + ππ β πππ π + π + πππ π + π + π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
45
EXERCISE 10.11
23.
INTEGRATION OF RATIONAL FUNCTIONS
π₯ 2 βπ₯β8 (2π₯β3)(π₯ 2 +2π₯+2)
π΄ π΅ 2π₯ β 2 + πΆ + 2 2π₯ β 3 π₯ + 2π₯ + 2
A(π₯ 2 + 2π₯ + 2) + π΅ 2π₯ + 2 2π₯ β 3 + πΆ(2π₯ β 3) A(π₯ 2 + 2π₯ + 2) + π΅ 4π₯ 2 β 2π₯ β 6 + πΆ(2π₯ β 3) A+4B=1 2A-2B+2C=-1 2A-6B-3C=-8 1
A=-2 1
A=2 C=1 ππ₯ 1 + (2π₯ β 3) 2
β1
2π₯ + 2 ππ₯ + π₯ 2 + 2π₯ + 2
1 βππ(2π₯ β 3) + ππβπ₯ 2 + 2π₯ + 2β + 2
ππ₯ π₯ 2 + 2π₯ + 2 ππ₯ π₯ + 1 2 + 12
1
= β 2 β ππ 2π₯ β 3 + ππβπ₯ 2 + 2π₯ + 2β + ππππ‘ππ π₯ + 1 + πΆ =
π ππ + ππ + π ππβ β + ππππππ π + π + π π ππ β π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
46
EXERCISE 10.11
25. Κ
INTEGRATION OF RATIONAL FUNCTIONS
π₯ 5 +2π₯ 3 β3π₯ π₯ 2 +1 3
=Κ
π₯ 5 + 2π₯ 3 β 3π₯ π₯ 6 + 3π₯ 4 + 2π₯ 2 + 1
=
π΄ 2π₯ + π΅ πΆ 2π₯ + π· πΈ 2π₯ + πΉ + 2 + 2 π₯2 + 1 π₯2 + 1 π₯ +1 2 π₯ +1 3
= π΄ 2π₯ π₯ 2 + 1
2
+ π΅ π₯2 + 1
2
3
+ πΆ 2π₯ π₯ 2 + 1 + π· π₯ 2 + 1 + πΈ 2π₯ + πΉ
= π΄ 2π₯ π₯ 4 + 2π₯ 2 + 1 + π΅ π₯ 4 + 2π₯ 2 + 1 + πΆ 2π₯ 3 + 2π₯ + π· π₯ 2 + 1 + πΈ 2π₯ + πΉ = π΄ 2π₯ 5 + 4π₯ 3 + 2π₯ + π΅ π₯ 4 + 2π₯ 2 + 1 + πΆ 2π₯ 3 + 2π₯ + π· π₯ 2 + 1 + πΈ 2π₯ + πΉ 1 2
π₯ 5 : 2π΄ = 1
; π΄=
π₯4: π΅ = 0
; π΅=0
π₯ 3 : 4π΄ + 2πΆ = 2
; πΆ=0
π₯ 2 : 2π΅ + π· = 0
; π·=0
π₯: 2π΄ + 2πΆ + 2πΈ = β3 ; πΈ = 0 π: π΅ + π· + πΉ = 0 =
; πΉ=0
π π ππ ππ + π + π π π +π
π
+πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
47
EXERCISE 10.11
INTEGRATION OF RATIONAL FUNCTIONS
27. π₯ 4 + 2π₯ 3 + 11π₯ 2 + 8π₯ + 16 π₯(π₯ 2 + 4)2 π΄ π΅ 2π₯ + πΆ π· 2π₯ + πΈ [ + + 2 ][(π₯ 2 + 4)2 ] π (π₯ 2 + 4) (π₯ + 4)2
A π₯2 + 4
2
+ π΅ 2π₯ π₯ (π₯ 2 + 4) + πΆ π₯ 2 + 4 (π₯) + π·(2π₯)(π₯) + πΈ(π₯)
A(π₯ 4 + 8π₯ 2 + 16) + π΅ 2π₯ 4 + 8π₯ 2 + πΆ π₯ 3 + 4π₯ + π·2π₯ 2 + πΈπ₯ π₯ 4 : π΄ + 2π΅ = 1
A=1
π₯3: πΆ = 2
B=0
π₯ 2 : 8A+8B+2D=11
C=2
X: 4C + E=8
D = 3/2
C : 16A = 16
E=0
=
ππ₯ 2ππ₯ 3 + 2 + π₯ π₯ +4 2
= πππ₯ + 2
1 2
ππππ‘ππ
2π₯ππ₯ (π₯ 2 + 4)2 π₯ 2
β
3 2 π₯ 2 +4
+πΆ
π π = πππ + ππππππ β +πͺ π π ππ + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
48
EXERCISE 11.1
SUMMATION NOTATION
π=10
β π = 10
π.
π
12π 3
π.
π
π=1
π=1
π3
π=10
π=1 2
10 10 + 1 4
= 12
π3 β π
=
π=10
= 12
π(π β 1)(π + 1) π=1
π3 β π
=
2
π=1 π=10
=
= 3(100 121 )
π=10 3
π + π=1
= πππππ
=
π π=1
10 2 10+1 2 4
β
10 10+1 2
= ππππ π=10
(12π 2 + 4π )
π.
π=ππ
π.
π=1 π=10
π=1
π
π=10
9π 2 + 6π + 1
=
π=1
10(10 + 1)(2 10 + 1) 10(10 + 1) = 12 +4 6 2
π=1
π2 + 6
=9
= 2 110 21 + 2 110 = ππππ
π
π=π
π=10
π2 + 4
= 12
ππ + π
=9
π+
10(10+1)(2 10 +1) 6
+6
1 10(10+1) 2
+ 10
= ππππ
π. ππ β ππ + ππ β ππ + β― + (ππ β ππ ) π
=
ππ β ππ π=π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
49
EXERCISE 11.1
SUMMATION NOTATION
ππ. π π₯1 βπ₯1 + π π₯2 βπ₯2 + β― + π π₯π βπ₯π π
=
π(ππ ) βππ π=π
ππ. 14 + 24 + 34 + β― + π4 π
ππ
= π=π
ππ. π1 π1+π2 π2+π3 π3 + β― + ππ ππ π
=
ππ ππ π=π
ππ. π’13 + π’23 + π’33 + β― + π’π3 π
πππ
= π=π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
50
EXERCISE 11.2
π.
THE DEFINITE INTEGRAL
2 3π₯ 2 ππ₯ 1
π = 0 ;π = 2 βπ₯ = =
(π₯ β 1)ππ₯
π=0
;
2β0 π
βπ₯ =
1β0 π
2 π
= πππ 2
= πππ 3
4π 2 2 ( ) π2 π
= πππβ‘3
8π 2 π3
πββ
= πππ 24 πββ
= ππππββ 24
π 2 +1 2π+1 6 π3 1 0 0 0 2π 3 +π 2 +2π 2 +π 6π 3
} 1 π
β
1 2π3 + π2 + 2π2 + π π3 6
1 0 π 2 βπ π2
β1
2 3
= β1 =β
π π
5 2π₯ 1
5.
π(π + 1)(2π + 1) 1 6 π3
π π3
]β
1 π π+1 2π+ π2 6
= ππππββ 2
πββ
πββ
1 π
πββ
2π 2 ( ) π π
π π
π₯ 2 β π₯ππ₯
= πππ 2
π=1
ππ =
π2
2π π 3
;
= { πππ 2 [ ( 2 )
2 =0+π π
= πππ π=β
π=1
πββ
ππ = π + πβπ₯
=
1 2π₯ 0
3.
βπ₯ =
+ 3ππ₯
5β1 π
;
ππ = 1 + π
4 π
4
=π
1
= ππππββ 24 =π
4π
4
=ππππ=β (1 + π ) β π + 3π 4 π
=ππππ=β =ππππ=β
4π π
+ 16
16π π2
+ π2 +
+
3π
π(π+1) 2
+ 3π
= ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
51
EXERCISE 11.2
2
π.
THE DEFINITE INTEGRAL
π₯ 3 ππ₯
0
βπ₯ =
2 2π ; ππ = π π 3
= πππ
2π π
= πππ
8π 3 π3
πββ
πββ
2 π 2 π
16 π2 π + 1 πββ π 4 4
2
= πππ
π3
4 2 2 (π (π + 2π + 1) πββ π 4
= πππ
= ππππββ
4π 4 π4
8
+ π3 +
4π 2 π3
=4+0+0 = 4
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
52
EXERCISE 11.3
2
π.
SOME PROPERTIES OF THE DEFINITE INTEGRAL
3π₯ 2 β 2π₯ + 1 ππ₯
3
π.
1
2
3π₯ 3 2π₯ 2 + +π₯ 3 2
=
π’ = π₯2 + 1 ππ’ = 2π₯ππ₯
=8β4+2β1+1β1 =
=5
π₯ππ₯ π₯2 + 1
=
1 2 1 2
3 2
ππ’ π’
ππ 10 β ππ 5
= π. πππ 3
3π₯ 2 +
π. 1
4 ππ₯ π₯2 9.
3
3π₯ 4 + 3 π₯
=
0
=
4
= 27 β 3 β 1 + 4 =
0 ππ¦ β1 β(π₯ 2 +2π₯β1)
β1
ππ¦ β(π₯ + 2π₯ + 1 β 1 β 1)
0
ππ π
= β1
ππ¦ β[ π₯ + 1
0
= β1 7
π.
3
1+
π₯2
ππ₯
π’ = 1 + π₯2 ππ’ = 2π₯ππ₯ 1
=2 =
3
4 1+π₯ 2 3
4
+ 2]
ππ¦ β π₯+1
2
0
ππ¦
β1
2β π₯+1
=
0
2
+2
2
πππ‘ π = 2 ; π’ = (π₯ + 1) = π΄πππ ππ =
π₯+1 2
+π
π π
ππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
53
EXERCISE 11.3 π
ππ. 0
SOME PROPERTIES OF THE DEFINITE INTEGRAL 1
π₯ππ₯ π₯2 + π
0
πππ‘ π’ = π₯ 2 + π ; ππ’ = 2π₯ππ₯ ; ππ’ π 2
= 1 2
ππ’ = π₯ππ₯ 2
1
π₯ β 2 β π₯ ππ₯ 0
π’
π
=
2π₯ β π₯ 2 ππ₯
ππ.
π π
ππ’ π’
=
1 π πππ’ 0 2
=
1 π ln π₯ 2 + π 0 2
1 = ln π 2 + π β ππ 0 + π 2
π ; πππ β πππ = ππ π
1 π2 + π 1 π π+1 = ππ = ππ 2 π 2 π 1
= 2 ππ π + 1 = ππ π + 1
π πππ =
π
4
=
π
=8
1 ππππ‘ππ 1 2
=
π π
4
4
0
u= x; du=dx; a=2
=
2πππ π β 2π πππ β 4π ππππππ πππ
π ππ2 ππππ 2 π ππ
0 π
1 π₯ = ππππ‘ππ 2 2
2π πππ = π₯
π
=8 2 ππ₯ 0 π₯ 2 +4
; 2πππ π = 2 β π₯
π₯ = 2π ππ π ; ππ₯ = 4π ππππππ πππ π΄π‘ π₯ = 1, π = π 4 ; π₯ = 0, π = 0
1 2
= ππ π + π
ππ.
2βπ₯ 2 π₯ ; 2 2
cos π =
=2
π 0
4
1 β πππ 2π 2
1 + πππ 2π ππ 2
1 β πππ 2 2π ππ
=π π
1
ππ.
π₯π π₯ ππ₯
0
π’=π₯ ; ππ’ = ππ₯ ; = π₯π π₯ β
ππ£ = π π₯ ππ₯ π£ = ππ₯
1 π₯ π ππ₯ 0
= π₯π π₯ β π π₯
= 1β1+0β1 = 1
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
54
EXERCISE 11.3
π 2
ππ.
SOME PROPERTIES OF THE DEFINITE INTEGRAL
π
π ππ π₯πππ₯ππ₯
0
πππ‘ π’ = π πππ₯ ; ππ’ = πππ π₯ππ₯ π’2 ππ’
=
6β1 6β3 6β5 2β1 6+2 6+2β2 6+2β4 6+2β6
=2
π’3 3
=
π ππ 3 π₯ 3
=
π ππ6 π’ πππ 2 π’ ππ’
π
π
=
π
=2
π 2
=
π₯ π₯ π ππ6 πππ 2 ππ₯ 2 2 0 π₯ ππ₯ π’ = ; ππ’ = 2 2 ππ.
2
π π
π 2
ππ πππ
π 4
ππ.
π ππ2 4π₯ πππ 2 2π₯ ππ₯
8
π 2
ππ.
π ππ6 π₯πππ 4 π₯ ππ₯
π
=
2β1 2β1 2+2 2+2β2
=4
1 2
π
=
6β1 6β3 6β5 (4β1)(4β3)( 2 ) (6+4)(6+4β2)(6+4β4)(6+4β6)(6+4β8)
=
ππ πππ
=
π 2
π ππ
2
ππ.
4 β π₯2
3 2
ππ₯ ; πππ‘ π₯ = 2π ππβ
0
ππ₯ = 2πππ β π ππβ π 2
ππ.
2
π ππ7 π₯
=
4 β 2π ππβ
=
ππ = ππ
(2πππ β πβ )
0
π (4β1)(7β3)(7β5) 7(7β2)(7β4)(7β6)
3
2 2
2
=
3
(4 πππ 2 β )2 2πππ β πππ β πβ
0 2
=
8 πππ 3 β 2πππ β πβ
0
=( =
4β1
4β3
4 4β2
π 2
ππ ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
55
EXERCISE 12.1
1. π¦ = 3π₯ 2 ;
AREA UNDER A CURVE
ππππ π₯ = 1 π‘π π₯ = 2
3. π₯π¦ = β1 ; ππππ π₯ = 1 π‘π π₯ = 2 π¦=β
1 π₯
2
π΄=
2
π¦ππ₯
π΄=
1
π΄=
2 3π₯ 2 ππ₯ 1 2 3
π΄= π₯
π΄= 2
π¦ππ₯ 1
1 3
β 1
π΄=
2 1 β ππ₯ 1 π₯ 2
π΄ = [β ππ π₯] 3
π¨ = π ππ. πππππ
1
π΄ = {[β ππ 2] β [β ππ 1]} π΄ = β ππ 2; ππ’π‘ π‘ππππ ππ ππ πππππ‘ππ£π ππππ, πππππ, π¨ = πππ ππ. πππππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
56
EXERCISE 12.1
AREA UNDER A CURVE
5. π¦ = 3πππ₯, π₯ = 2 π‘π π¦ = 4
9. π₯ + π¦ = 3 & π‘ππ πππππππππ‘π ππ₯ππ
π
ππ΄ =
π¦ππ₯
0 4
π΄= 3
πππ₯ππ₯ 2
= 3[π₯πππ₯ β π₯] = 3[4 ππ 4 β 4] β 3[2 ππ 2 β 2] = 3[4 ππ 4 β 4 β 2ππ 2 + 2] = 3[8ππ2 β 2ππ2 β 2] π΄=
= 3[6ππ2 β 2] = 6[3ππ2 β 1]
3 0
3 β π₯ ππ₯
π΄ = 3π₯ β
π¨ = π[πππ β π] ππ. πππππ
π₯3 2
π΄= 3 3 β π¨= 7. π¦ = 9 β π₯ 2 π΄=
3 β3
; π₯ = β3 π‘π π₯ = 3
3 0 3 2 2
π ππ. πππππ π
4 β π₯ 2 ππ₯
π¨ = π ππππππ πππππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
57
EXERCISE 12.1
AREA UNDER A CURVE
11. π¦ 2 = 4π₯, π₯ = 1 πππ π₯ = 4
ππ. π₯π¦ = 1, π¦ = π₯, π₯ = 2, π¦ = 0
4
π΄=
4π₯ππ₯ 1 4
π΄= π΄=
1
4π₯ 2 ππ₯ 1
π₯π¦ = 1; π¦ = π₯
8 3 π₯4 3
π₯(π₯) = 1
8(4)3/2 8(1)3/2 π΄= β 3 3 π΄=
64 3
β
8 3
ππ π¨= ππ. πππππ π
π₯=1 ; π¦=1 2
π΄1 = 1
; (1,1)
1 ππ₯ π₯
= (ππ π₯) = ππ 2 β ππ 1 π΄1 = ππ 2 π π. π’πππ‘π π΄2 = =
1 ππ 2
1 1 1 2
π΄2 =
1 π π. π’πππ‘π 2
π΄π‘ = π΄1 + π΄2 π π¨ = (ππ π + )ππ. πππππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
58
EXERCISE 12.2
AREA BETWEEN TWO CURVES
5. y = x 2 ; y = 2 β x 2
1. π¦ = π₯ 2 ; π¦ = 2π₯ + 3 π¦ = 2π₯ + 3 π₯ 2 = 2π₯ + 3 π₯ 2 β 2π₯ β 3 = 0 π₯β3 π₯+1 = 0 π₯ = 3, π₯ = β1 π΄=
3 β1
ππ¦ = 2π₯ ; (0,0) ππ₯ π₯ = 0 ,π¦ = 0 π2 π¦ = 2 (ππππππ£π π’ππ€πππ) ππ₯ 2
2π₯ + 3 β π₯ 2 ππ₯
= [π₯ 2 + 3π₯ β
π₯3 ] 3 3 -1
= 32 + 3(3) β
(3)3
πππππ‘ ππ πππ‘πππ πππ‘πππ: β (β1)2 + 3(β1) β
3
(β1)3 3
5
= 9+3 π¨=
3. π₯ 2 = π¦ β 1
(2π₯ + 2)(π₯ β 1) 2π₯ + 2 = 0π₯ β 1 = 0 2 2π₯ = β π₯ = 1 2
; π₯ =π¦β3
π₯ = β1 π¦ = 1
Y1=Y2 π¦β3 2 =π¦β1 π¦ 2 β 6π¦ + 9 = π¦ β 1 π¦β5 π¦β2 =0 π¦ = 5 ,π¦ = 2 π₯ =5β3=2 2
π₯2 = 2 β π₯2 π₯2 β 2 + π₯2 = 0
ππ ππ. πππππ π
π΄=
y1= y2
ππ΄ = [π1 β π2]ππ₯ 1 β1 1
=
π₯ + 3 β (π₯ 2 + 1) ππ₯
2 β1
=
π₯2 π₯3 + 2π₯ β 2 3
π₯ + 2 β π₯ 2 ππ₯
=
22 2
+ 2(2) β
=
10 3
+6 =A=
7
23 3
= 2π₯ β 2 3
2
2π₯ 3 3
β =
β1 2 2 π π
+ 2(β1) β
(β1)3 3
=
2
2
= 2 β 3 β [β2 + 3]
=2β +2β
-1
27 6
(2 β 2π₯ 2 ) ππ₯
β1
β1
=
(2 β π₯ 2 β π₯ 2 ) ππ₯
ππ΄ =
2 3
=
12β4 3
π ππ. πππππ π
ππ. πππππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
59
EXERCISE 12.2
AREA BETWEEN TWO CURVES
7. π¦ = π πππ₯ ; π₯ = πππ π₯ ; π₯ = x 0 90 180 270 360
y 0 1 0 -1 0
π΄2 =
x 0 90 180 270 360
π 2 π 4
πππ π₯ =
π 2
11. π¦ = π₯ 3 , π¦ = 8, π₯ = 0 ππ¦ = 3π₯ 2 ππ₯
y 1 0 -1 0 1
, 0 = 3π₯ 2
π¦ = 0 ,π₯ = 0 π2 π¦ = 6π₯(ππππππ£π π’ππ€πππ) ππ₯ 2 πππππ‘ ππ πππ‘πππ πππ‘πππ:
π πππ₯ππ₯ = [-πππ π₯]
y1= y2
π 4
π 2
= [-πππ ] β [-πππ ] = π΄1 =
π 4
π 2 π 4
2 2
π₯3 = 8
π π πππ π₯ππ₯ = π πππ₯ = π πππ₯ β π ππ 2 4
π₯3 β 8 = 0 π₯3 = 8 π₯=
2 = 1β 2
3
8
π₯=2
π¨π β π¨π = π β π ππ. πππππ
π€πππ π₯ = 2 π¦ = 8 , (2,8) π€πππ π₯ = β2
9. π₯ 2 = 4π¦ , π¦ = π¦=
π₯2
8 π₯ 2 +4
4
π₯ 2 π₯ 2 + 4 = 32 2 8 π₯2 π΄= β ππ₯ 2 4 β2 π₯ + 4 π΄ = 4.95
π¦ = β2
3
, π¦ = β8
(-2,-8) ππ΄ = [π1 β π2]ππ₯ 2
ππ΄ =
(8 β π₯ 3 ) ππ₯
0
= ππ β π = ππ ππ. πππππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
60
EXERCISE 12.2
AREA BETWEEN TWO CURVES
13. π¦ = 2π₯ + 1 , π¦ = 7 β π₯ , π₯ = 8
π
8
π΄=
15. π¦ = πππ₯ 3 , π¦ = πππ₯; π₯ = π
2π₯ + 1 β 7 β π₯ ππ₯
π΄= 1
2
π
8
=
2π₯ + 1 β 7 + π₯ ππ₯
=
8 2
3π₯ β 6 ππ₯
=
3π₯ 2 = β 6π₯ 2 =
3(8)2 2
π πππ₯ 3 1
π’ = πππ₯ 3
8 2
β 6(8) β
[(πππ₯ 3 ) β (πππ₯)]ππ₯
1
2
=
(π¦2 β π¦1 )ππ₯
ππ’ = 3(2)2 2
β 6(2)
= ππ₯ ; ππ’ = π
1
= ππ ππ πππππ
πππ₯
; π£ = π₯ ; π’ = πππ₯ ; ππ£ = ππ₯
3π₯ 2 ππ£ π₯3
= π₯πππ₯ 3 β
π 1
β
= π₯πππ₯ 3 β
3π₯ 2 π₯( 3 ) π₯ 3π₯
= π₯πππ₯ 3 β 3π₯
π 1
ππ₯ π₯
π β [π₯πππ₯ β 1
;
π£=π₯
ππ₯ π π₯( )] π₯ 1
β [π₯πππ₯ β π₯] π1
= π ππ. πππππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
61
EXERCISE 12.2
AREA BETWEEN TWO CURVES
17. π¦ 2 = 2ππ₯ , π¦ 2 = 4ππ₯ β π2
π€πππ π₯ = β4π
π¦ 2 = 2ππ₯π¦ 2 = 4ππ₯ β π2
π₯ = (β4π)2
π₯= ππ₯ ππ¦
π¦2 2π
;x=
π¦ 2 +π 2 4π
2π¦
=
16π2 2π
= 8π
= 2π
ππ₯ π¦ = ππ¦ π
π΄
π
ππ΄ =
0 = 0 ; (0,0)
π
[ βπ
π¦ 2 + π2 π¦ 2 β ]ππ¦ 4π 2π
π2 π₯ 1 = ππππ π‘π π‘ππ πππππ‘ ππ¦ 2 π
=
π π¦ 2 +π 2 βπ¦ 2 ( 4π )ππ§ βπ
πππππ‘ ππ πππ‘πππ πππ‘πππ:
=
π¦3 π2 π¦ 2π¦ 3 + β 12π 4π 12π
=
π3 π2 π 2π3 (βπ)3 π2 (βπ) 2(βπ)3 + β β + β 12π 4π 12π 12π 4π 12π
=
π3 β 2π3 + π3 β 2π3 π3 + π3 + 12π 4π
=
β2π 3 12π
X1 = X2 π¦ 2 π¦ 2 + π2 = 2π 4π 4ππ¦ 2 = 2ππ¦ 2 + 2π3 4ππ¦ 2 β 2ππ¦ 2 β 2π3 = 0 2ππ¦ 2 β 2π3 = 0 2ππ¦ 2 = 2π3 2π3 π¦2 = 2π 2 π¦ = π2 π¦ = π2 π¦ = Β±π π2
+
2π 3 4π
=
a -a
β2π 3 +6π 3 12π
4π 3
= 12π A=
a2 sq. 3
units
π
X1 = X2=2π = 2 π€πππ π₯ = 4π π₯= =
(4π)2 2π
16π2 2π
= 8π DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
62
EXERCISE 12.2
AREA BETWEEN TWO CURVES
ππ. π¦ 2 = π₯ + 1 ; π¦ = 1 β π₯
ππ. π¦ 2 = 4π₯ ; π¦ = 4π₯ β 4
π£1= π¦ 2 β 1; π¦π₯ = 1 ππ₯ = 2π¦ ; π₯2 = 1 β π¦ ππ¦ π₯ = 0; π¦ = 0 π2 π₯ = 2 (ππππππ£π π‘π π‘ππ πππππ‘) ππ¦ 2 πππππ‘ ππ πππ‘πππ πππ‘πππ 1 π₯1= π¦2 ; π¦ 2 β 1 = π¦ π¦2 + π¦ β 2 = 0 (π¦ β 1)(π¦ + 2) π¦β1=0 π¦+2=0
4π₯ = π¦ 2 2π₯ = π¦ + 4
y=1
π₯=
ππ₯ 1 = 2π¦ ππ¦ 4 1 0 = 2π¦ 4 0=0 0,0
y=-2
π£=0
π¦2 π¦+4 π₯= 4 2
π2π₯ ππ¦ 2
π¦=3
π€πππ π₯ = 1, π¦ = 2
= (concave to the right)
πππππ‘ ππ πππ‘πππ πππ‘πππ π¦2 π¦ + 4 = 4 2
π€πππ π₯ = 2, π¦ = 5 π€πππ π¦ = 1, π₯ = 0 π€πππ π¦ = 2, π₯ = 3 π€πππ π¦ = 3, π₯ = 8 π‘πππ;
2π¦ 2 β 4π¦ + 4(4) 2π¦ 2 β 4π¦ β 16 = 0
ππ΄ = π2 β π1 ππ¦ 1
1
1 β π¦ β π¦ 2 β 1 ππ¦
ππ΄ = β2
2π¦ β 8 π¦ + 2 2π¦ β 8 = 0π¦ + 2 = 0 π¦ = 4; π₯ = 4(1, 2)
β2
π΄ = 1 β π¦ β π¦2 + 1 π΄ = 2βπ¦β
1 β2
(4, 4)
π¦ 2 1β2
ππ΄ = (π₯2 β π₯1 )ππ¦
3 1
π΄ = 2π¦ β
π΄ = 2(1) β
π¦2 π¦ β 2 3
(1)2 (1)3 β 2 3
1
1
π΄= β2 β π΄ = 2(β2) β
(β2)2 (β2)3 β 2 3
4 π¦+4 β2 2
β
π¦2 4
ππ¦
π¨ = π ππ. πππππ
8
π΄=2β2β3+4+2β3 π¨=
π ππ. πππππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
63
EXERCISE 12.2
AREA BETWEEN TWO CURVES
23. π¦ 2 = π₯ + 4 , π₯ β 2π¦ + 1 = 0
3
π΄=
2π¦ β 1 β π¦ 2 β 4 ππ¦
β1 3
3 + 2π¦ β π¦ 2 ππ¦
= β1
= 3π¦ + π¦ 2 β =
3 π¦3 3 β1
ππ π
25. π¦ = π 2π₯ , π¦ = π , π₯ = 2
2
π΄=
π 2π₯ β π π₯ ππ₯
0
= =
π 2π₯ β ππ₯ 2 π4 2
2 0 1
β π2 β 2 + 1
= π π ππ β π
π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
64
EXERCISE 12.4
VOLUME OF A SOLID OF REVOLUTION
1. π¦ = π₯ 2 β 2π₯ , π₯ β ππ₯ππ , ππππ’π‘ π‘ππ π₯ β ππ₯ππ ππ¦ = 2π₯ β 2 , πππ’ππ‘π π‘π π§πππ ππ₯ 0 = 2π₯ β 2 ; π¦ = 12 β 2(1) π₯=1
;
π¦ = β1
π2π¦ =2 ππ₯ 2
π¦ = π₯ 2 β 2π₯
π£ 1, β1 x y
0 0
1 -1
2 0
3 3
1 -1
ππ£ = ππ¦ 2 ππ₯
dx
ππ£ = π π₯ 2 β 2π₯ 2 ππ₯
2
y (1,-1)
-2
Κππ£ = πΚ π₯ 4 β 4π₯ 3 + 4π₯ 2 ππ₯ π£=π =π
π₯5 5
β
1 2 5
4π₯ 4 4
5
+
4π₯ 3 3
β 24 +
4 2 3
=π
32 32 β 16 + 5 3
=π
96 β 240 + 160 15
=π
16 15
π½=
3
β 0
πππ ππππππ ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
65
EXERCISE 12.4
VOLUME OF A SOLID OF REVOLUTION
π. π₯ + π¦ = 5 ; π¦ = 0 ; π₯ = 0 ; ππππ’π‘ π¦ = 0
π. π₯ + π¦ = 6 ; π¦ = 3 ; π₯ = 0 ; ππππ’π‘ π¦ β ππ₯ππ
π€πππ π₯ = 0 ; π¦ = 5
π₯ = (6 β π¦)
π€πππ π¦ = 0 ;
π₯=5
ππ£ = ππ₯ 2 ππ¦
ππ£ = ππ¦ 2 ππ₯
; ππ’π‘ π¦ = 5 β π₯
ππ£ = π 6 β π¦ ππ¦
π¦2 = 5 β π₯
2
ππ£ = π 36 β 12π¦ + π¦ 2 ππ¦
ππ£ = π 5 β π₯ 2 ππ₯ π£
5
ππ£ = π 0
π£ 0
25 β 10π₯ + π₯ 2 ππ₯
π=π
10π₯ 2 π₯ 3 + 2 3
25 5 β 5 5
π = π 125 β 125 + π½=
2
125 3
0
π=
+
1 5 3
3
36 β 12π¦ + π¦ 2 ππ¦
π¦2 π¦3 π = π 36π¦ β 12 + 2 3
0
π = π 25π₯ β
3
ππ£ =
β 0
36 3 β 6 3
2
+
π = π 36 3 β 6 9 +
β0
1 3 3
2
β 0
1 3 27
π = π[ 36 3 β 6 9 + 9]
ππππ ππππππ π
π = π(9)(12 β 6 + 1) π = π(9)(7)
y
π½ = πππ ππππππ
π₯=0
π¦π₯ = 0 5
(0,6)
3 2 1 ππ₯
π₯+π¦ =6
5 y
π¦=3
3 3
5
x π¦=0
ππ¦ 1 0
(6,0) π₯ 1
3
5
π₯
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
66
EXERCISE 12.4
VOLUME OF A SOLID OF REVOLUTION
π. π₯π¦ = 4, π₯ = 2, π¦ = 4; ππππ’π‘ π¦ = 4
9. π¦ 2 = 4ππ₯, π₯ = π; ππππ’π‘ π₯ = π
π = ππ 2 π
2
π = ππ π π = π(4 β π¦)2 ππ₯
π = π(π β π₯)2 π π£
π = π 4β π£
4 π₯ 2
ππ£ = π 0
0
ππ₯ 32 16 (16 β + 2 ) ππ₯ π₯ π₯
2π
ππ£ =
2 0
β2π 2π
π= π
π(π β
(π2 β
β2π
16 π = π 16 2 β 32ππ2 β β0 2
π = π π2 π¦ β
π = 8π 4 β 4 ππ 2 β 1
π = π2 2π β
π½ = ππ π β π ππ π ππ. πππππ
π¦2 2 ) ππ¦ 4π
π¦2 π¦4 + ) ππ¦ 2 16π2
π¦3 π¦5 + 6 16 5 π2
2π3 2π5 β2π3 β2π5 + β π2 β2π β + 6 16(5)π2 16 16(5)π2 2
1
π = 4π3 π 1 β 3 + 5 π½=
ππππ π ππ. πππππ ππ
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EXERCISE 12.4
VOLUME OF A SOLID OF REVOLUTION
ππ. π¦ = π ππ π₯, π₯ = 0, π¦ = 1; ππππ’π‘ π¦ = 1
π = ππ 2 π π = π(1 β π¦)2 ππ₯ π 2
π£
π£= 0
π (1 β π ππ π₯ )2 ππ₯
0 π 2
π= π
0
1 β 2π πππ₯ + π ππ2 π₯ ππ₯
π = π[π₯ + 2 πππ π₯ +
π₯ π ππ2π₯ β ] 2 4
π= π
3π₯ π₯ π ππ 2π₯ + 2 πππ π₯ + β 2 2 4
π= π
3π₯ π ππ 2π₯ + 2 πππ π₯ β 2 4
π= π
3π + 0 β 4(0) β 0 + 2 + 0 4
π= π½=
3π 2 4
β 2π
π ππ β π ππ. πππππ π
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EXERCISE 12.5
THE WASHER METHOD 9
1. π¦ = π₯ 2 , π₯ = 3, π¦ = 0; ππππ’π‘ π‘ππ π¦ β ππ₯ππ
π=π
32 β π₯ 2 ππ¦
0 9
π=π
9 β π¦ ππ¦ 0
π = π 9π¦ β
π¦2 9 2 0
π = π 9(9) β π½=
(9)2 9 2 0
πππ πͺπΌπ©π°πͺ πΌπ΅π°π»πΊ π
3. π¦ 2 = 4ππ₯, π₯ = π; ππππ’π‘ π‘ππ π¦ β ππ₯ππ x 0 a
π¦ 2 = 4ππ₯
y 0 2a
dy x 2π
X=a
π2 β π₯ 2 ππ¦
π= π β2π 2π
π¦2 = π (π β 4π β2π 2π
= π
2
π2 β
β2π
2
)ππ¦
π¦4 ππ¦ 16π2
π¦5 2π = π π π¦β 80π2 β2π 2
= π (2π3 β
32π5 32π5 3 ) β (β2π + ) 80π2 80π2
= π (2π3 β
2π 3 )β 5
π½=
(2π3 +
2π 3 ) 5
πππ ππ π
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EXERCISE 12.5
THE WASHER METHOD
5. π₯ 2 +π¦ 2 = π2 , π₯ = π π π = 4π 0 π2 β π¦ 2 + π ππ¦ π = 4π π2 π¦ β π = 4π π = 4π π½=
π¦3 3
+ ππ¦
π3 π3 β 3 β 2π 3 β ππ 3
ππ
a
(-a,0)
o
(a,0)
x=b
ππ ππ π
7. π₯ 2 + π¦ 2 = 25 , π₯ + π¦ = 5 ; π¦ = 0 π= π π½=
5 0
25 β π₯ 2 β 5 β π₯
2
ππ₯
ππππ ππ. πππππ π
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EXERCISE 12.5
THE WASHER METHOD
9. π¦ 2 = 4π₯, π₯ 2 = 4π¦; ππππ’π‘ π‘ππ π₯ β ππ₯ππ
π¦2 = π¦1
2
π₯2 4π₯ = 4 3 64π₯ = π₯ π₯ = 4, π¦ = 4: πππΌ (4,4) 4
π=π
4π₯
2
β
0 4
π₯2 4
2
ππ₯
π₯4 ππ₯ 16 0 π₯5 4 π = π 2π₯ 2 β 80 0 (4)5 π = π 2(4)2 + 80 πππ π½= πͺπΌπ©π°πͺ πΌπ΅π°π»πΊ π π=π
11. π¦ 2 = 8π₯, π = 2π₯; ππππ’π‘ π¦ = 4
4π₯ β
π¦2 = π¦1 8π₯ = 2π₯
2
8π₯ = 4π₯ 2 π₯ = 2, π¦ = 4: πππΌ (2,4) 4π₯ 3 2 π = π 4π₯ 2 β 3 0 4(2)3 π = π 4(2) + 3 2
π½=
πππ πͺπΌπ©π°πͺ πΌπ΅π°π»πΊ π
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EXERCISE 12.6
THE CYLINDRICAL SHELL METHOD
π. 4π¦ = π₯ 3 , π¦ = 0, π₯ = 2, ; ππππ’π‘ π₯ = 2
V = 2Ο
2 π₯π¦ππ₯ 0
V = 2Ο
2 0
V = 2Ο
2 π₯2 [ 0 2
V = 2Ο V = 2Ο V = 2Ο V=
2βπ₯
π₯4 4
β
β
(2)4 4
β
π₯3 4
dx
π₯4 ]ππ₯ 4
π₯5 20
2 0
(2)5 20
3 5
ππ cubic units π
2 0
3. π₯ = 4π¦ β π¦ 2 , π¦ = π₯ , ππππ’π‘ π¦ = 0
V = 2Ο
3 π₯π¦ππ¦ 0
V = 2Ο
3 0
V = 2Ο
3 0
V = 2Ο
4π¦ β π¦ 2 β π¦ π¦ππ¦ 4π¦ 2 β π¦ 3 β π¦ 2 ππ¦
4 2 π¦ 3
V = 2Ο π¦ 3 β
π¦4 4
V = 2Ο (3)3 β V=
πππ πππππ π
1 4 π¦ 4
β
1
β 3 π¦3
3 0
(3)4 4
3 0
πππππ
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3 0
EXERCISE 12.6
THE CYLINDRICAL SHELL METHOD
π
5. π¦ = π πππ₯, π¦ = πππ π₯, π₯ = 2 π = 2π π½=
π 2 π 4
π₯ π πππ₯ β πππ π₯ ππ₯
π π + ππ β ππ ππ. πππππ π
7. π₯ = 2 π¦ π = 2π π½=
9 0
π 2
,π₯ = 0 ,π¦ = 0
Y=9
9 β π¦ 2 π¦ ππ¦
πππππ ππ. πππππ π
9.π¦ = πππ₯ , π₯ = π , π¦ = 0 π = 2π
π π₯ 1
πππ₯ ππ₯ (e,1)
π½ = ππ. ππ ππ. πππππ
(1,0) X=e
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EXERCISE 12.6
THE CYLINDRICAL SHELL METHOD
11. π¦ 2 = 8π₯ , π₯ = 0 , π¦ = 4 ; about π¦ = 4
4
π = 2π
4βπ¦ 0
=
π 4
4
π¦2 ππ¦ 8
4π¦ 2 β π¦ 3 ππ¦
0
=
π 4π¦ 3 4 3
=
πππ π
β
4 π¦4 4 0
13. ( π₯ β 3 ) 2 + π¦ 2 = 9; ππππ’π‘ π‘ππ π¦ β ππ₯ππ . 3
π = 8π
2 x 9 ο ( x ο 3) dx
0 π = 8π(
β ( 9 β ( x β 3 ) 2 3 27 π πππ₯ β 3 9 )2 + + (π₯ β 3)( 9 β π₯ β 3 3 2 3 2
π = 8π(27π πππ β
2
3 0
27 π ππ β 1) 2
27 π = 8π( )(βπ ππ β 1 + π πππ) 2 π π = 108π( ) 2 π½ = πππ π
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74
EXERCISE 12.6
THE CYLINDRICAL SHELL METHOD
15. π₯ 2 + π¦ 2 = π2 ; ππππ’π‘ π₯ = π π > π π
π= π
πβπ₯ βπ π
π= π
2
β πβπ₯
2
ππ¦
π 2 β ππ₯ + π₯ 2 β π 2 β ππ₯ + π₯ 2 ππ¦
βπ π
π= π
4ππ₯ππ¦ βπ
πππ‘π: π₯ 2 + π¦ 2 = π2 π = 4ππ π = 4ππ
π βπ
π¦ 2 β π2
=π₯=
π¦ 2 β π2 ππ¦
π¦ π2 π π¦ 2 β π2 β ln π¦ + π¦ 2 β π2 + π βπ 2 2
π½ = ππ π ππ π π₯ 2 + π¦ 2 = π2
a
a
a
a
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75
EXERCISE 12.7
π.
VOLUME OF SOLIDS WITH KNOWN CROSS SECTIONS
π₯ 2 + π¦ 2 = 36
π. 9π₯ 2 + 16π¦ 2 = 144
π2 π΄ π₯ = , 2
π = 2π¦
π΄ π₯ = 2π¦ 2 ,
π¦=
36 β π₯ 2
6
π£=
π΄ π₯ ππ₯ β6
1 π΄ π₯ = (2π¦)(π¦) 2 π΄ π₯ = π¦2 8
0
2π₯ 2 ππ₯
π£=
π¦ 2 ππ₯
π=2
6
π=2
β6
π£=
1 π΄ π₯ = ππ 2
6 2(3π₯ β6
β π₯ 2 )ππ₯
8 144β9π₯ 2 0 16
ππ₯
π½ = ππ ππ. πππππ
π = πππ ππ. πππππ
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EXERCISE 12.7
VOLUME OF SOLIDS WITH KNOWN CROSS SECTIONS
π.
π΄ π¦ = (1 β π₯)(2π¦ 2 ) 2
π = 2 (1 β π₯ )2π¦ 2 ππ¦ 0 2
π = 2 (1 β 0
π¦2 2 )π¦ ππ¦ 4
64
π = 15 π½ = π. ππππ ππ. πππππ
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77
EXERCISE 12.8
LENGTH OF AN ARC
3
π. π¦ = π₯ 2 ππππ π₯ = 0 π‘π π₯ = 5
2
2
3
π¦ = π₯2 3 1 ππ¦ = π₯ 2 ππ₯ 2 ππ¦ 3 1 = π₯2 ππ₯ 2 π = 0
π=
=
ππ¦ 1 + ( )2 ππ₯ ππ₯
=
5 0
1+
1+ β 0
2
π=
π₯
0
π¦3 1
π₯3
2
ππ₯
1
π₯3 + π¦3
2
5 0
1
9
9 5
2
3. π‘ππ πππ‘πππ ππ¦ππππ¦πππππ π₯ 3 + π¦ 3 + π3
2 3
2
ππ₯
2
πππ‘π: π3 = π₯3 + π¦3
3 1 ( π₯ 2 )2 dx 2
2
9
9 4
π3
π=
1 + π₯ ππ₯
2
π₯3
0
π = ππ. πππ πππππ
9
π= 0
ππ₯
1
π3 1
π₯3
ππ₯ 2
3π₯ 3 9 π= π 2 0 1 3
π=
3π 2
π=4
3π 2
πΊ = ππ
X=0
x=5
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EXERCISE 12.8
LENGTH OF AN ARC
5. π¦ = π΄πππ πππ π₯ , ππππ π¦ =
π 6
π‘π π¦ =
π 2
π¦ = π΄πππ πππ π₯ ; ππ π πππ¦ = π₯ 1 πππ π¦ππ¦ = ππ₯ π πππ¦
7. πππ ππππ ππ π‘ππ ππ¦πππππ π₯ = π π β π πππ , π¦ = π(1 β πππ π) π₯ = π(π β π πππ) ππ₯ = π(ππ β πππ πππ)
π¦ = π(1 β πππ π) ππ¦ = π(π πππππ)
ππ₯ ππ
ππ¦ ππ
= π(1 β πππ π)
ππ₯ πππ π¦ = ππ¦ π πππ¦
2π
π =
ππ₯ = πππ‘π¦ ππ¦ π= =
π 2 π 6
π 2 π 6
2
+ π2 sin2 π
0
π =π 1+
π2 1 β πππ π
= ππ πππ
ππ₯ 2 ππ¦
ππ¦
2π 0
1 β πππ π
2
+ sin2 π
π = ππ
1 + πππ‘ 2 π¦ ππ¦
πΊ = π. πππππ πππππ
9. πππ πΆπππππππ π = 2 1 β πππ π
π 2
π = 2 1 β πππ π ππ = 2 π πππ ππ ππ = 2π πππ ππ π 2 = 4(1 β πππ π)2 2π
π=
4(1 β πππ π)2 + 4π ππ2 π ππ
0
π=2 π 6
2π 0
(1 β πππ π)2 + π ππ2 π ππ
πΊ = ππ πππππ
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79
EXERCISE 12.9
AREA OF A SURFACE OF REVOLUTION
1. π₯ 2 + π¦ 2 = 16 ; ππππ π₯ = 2 π‘π π₯ = 4
3. π¦ 2 = 12π₯ ; ππππ π₯ = 0 π‘π π₯ = 3
4
π = 2π
3
π¦ππ
π = 2π
2
π¦=
π¦ = 12π₯
16 β π₯ 2
ππ¦ 1 = 16 β π₯ 2 ππ₯ 2
1 2
β
ππ¦ 1 = 12π₯ ππ₯ 2
(β2π₯)
2
ππ =
ππ¦ 1+ ππ₯
ππ =
π₯2 1+ ππ₯ 16 β π₯ 2
ππ =
16 β + 16 β π₯ 2
16 β π₯ 2
π = 2π
π₯2
(12)
ππ =
16 β π₯ 2
36 ππ₯ 12π₯
12π₯ + 36
ππ =
12π₯ 2 3π₯ + 9
ππ₯
12π₯
ππ₯
ππ₯
3
12π₯
2 3π₯ + 9
0
ππ₯
2
π = 2π
1+
π = 2π
4 4
ππ =
ππ₯
π₯2
1 2
ππ¦ 6 = ππ₯ 12π₯
ππ¦ π₯ =β ππ₯ 16 β π₯ 2
ππ =
π¦ππ 0
4 16 β π₯ 2
π = 4π ππ₯
3 0
12π₯
ππ₯
3π₯ + 9 ππ₯
πΊ = πππ. πππ ππ. πππππ
4 4ππ₯ 2
πΊ = πππ ππ. πππππ
5. π¦ = π₯ 3 ; ππππ π₯ = 0 π‘π π₯ = 1 ππ¦ = 3π₯ 2 ππ₯ ππ = π = 2π
1 + 9π₯ 4 ππ₯ 1 3 π₯ 0
1 + 9π₯ 4 ππ₯
πΊ = π. ππππ ππ. πππππ
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80
EXERCISE 12.9
AREA OF A SURFACE OF REVOLUTION
7. π₯ = πππ 2π¦ ; ππππ π¦ = 0 π‘π π¦ = π 4
π = 2π
π 4
π₯ππ
0
ππ₯ = β π ππ 2π¦(2) ππ¦ ππ = π = 2π
1 + 4 π ππ2 2π¦ π 4
0
πππ 2π¦ 1 + 4 π ππ2 2π¦ ππ¦
πΊ = π. πππππ ππ. πππππ
9. 4 β π₯ 2 ππππ π₯ = 0 π‘π π₯ = 2 2
π = 2π
π₯ππ 0
ππ¦ = β2π₯ ππ₯ ππ = 1 + 4π₯^2 ππ₯ π = 2π
2 π₯ 0
1 + 4π₯ 2 ππ₯
πΊ = ππ. ππππ ππ. πππππ
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81
EXERCISE 12.9
AREA OF A SURFACE OF REVOLUTION
13. π¦ = ππ₯ ; π₯ = 0 ; π₯ = 1 ; ππππ’π‘ π‘ππ π₯ β ππ₯ππ 1
π = 2π
ππ₯
1 + π2
ππ₯
0 1
π = 2π π
1 + π2 =
π₯ ππ₯ 0
π = 2π π π = 2π π
1 + π2( π₯2/2 )
1 0
1 + π2( Β½ )
πΊ = π π π + ππ
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82
EXERCISE 13.1
1.
FORCE OF FLUID PRESSURE
πΉ = π€π΄π₯ = (62.5ππ/ππ‘ 3 )(96ππ‘ 2 )(4ππ‘) = 24000ππ
π=
πΉ π΄
π=
π€π΄π₯ π΄
3. πΉ = π€π΄π₯ 1 πΉ=π€ 5 3 2
2 2 + ( )(3) 3
π = πππ ππ
2 3
π = π€π₯
5
62.5ππ3 1ππ‘ 2 π=( )(4ππ‘)( ) ππ‘ 144ππ 2 π=
3 5
(625)(4) 144
5 3
π· = π. ππ πππ
5
5. πΉ = 50π€
12ft
πππ π = 3ππ‘ 8ft
π₯
πΉ = π€π΄π₯ 50 =
1 π 3 2
50 =
π2 2
1 π 3
100 = π2 π = ππππ
3 5
h 5
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83
EXERCISE 13.1
FORCE OF FLUID PRESSURE
7. πΉ = π€π΄π₯ = π€[(π)(3)(2)](2) π = πππ π π = 6 = major axis π = 4 = πππππ ππ₯ππ
0
y π₯
b a
A=πππ
x
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
84
EXERCISE 13.2
WORK
1.
π
π€=
π π₯ ππ₯ π
π π₯ = ππ₯ 40 ππ = π π€=
;
π€ππππ π₯ =
1 ππ‘, 2
π π₯ = 40 ππ ; π = 0,
π = 14 β 10 = 4
1 ππ‘ , π = 80 2
4 80π₯ππ₯ 0
π = πππ ππ β ππ
3.
π
π€=
π π₯ ππ₯ π
π π₯ = ππ₯ ; π€ππππ π₯ = π€=
1 πΏ ππ‘, 10
π π₯ = 5 ππ
π = 0, π = πΏ
πΏ 50 π₯ππ₯ 0 πΏ
π = πππ³ ππ β ππ
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85
EXERCISE 13.2
WORK
5. π = πΉπ ππ€ = π€ ππ£ 60 β π₯ ππ€ = ππ 2 π€ 60 β π₯ ππ₯ ππ€ = 9ππ€(60 β π₯)ππ₯ π€
10
ππ€ = 9ππ€ 0
60 β π₯ ππ₯ 0
π€ = 9ππ€ 60π₯ β π₯ 2 π€ = 9ππ€ 60π₯ β
10 0
π₯ 2 10 2 0
π€ = 9ππ€ 600 β 50 π = ππππππ ππ. ππ
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86
EXERCISE 13.2
WORK
9.
π
π€=π€
πππ π
πππππ‘πππππ = π π₯ π€ π₯ π; π€ππππ π = 10 ππ‘, π€ = 2π₯, π = ππ¦ π₯2 + π¦2 = π2 ; π₯ =
π 2 β π¦ 2 ; π€ππππ π = 2
2
π€=π
6βπ¦
10 ππ‘ 2π₯ ππ¦
β2
2
π€ = 20π
6βπ¦
22 β π¦ 2 ππ¦
β2
π = ππππ π ππ β ππ
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87
EXERCISE 13.3
FIRST MOMENT OF A PLANE AREA
1. π¦ 2 = 4π₯, π‘ππ π₯ β ππ₯ππ πππ π₯ = 4 ππ₯ =
5. π¦ 2 = 4π₯ πππ π₯ 2 = 4π¦
1 4 4π₯ππ₯ 2 0
π΄π = ππ ππ¦ =
4 π₯ 0
4π₯ ππ₯
π΄π = ππ. π
π¦2 = 4π₯
4π₯ =
π₯4 16
64π₯ β π₯ 4 = 0
π₯=4
π₯ 64 β π₯ 3 = 0 π₯1 = 0, π₯2 = 4 1
ππ = 2 3. π₯ = 4
π΄π =
4 π
4π₯ β
π₯4 16
ππ¦
ππ π
π
ππ = ππ = ππ = ππ = π΄π =
πππ΄ π 4
π₯ 4 β π₯ ππ¦ π 4 π 4
4π₯ β π₯ 2 ππ¦ π¦2 β
π
π¦4 ππ¦ 16
πππ ππ DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
88
EXERCISE 13.3
FIRST MOMENT OF A PLANE AREA
7. π =
3 0
3βπ¦ [
=
3 0
3βπ¦
27π β9β9 4 27π π= β 18 4 27πβ72 π= 4
9 β π¦ 2 β 3 β π¦ ππ¦
π=
3 + π¦ 3 β π¦ β (3 β π¦)2 ππ¦
3 1 3 (3 β π¦)2 (3 + π¦)2 β (3 β π¦)2 ππ¦ 0 3 3 3 = 0 3 9 β π¦ 2 ππ¦ β 0 π¦ 9 β π¦ 2 ππ¦ β 0 (3 β π¦)2 ππ¦
=
3 9 0 9βπ¦ 2 3
*π΄ =3 πππ π =
β π¦2
ππ¦
π [ππ β π] π
π¦
π πππ = 3
9 β π¦2
3πππ π =
π΄=
π¦
3π πππ = π¦; π = ππππ ππ 3
3πππ πππ = ππ¦ π π¦ = 3; π = 2
π. π₯ = 4π¦ β π¦ 2 , π¦ = π₯
π¦ = 0; π = 0 =3 = 27
π 2
0
π 2
3πππ ππππ πππ
0
= 27
π 2
πππ 2 πππ = 27
0
π
1 + πππ 2π ππ 2
π π ππ2π 2 π 27π + = 27 = 2 4 4 4 0 3
*π΅ = β
π¦ 9 β π¦ 2 ππ¦
0
π’ = 9 β π¦2 ππ’ = β2π¦ππ¦ β = = =
1 2
ππ’ 2
9βπ¦ 2 3 2
= π¦ππ¦ | 30 3 2
1 2 ( )[(9 β 9) β 2 3 1 β27 = β9 3
*πΆ =β
@ π¦ = 3; π’ = 0 π¦ = 0; π’ = 9
3 (3 β 0
3 2
π
ππ¦ =
1 2
ππ¦ =
1 3 2 0
π΄π =
ππ π
(9 β 0) ]
π
π₯π2 β π₯π2 ππ¦ 4π¦ β π¦ 2
2
β π¦ 2 ππ¦
π¦)2 ππ¦
π’ =3βπ¦ ππ’ = βππ¦ = =
(3βπ¦)3 3 |0 3 3 0 3 β 3 3
= β9
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89
EXERCISE 13.4
CENTROID OF A PLANE AREA
1. π₯ + 2π¦ = 6, π₯ = 0, π¦ = 0 Solving for A ππ΄ = π¦ππ₯ 6 ππ΄ 0
=
6 0
π΄ = [3π₯ β
π₯
3 β 2 ππ₯
π₯2 6 ] 4 0
π΄= 3 6 β
36 4
π¨ = π ππ. πππππ
Solving for π₯
Solving for π¦
π΄π₯ =
6 ππ 0
ππ΄
π΄π₯ =
6 π₯ 0
3β
π₯2 2
ππ₯
π΄π¦ = 2
π΄π₯ =
6 0
3π₯ β
π₯2 2
ππ₯
π΄π¦ =
3π₯ 2 2
π΄π₯ = [
β
π΄π¦ =
6 ππ 0 1
π₯3 6 ] 3 0
ππ΄
6 π₯ π₯ (3 β 2 ) (3 β 2 )ππ₯ 0
1 6 (9 β 2 0 1
3
3π₯ +
π₯2 )ππ₯ 4 π₯3
π΄π¦ = 2 [9π₯ β 2 π₯ 2 + 12 ] 60 1
9π₯ = 18
π¦ = 3 (3)
π = π πππππ
π = π ππππ
Centroid: (2,1)
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
90
EXERCISE 13.4
CENTROID OF A PLANE AREA
3. π¦ = π πππ₯, π¦ = 0 ππππ π₯ = 0 β π
A= π¦ππ₯ =
π 0
π πππ₯ππ₯
= βπππ π₯ A=2 π¦
ππ₯ = =
1
π 0
1
π 0
=2
π₯πππ; π₯π = π₯
=
π 0
π¦ 2 ππ₯
=
π 0
π ππ2 π₯ ππ₯
π’ = π₯ ; ππ£ = π πππ₯
1 π 1βπππ 2π₯ ( ) ππ₯ 2 0 2 1 π₯
= 2 (2 β 2 1 π₯
= 2 (2 β
π ππ 2π₯ 2
π₯π¦ππ΄ π₯π πππ₯ππ₯
ππ’ = ππ₯ ; π£ = βπππ π₯ ππ₯
= βπππ π₯ β βπππ π₯ππ₯
π ππ 2π₯ ) 4
= [βπ₯πππ π₯ + π πππ₯]
π 4
ππ₯ = (2) π₯ =
π 0
ππ¦ =
π π¦ ( )π¦ππ₯ 0 2
=2
=
π¦πππ; π¦π = 2
= βππππ π + π πππ + 0 β π ππ0
π 2
=π π
π¦ = ( 4 )(2) =
Centroid:
π 8
π π , π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
91
EXERCISE 13.4
CENTROID OF A PLANE AREA
7. π¦ 2 = π₯ 3 , π¦ = 2π₯ 4
π΄=
3
(2π₯ β π₯ 2 )ππ₯ 0
2 5 π΄ = [π₯ 2 β π₯ 2 ] 5 π΄ = [16 β π΄=
64 ] 5
16 π π. π’πππ‘π 5
4
π΄π₯ =
π¦π₯ππ₯
π΄π¦ =
0 4
π΄π₯ =
3
(2π₯ β π₯ 2 )π₯ππ₯
π΄π¦ =
0 4
π΄π₯ =
5
(2π₯ 2 β π₯ 2 )ππ₯
π΄π¦ =
0
2 2 7 π΄π₯ = [ π₯ 3 β π₯ 2 ] 3 7 π₯ =
π΄π¦ =
7 5 2 2 [ (4)3 β (4)2 16 3 7
π₯=
5 128 257 [ β ] 16 3 7
π₯=
40 π’πππ‘π 21
π¦=
πͺπππππππ :
1 2 1 2 1 2
4
π¦ 2 ππ₯
0 4
5
[(2π₯)2 β π₯ 2 ]ππ₯
0 4
(4π₯ 2 β π₯ 3 ) ππ₯
0
1 4 3 π₯4 4 π₯ β 2 3 4 0
10 π’πππ‘π 3
ππ ππ , ππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
92
EXERCISE 13.4
CENTROID OF A PLANE AREA
9. π₯ 2 + π¦ 2 = 25,
π₯+π¦ =5
25π β 50 4 25 π΄= (π β 2) 4 π΄=
5
ππ¦ = 5
=
25 β π₯ 2 β 5π₯ ππ₯
π₯ 0
π₯ 25 β
π₯ 2 ππ₯
5
β
0
5 x 25 β π₯ 2 β 5π₯ ππ₯
π΄= π΄=
0 5 0
25 β π₯ 2 ππ₯ β 5
π΄βΆ
5 0
A 25 β π₯ 2 ππ₯
5 ππ₯ 0
+
B
25 β 5 π₯ππ₯ 0
C
25 β π₯ 2
5 cos π =
π₯ 5 sin π = π₯ ; π = arcsin 5 π 5 cos π = ππ₯ @π₯ = 5 ; π = 2 π₯ =0; π =0 =
π 2
5 cos π β 5 cos π
0
= 25
π 2
0
1 = 25 2
1 + πππ 2π cos πππ β cos π = 2 2
π 2
0
1 ππ + 2
2
π 2
0
β/2 π +0 4 0 25π = 4 5 π΅ βΆ β 5 0 ππ₯ 5 = β5π₯ 0 = β25
= 25
π₯2
πΆβΆ 2 25 = 2
5 0
25π 25 β 25 + 4 2 25 π΄ = 25π β 2 β΄π΄=
cos 2πππ
π₯2
π’ = 25 β π₯ 2 ππ’ = β2π₯ππ₯ =
1 25βπ₯ 2 2 β2 3
β
2
3
=β
25βπ₯ 2 2
5π₯ππ₯ + 0
3
5
5
5π₯ 2 2
+
π₯3 3
5
5π₯ 2 2
π₯ 2 ππ₯
0
5 0
π₯3 3 0
β + 0 125 125 125 = β β + β β β0+0 3 2 3 3 125 250 375 + 500 =β + = β 2 3 6 125 ππ¦ = 6 2 1 5 ππ₯ = 25 β π₯ 2 β 5π₯ 2 ππ₯ 2 0 5 1 1 5 = 25 β π₯ 2 ππ₯ β 5 β π₯ 2 ππ₯ 2 0 2 0 1 π₯3 π₯3 = 25π₯ β β 1/2 25π₯ + β 5π₯ 2 3 3 =
3
1 125 1 125 125 β β 125 + β 125 2 3 2 3
β 0
125 6 β΄ ππππ£πππ πππ ππππ‘ππππ: ππ₯ =
ππ₯ π₯= = π΄ π₯=
125 6 25 πβ2 4
=
125 4 β 6 25 π β 2
10 3 πβ2
ππ₯ π¦= = π΄ π¦=3
10 πβ2
125 6 25 πβ2 4
πͺπππππππ ππ ππ
ππ ππ , π π βπ π π βπ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
2
93
EXERCISE 13.5
CENTROID OF A SOLID OF REVOLUTION
1. π¦ 2 = π₯ ; π¦ = 3 ; π₯ = 0 ; ππππ’π‘ π‘ππ π¦ β ππ₯ππ
π. π₯ 2 π¦ = 4, π₯ = 1, π₯ = 4, π¦ = 0 ππππ’π‘ π₯ β ππ₯ππ π
3βπ¦
ππ₯π§ =
ππππ£ π
π¦
π¦2 = π₯ ππ₯
4
π¦ 2π π₯π¦ππ₯ = 2 2
= 2π 1 4
=π 1
= 16π ππ₯π§ =
ππ ππ
;
1
=
π₯ π¦ 0 0 9 3
4
π¦ π₯ππ₯ = π 1
1
ππ₯ π₯ β2 = 16π π₯3 β2
4
1
2
4 π₯2
2
16 π₯ 4 π₯ππ₯ = 16π π₯ 4
ππ₯π§ π¦= π
4
π₯ππ₯
π₯ π₯ππ₯ π₯4
4
β8π π₯2
= 1
4 1
15π 2 π
π 2 ππ₯
π=π 9
π = 2π
π₯π¦ππ₯ = 2π 0
ππ ππ = 2π
ππ₯π§ = 381.70
4
π₯ 3 β π¦ ππ₯ 0
9
ππ₯π§ =
π
9
0
3+π¦ 2
4 π₯2
=π 1
π₯
π₯ ππ₯
π₯ 1
ππ₯π§ 381.70 π¦= = π 152.68 π¦ = 2.5 π, π. π, π
ππ₯ = π 1
4
= 16π
π=
21π 4
π¦=
ππ₯π§ π
= 0,
10 , 7
=
4
2
β4
16 ππ₯ = 16π π₯4
π₯ β3 ππ₯ = 16π β3
4
= β 1
4
1
ππ₯ π₯2
16π 3π₯ 3
4 1
15π 2 21π 4
0
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
94
EXERCISE 13.5
π. π₯ 2 = 4π¦ x 0 1 2 4
,
4
4π₯
2
0 4
=π 0
=
π₯2 β 4
y 0 1 2 4
2
ππ₯
π₯4 4π₯ β ππ₯ 16
96π 5 4
ππ₯π§ = 2π 0
=
π¦ 2 = 4π₯ ππππ’π‘ π₯ β ππ₯ππ x 0 1/4 1 4
y 0 ΒΌ 1 4
=π
CENTROID OF A SOLID OF REVOLUTION
4π₯ + 2
π₯2 4
π₯
4π₯ β
π₯2 ππ₯ 4
128π 3
ππ₯π§ π
=π¦=
π = π,
128 π 3 96π 5
ππ ,π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
95
EXERCISE 13.5
CENTROID OF A SOLID OF REVOLUTION
ππ. π¦ 2 = 4π₯, π¦ = π₯ ππππ’π‘ π₯ = 0
X 0 1/4 1 4
Y 0 1 2 4
X 1 2 3 4
Y 1 2 3 4
4
π = 2π
π( 4π β π)ππ₯ 0
π = 26.80829731 ππ’. π’πππ‘π ππ₯π§ = 2π
π¦π π₯ππ₯ 4
ππ₯π§ = 2π ( 0
4π₯ + π₯ )π₯ 2
4π₯ β π₯ ππ₯
ππ₯π§ = 64π/3 π¦=
ππ₯π§ π
= 2.5
y=(0, 2.5, 0)
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
96
EXERCISE 13.6
MOMENT OF INERTIA OF A PLANE AREA
1. 2π₯ + π¦ = 6 , π₯ = 0 , π¦ = 0 ; ππππ’π‘ π₯ β ππ₯ππ
5. π₯ = 2 π¦ , π₯ = 0, π¦ = 4
dy
x
y
4-y
(4,4)
dx π₯ 0 3
π¦ 6 0
πΌπ₯ =
6 2 π¦ π₯ππ¦ 0
= 6
6 2 6βπ¦ π¦ 0 2
π₯ 0 4
ππ¦
1 6π¦ 2 β π¦ 3 ππ¦ 2 0 1 π¦4 = 2π¦ 3 β 2 4
π¦ 0 4
=
1 = 2 6 2
6 3 β 4
4
πΌπ¦ =
0 4
4
π 2 ππ΄
=
π₯ 2 4 β π¦ ππ₯
0
= ππ
=
3
3. π¦ = π₯ , π₯ = 8 , π¦ = 0 ; π€ππ‘π πππ ππππ‘ π‘π π¦ = 0
4 2 π₯ 0
π°π =
π₯2
4β2
ππ₯
πππ ππ
dy x
πΌπ₯ = =
2 2 π¦ π₯ππ¦ 0 2 5
π¦ ππ¦
0
π°π =
= =
2 2 3 π¦ (π¦ )ππ¦ 0 6 6
π¦ 6
=
2 6
ππ π DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
97
EXERCISE 13.6
MOMENT OF INERTIA OF A PLANE AREA
7. π¦ 2 = 8π₯ , π¦ = 2π₯
9. π¦ = 4π₯ 2 , π¦ = 4π₯ ; π€ππ‘π πππ ππππ‘ π‘π π¦ β ππ₯ππ π¦ = 4π₯ 2 π¦ = 4π₯
X1 X2
(1,4) dy
dx
y (0,0)
π₯ 0 1
π₯ π¦ π₯ π¦ 0 0 0 0 1 2 2 1 2 2 4 2 4
π¦ 0 4
π
πΌπ¦ = 4
πΌπ₯ =
0
π¦ 2 (π₯π β π₯π ) ππ¦
4
πΌπ₯ = πΌπ₯ = π°π =
0
2
π¦ π¦ π¦ 2 ( β ) ππ¦ 2 8
4 π¦3 ( 0 2
β
π₯ π¦ 0 0 1 4
πΌπ¦ = Iy =
π
π₯ 2 (π¦π’ β π¦π ) ππ₯
1 2 π₯ (4π₯ 0
β 4π₯ 2 ) ππ₯
1 5
π¦4 ) ππ¦ 8
ππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
98
EXERCISE 13.6
MOMENT OF INERTIA OF A PLANE AREA
11. π¦ 2 = 8π₯ , π₯ = 0 , π¦ = 4 , with respect to π¦ = 4
13. π¦ = π₯ , π¦ = 2π₯ , π₯ + π¦ = 6, π€ππ‘π πππ ππππ‘ π‘π π₯ = 0 π₯+π¦ = 6
(6 β π₯ β 2π₯) 6 β 3π₯ =
4
πΌπ₯ = =
1 8
4βπ¦
2
0 4
π¦2 ππ¦ 8
16π¦ 2 β 8π¦ 3 + π¦ 4 ππ¦
0
=
1 16π¦3 8 3
=
ππ ππ
β 2π¦ 4 +
4 π¦5 5 0
π₯ π¦ 0 0 1 1 2 2
π¦ = 2π₯
π₯ π¦ 0 0 1 2 2 4
π₯ 0 1 2
π¦ 0 5 4
π
π°π = π°π = π°π =
π
ππ ππ β ππ π π
π π π π
πβπ β
π π
π π
ππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
99
EXERCISE 13.7
MOMENT OF INERTIA OF A SOLID OF REVOLUTION
1. π¦ = 2 π₯ , π¦ = 0 , π₯ = 4 ;about π₯ = 0
3. ππ₯ + ππ¦ = ππ , π₯ = 0 , π¦ = 0 ;about the y-
axis
4
πΌπ¦ = 2π 4
= 4π
π₯ 3 2 π₯ β 0 ππ₯
0
π₯
7
2 ππ₯
=
0 9
2π₯ = 4π 9
= 4π
= 4π =
2 4 9 1024 9
πππππ π
2
π
πΌπ¦ = 2π
4
= 0
9
2
β
2 0 9
9
2
4
2ππ π
0 π
π₯ 3 π β π₯ 4 ππ₯
0
2ππ π π
π
π
π₯ 3 ππ₯ β
0
2ππ π₯4 = π π 4
0
ππ β ππ₯ β 0 ππ₯ π
π₯3
π₯ 4 ππ₯
0 π 0
=
2ππ π5 π5 β π 4 5
=
2ππ π
=
ππ ππ ππ
π₯5 β 5
π 09
π5 20
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
100
EXERCISE 13.7
MOMENT OF INERTIA OF A SOLID OF REVOLUTION
5. 2π₯ + 3π¦ = 6 , π₯ = 0 , π¦ = 0 ; about the x-
9. π₯π¦ = 4 , π¦ = π₯ , π¦ = 1 ; aboutπ¦ = 0
axis
πΌπ₯ =
3
π 2
4
6 β 2π₯ 3
0
β 0 ππ₯ 2
3 16π₯ 4 β192π₯ 3 +864π₯ 2 β1728π₯+1296
=
π 2 0
=
πππ π
81
ππ₯
πΌπ₯ = 2π 2
= 2π
π¦3
1
4 β π¦ ππ¦ π¦
4π¦ 2 β π¦ 4 ππ¦
1
7. π¦ 2 = 3π₯ , π¦ = π₯ ; aboutπ₯ = 0
π¦3 = 2π 4 3 = 2π =
3
πΌπ¦ = 2π
28 3
β
2
π¦5 β 5 1
2 1
31 5
πππ ππ
π₯ 3 π₯ 3 β π₯ ππ₯
0
3
= 2π
π₯
7
3 β π₯ 4 ππ₯
2
0 3
= 2π
3
π₯
7
0
= 2π 54 β =
3 2 ππ₯
β
π₯ 4 ππ₯
0
243 5
πππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
101
EXERCISE 13.7
MOMENT OF INERTIA OF A SOLID OF REVOLUTION
11. π¦ = π₯ 2 , π¦ = 2π₯ ;about the y-axis
2
πΌπ¦ = 2π 2
= 2π
π₯ 3 2π₯ β π₯ 2 ππ₯
0 1
2π₯ 4 β π₯ 5 ππ₯
πΌπ¦ = 2π
0 2
= 2π 2
π₯ 4 ππ₯ β
0
=
64 5
β
2 0
π₯5 = 2π 2 5 = 2π
13. π¦ = π₯ 3 , π₯ = 1 , π¦ = 0 ; about π₯ = β1
2
π₯6 β 6 0
32 3
2
π₯ 5 ππ₯
1
= 2π
π₯+1
3
π₯ 3 β 0 ππ₯
0
π₯ 6 + 3π₯ 5 + 3π₯ 4 + π₯ 3 ππ₯
0
= 2π
π₯7 7
0
=
+
π₯6 2
+
3π₯ 5 5
+
1 π₯4 4 0
ππππ ππ
πππ ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
102
EXERCISE 13.7
MOMENT OF INERTIA OF A SOLID OF REVOLUTION
15. π¦ = 2π₯ , π₯ = 1 , π¦ = 0 ; about π₯ = 2
1
πΌπ¦ = 2π 1
= 4π
2βπ₯
3
2π₯ ππ₯
0
8π₯ β 12π₯ 2 + 6π₯ 3 β π₯ 4
0
= 4π 4π₯ 2 β 4π₯ 3 + =
3π₯ 4 2
β
1 π₯5 5 0
πππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
103
1.
BASIC INTEGRATION FORMULAS
6π₯ 2 β 4π₯ + 5 ππ₯ =
6π₯ 3 3
β
4π₯ 2 2
7.
+ 5π₯ + π
= Factor, (x-c), c = 2 P(c) = 0 β the (x-c ) is the factor P(c) = 0 2 1 0 0 -8 2 4 8 12 4 0
= πππ β πππ + ππ + π
3.
=
π₯( π₯ β 1)ππ₯ =
= π₯ ππ₯ β π₯ππ₯
5.
π π ππ π
π π π π
β
(π 2 +2π+4)(πβ2) (πβ2)
= (π₯ 2 + 2π₯ + 4)ππ₯
π₯ π₯ β π₯ ππ₯ 3 2
=
π₯ 3 β8 ππ₯ π₯β2
+π
9.
2π₯ 2 +4π₯β3 ππ₯ π₯2
=
π₯3 3
=
ππ + π
+
2π₯ 2 2
+ 4π₯ + π
ππ + ππ + π
π₯ 4 β 2π₯ 3 + π₯ 2 ππ₯ 2
=
2+
4 π
= 2ππ₯ + = 2π₯ + 4
β
3 π2
4 ππ₯ π₯ ππ₯ π₯
ππ₯ β
3 ππ₯ π₯2
=
ππ π
π
β
ππ π
+
ππ π
+πͺ
3π₯ β1 ππ₯ β1
β
= ππ + ππππ +
= π₯ 2 ππ₯ β 2π₯ 3 ππ₯ + π₯ππ₯
π π
+π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
1
EXERCISE 9.2
1.
INTEGRATION BY SUBSTITUTION
2 β 3π₯ ππ₯
(2π₯+3)ππ₯ π₯ 2 +3π₯+4
5.
ππ’
Let u = 2 - 3xππ₯ = β3
Let u = π₯ 2 + 3π₯ + 4 ππ₯ = 2π₯ + 3
βππ’ = ππ₯ 3
ππ’ = (2π₯ + 3)ππ₯
= π’
1 2
ππ’ π’
=
ππ’ (β 3 )
1 =β 3
ππ’
= πππ’ + π
1 2
π’ ππ’
= π₯π§ ππ + ππ + π + π
3
1 2π₯ 2 3
= β3
+π π
=β
π πβππ π π
2
π₯ 2 ππ₯ (π₯ 3 β1)4
7. +π
3
ππ’
Let u = π₯ 3 β 1 ππ₯ = 3π₯ 2 4
3. π₯ (2π₯ β 1) ππ₯ Let u = 2π₯ 3 β 1
ππ’ = π₯ 2 ππ₯ 3 ππ’ 3 π₯4
=
ππ’ = 6π₯ 2 ππ₯
=
1 3
ππ’ = π₯ 2 ππ₯ 6
=
1 π’ β3 3 β3 π’ β3 β9
π’β4 +π
=
π₯ 2 (2π₯ 3 β 1)4 ππ₯
=
=
ππ’ (π’ )( ) 6
= β π(ππβπ)π + π
=
1 (π’4 )ππ’ 6
=
1 π’5 +π 6 5
4
+π π
π’5
= 30 + π =
(πππ β π)π +π ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
2
EXERCISE 9.2
INTEGRATION BY SUBSTITUTION 13. cos4 π₯ sin π₯ππ₯
ππ₯ π₯ππ 2 π₯
9.
ππ’
1
Let u = πππ₯ ππ₯ = π₯
= π’4 βππ’
=
1 ππ₯ ( ) ππ 2 π₯ π₯
= - π’4 ππ’
=
1 ππ’ π’2
=β
π’5 5
=
π’β2 ππ’
=β
ππ¨π¬ π π π
=
π’ β1 β1
= βsin π₯
ππ’ = βsin π₯ ππ₯
ππ₯ π₯
ππ’ =
ππ’ ππ₯
Let u = cos π₯
+π +π
+π
π
= β πππ + π 15. ππ₯ π π₯ β1
11.
Let u = 3π₯
Let u = π π₯ ππ’ = π π₯ ππ₯ 1 π’β1
= ( =
1 π’
β )ππ’
1 ππ’ π’β1
1 π’
β
ππ’
ππ’ =3 ππ₯ ππ’ = ππ₯ 3
=
ππ£ = ππ’ π’ 1 π’
ππ£ β
ππ’
=
Let v = π’ β 1
=
1 + 2 sin 3π₯ πππ 3π₯ππ₯
1 + 2 sin π’ πππ π’ ( 3 ) 1 3
1 + 2 sin π’ πππ π’ππ’
Let v =1 + 2 sin π’ 1 ππ£ π’
= πππ’ β πππ’ + π π’ =π’β1
ππ£ ππ’
=
; π’ = ππ₯
= lnβ‘|π’ β 1| β lnβ‘|π π₯ | + π = lnβ‘ (1 β π π₯ )
ππ£ 2
= 2πππ π’ ;
= πππ π’ππ’
1 + 2 sin 3π₯ πππ 3π₯ππ₯ 1
=
1 [ 3
ππ£
=
1 2π£ 2 6 3
=
(π+πππππ)π π
π£ 2 ( 2 )] 3
= π₯π§ π β ππ β π + π
+π π
+π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
3
EXERCISE 9.2
INTEGRATION BY SUBSTITUTION
π ππ 2 π₯ππ₯ ` π+π π‘πππ₯
17.
3π₯ 2 +14π₯+14 ππ₯ π₯+4
21.
π (π₯) π (π₯)
Let u = π + π π‘πππ₯
=
ππ’ ππ₯
* using synthetic division
=π
;
ππ’ π
= π ππ 2 π₯ππ₯
ππ’ π
=
π π₯ π π₯
π(π₯)
-4 3 14 13
π’
1 =π
=
sin π₯ πππ π₯
= π π₯ π π₯ +
-12 -8
ππ’ π’ π π₯π§ π
3 2 5 - R(x) π + πππππ + π
π π₯ = 3π₯ + 2 π₯ + 4 = πππππππππ‘ππ π(π₯) 5 ππ₯ π₯+4
= (3π₯ + 2) ππ₯ +
For the second integral : 2
19.
π‘ππ3π₯ π ππ 3π₯ππ₯ πππ‘ π’ = π₯ + 4
Let u = π‘ππ3π₯ ππ’ ππ₯
= 3π ππ 2 3π₯ ;
= π’
1 2
ππ’ 3
=
1 [ 3 3
3π₯ 2 2
=
1 3
=
π(πππ§ ππ)π π
=
]+π
ππ¦ = 1 ; ππ’ = ππ₯ ππ₯
= (3π₯ + 2)ππ₯ + 5 =[
ππ’ (3)
3 2π’ 2
= π ππ 2 3π₯ππ₯
;
πππ π
ππ’ π’
+ 2π₯ + 5πππ’ + π]
+ ππ + ππ₯π§β‘ (π + π) + π
3
2tan 3π₯ 2 3
+π
π
+π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
4
EXERCISE 9.2
INTEGRATION BY SUBSTITUTION
π₯ 5 β2π₯ 3 β2π₯ π₯ 2 +1
23.
ππ₯
π₯ 3 β 3π₯ π₯ 2 + 1 π₯ 5 β 2π₯ 3 β 2π₯ π₯5 + π₯3 β3π₯ 3 β 2π₯ β3π₯ 3 β 3π₯ π₯ π(π₯) dx = π(π₯)
π π₯ ππ₯ +
= π₯ 3 β 3π₯ ππ₯ + π₯4
=4 β
3π₯ 2 + 2
π (π₯) ππ₯ π(π₯) π₯ ππ₯ π₯ 2 +1
π₯ ππ₯ π₯ 2 +1
For the 2nd term Let u = x2+1 ππ’ = 2π₯ ππ₯ ππ’ = π₯ππ₯ 2 π₯4 =4
=
3π₯ 2 β 2
ππ π
β
+
πππ π
ππ’ 2
π’ π
+ π π₯π§ ππ + π + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
5
EXERCISE 9.3 1.
INTEGRATION OF TRIGONOMETRIC FUNCTIONS
π ππ5π₯π‘ππ5π₯ππ₯ πΏππ‘ π’ = 5π₯ ππ’ ππ₯
ππ’ 5
=5
=
cos 3 π₯ππ₯ 1+π πππ₯ . 1βπ πππ₯ 1+π πππ₯
7. =
(cos 3 π₯) 1+π πππ₯ ππ₯ (1βπ πππ₯ )(1+π πππ₯ )
=
(co s 3 π₯+cos 3 π₯π πππ₯ )ππ₯ 1βsin 2 π₯
=
cos 3 π₯ 1+π πππ₯ ππ₯ cos 3 π₯
= ππ₯
π πππ’π‘πππ’
ππ’ 5
1
= 5 π πππ’π‘πππ’ππ’ 1 =5
π πππ’ + π
π =π
πππππ + π
= πππ π₯ 1 + π πππ₯ ππ₯ = π πππ₯ + πππ π₯π πππ₯ππ₯ πΏππ‘ π’ = π πππ₯ ππ’ ππ₯
π πππ₯ +πππ π₯ π ππ 2 π₯
3.
ππ₯
=
π πππ₯ π ππ 2 π₯
=
1 π πππ₯
=
ππ πππ₯ +
= πππ π₯ ; ππ’ = πππ π₯ππ₯
= π πππ₯ + π’ππ’ πππ π₯ π ππ 2 π₯
ππ₯ +
ππ₯
ππ₯ + πππ‘π₯ππ ππ₯ππ₯
= π πππ₯ +
π’2 2
= ππππ +
π¬π’π§π π π
+π +π
πππ‘π₯ππ ππ₯ππ₯
= β ππ ππππ + ππππ β ππππ + π
9.
1 + π‘πππ₯ 2 ππ₯ = (1 + 2π‘πππ₯ + tan2 π₯)ππ₯
5.
ππ₯ 1 2
1
; Let u= 2 π₯
1 2
sin π₯ cot π₯ ππ’ ππ₯
=
= =2 =2 =2
1 2
2ππ’ = ππ₯
= [2π‘πππ₯ + (1 + tan2 π₯)]ππ₯ = 2 π‘πππ₯ππ₯ +
sec 2 π₯ππ₯
= 2 βππ|πππ π₯| + π‘πππ₯ + π
2ππ’ π πππ’πππ‘π’
= βπ ππ |ππππ| + ππππ + π
ππ’ π πππ’ πππ‘π’ ππ’ π πππ’ ( 1 πππ π’
πππ π’ π πππ’
)
(ππ’)
= 2 π πππ’ππ’ = πππ ππππ + ππππ + π DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
6
EXERCISE 9.3
INTEGRATION OF TRIGONOMETRIC FUNCTIONS
πππ 6π₯ππ₯ cos 2 3π₯
11.
Let u = 3x ; 2u = 6x ππ’ ππ₯
=3 ;
ππ’ 3
πππ 2π’
ππ’ 3
= =
= ππ₯
cos 2 π’ 2 3
4 sin 2 π₯ππ π 2 π₯ π ππ 2π₯πππ 2π₯
15.
1 πππ π’
πππ π’ πππ π’
ππ’
=
(4π πππ₯πππ π₯ )(π πππ₯πππ π₯ ) ππ₯ 2π πππ₯πππ π₯ πππ 2π₯
=
2π πππ₯πππ π₯ πππ 2π₯
=
π ππ 2π₯ πππ 2π₯
ππ’ ππ₯
2
ππ₯ πππ π₯
=
1 πππ π₯
= ππ₯
ππ’ 2
π‘πππ’ππ’
π = β π₯π§ πππππ + π π
π ππ 2π₯ππ₯ 2π πππ₯ππ π 2 π₯
=
.
1 2
=
2π πππ₯πππ π₯ππ₯ (2π πππ₯πππ π₯ )πππ π₯
ππ₯
ππ’ 2
=2
π πππ’ πππ π’
π
= π π₯π§ πππππ + πππππ + π
=
ππ₯
Let u = 2x
= 3 π πππ’ππ’
13.
ππ₯
ππ₯ π ππ 3π₯π‘ππ 3π₯
17.
Let u = 3x ππ₯
= π πππ₯ππ₯ = ππ ππππ + ππππ + π
ππ’ ππ₯
ππ’ 3
=3
= ππ₯
ππ’ 3
=
π πππ’π‘πππ’ 1
=3 ππ ππ’ + π π
= β π πππππ + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
7
EXERCISE 9.4
1.
ππ₯ π 2π₯
=
π β2π₯ dx
INTEGRATION OF EXPONENTIAL FUNCTIONS
ππ’ ππ’ πππ‘ π’ = 2π₯ ; = β2 ; β = ππ₯ ππ₯ 2 = π π’ (β =β
1 2
2ππ’ ) 3
2
π π’ ππ’
= 3 π π’ ππ’
1
1
π
=β π (πβππ ) + π
π π ππ 4π₯ πππ 4π₯ππ₯
2 3
=
π πππ π
ππ’ 4
1 4
=
π
π πππ’
1 4
)
1 =4
=
ππ£ + π
ππππππ π
53β2π₯ ππ₯
= 5π’ (β
πππ π’ππ’
π π£ ππ£
+π
πππ‘ π’ = 3 β 2π₯ ;
ππ£ πππ‘ π£ = π πππ’ ; = cos π’ ; ππ£ = πππ π’ππ’ ππ’ =
π2 +π
7.
ππ’ ππ’ πππ‘ π’ = 4π₯ ; =4; = ππ₯ ππ₯ 4 π π πππ’ πππ π’(
3π₯
=
=β 2π π’ + π
=
3π₯ 2ππ’ ; = ππ₯ 2 3
πππ‘ π’ = = ππ’ (
ππ’ ) 2
=β 2 π π’ + π
3.
3π₯
π 3π₯ ππ₯ = π 2 ππ₯
5.
ππ’ ππ’ = β2 ; β = ππ₯ ππ₯ 2
ππ’ ) 2
1
= β 2 5π’ ππ’ 1 53β2π₯ ππ 5
= β2 =β
ππβππ ππππ
+π
+π
+π 3π₯ 2π₯ ππ₯
9.
π π₯ π π₯ = (ππ)π₯ = 6π₯ ππ₯ ππ
= πππ + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
8
EXERCISE 9.5
1.
INTEGRATION OF HYPERBOLIC FUNCTIONS
π πππ 3π₯ β 1 ππ₯
π ππ π 2 πππ₯ ππ₯ π₯
5.
Let u = 3π₯ β 1 ππ’ ππ₯
πππ‘ π’ = πππ₯ ; ππ’ 3
= 3 ; ππ₯ =
=
ππ’
= 1 3
=
1 πππ π π’ππ’ 3
=
π ππππ π
π πππ2 π’ππ’
= π‘ππππ’ + π
π πππ π’ ( 3 )
=
ππ’ 1 ππ₯ = ; ππ’ = ππ₯ π₯ π₯
= ππππβ‘ (πππ) + π
π πππ π’ππ’ +c
ππ β π + π
7.
1
1
ππ ππ 2 π₯ πππ‘π 2 π₯ππ₯ 1
Let u = 2 π₯ ;
ππ’ ππ₯
=
1 2
; 2ππ’ = ππ₯
= 2 ππ πππ’ πππ‘ππ’ππ’ 3.
ππ ππ2 1 β π₯ 2 π₯ππ₯
= 2(βππ πππ’ + π)
Let u=1 β π₯ 2
= βπππππ π + π
ππ’ ππ₯
β
π π
= -2π₯
ππ’ = π₯ππ₯ 2 = ππ ππ2 π’(β =β
1 2
ππ’ ) 2
ππ ππ2 π’ππ’
1
=β 2 (βπππ‘ππ’ + π) =
π ππππ π
π β ππ + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
9
EXERCISE 9.6
APPLICATION OF INDEFINITE INTEGRATION
1. Given slope 3π₯ 2 + 4 ππ¦ = 3π₯ 2 + 4 ππ₯ ππ¦ = 3π₯ 2 + 4 ππ₯
π¦=
π = πππ + ππ + π
π₯+1 π¦β1
ππ¦ π₯ + 1 = ππ₯ π¦ β 1
ππ¦ = π¦2
π¦ 2 β 2π¦ =
ππ₯ π₯
β ln π₯ β
1 =π 4
β ln 1 β
1 =π 4
π=β
1 4
β ln π₯ β
π¦ β 1 ππ¦ = π₯2 2
through 1,4
1 β = πππ₯ + π 4
+ 4π₯ + π
3. Given slope
π¦2 , π₯
ππ¦ π¦ 2 = ππ₯ π₯
3π₯ 2 + 4 ππ₯
ππ¦ = 3π₯ 3 3
7. Given slope
π₯ + 1 ππ₯ +π₯+π 2
1 1 + = 0 4π¦ π¦ 4
β4π¦ ln π₯ β 4 + π¦ = 0 ππ π₯π§ π β π + π = π
π¦ 2 β 2π¦ = π₯ 2 + 2π₯ + 2π ππ β ππ + ππ + ππ + ππ = π
1 5. Given slope π₯π¦
ππ¦ 1 = ππ₯ π₯π¦
π
=
ππ¦ = ππ₯
π¦
1
π¦ β2 ππ¦ =
ππ₯
1
π¦2 ππ₯ π₯
π¦ππ¦ = π¦2 2
9. Given slope π¦, through 1,1
ln π₯ 2 2 π
+π 2
π = πππ + ππ
1 2
2π¦
=π₯+π 1
2
=π₯+π
When π₯ = 1 , π¦ = 1 2 1 =1+π ; π =1 2π¦
1
2
=π₯+π
ππ = π + π
2
π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
10
EXERCISE 9.6
APPLICATION OF INDEFINITE INTEGRATION
11. Given slope π₯ β2 , through 1,2
v = -32t + vo
ππ¦ 1 = 2 ππ₯ π₯
when t = 1 sec, s=h=48ft ππ₯ π₯2
ππ¦ =
1 π¦ =β +π π₯
h=-16t2+ vot + c1 48 = -16(1)2 + vo(1) + c2 64 - vo = c2 When t = 0, s = 0, c2 = 0
1 2=β +π 1
s = -16t2 + vot
2 = β1 + π
when t = 1 sec, s = 48
π=3
s = -16t2 + c1t 1 π₯
π¦ = β +3 x π₯π¦ = β1 + 3π₯ ππ β ππ + π = π
c1=64 s=-16t2 + 64t v = -32t + 64 @ max, v = 0
13. a=-32 ft/sec2 a=-2
0 = -32t + 64 32t=64 t = 2 sec
ππ¦ = β32 ππ‘ ππ£ =
48 = -16(1)2 + c1(1)
s = -16t2 + 64t β32ππ‘
s = -16(2)2 + 64(2) s = 64ft
v=-32t+c ππ = β32π‘ + π1 ππ‘ ππ =
(β32π‘ + π1 )ππ‘
s=16t2 + c1t + c2 when t = 0, v = vo v=-32t + c1 vo= -32(0) + c1 vo =c1 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
11
EXERCISE 9.6
APPLICATION OF INDEFINITE INTEGRATION
15. a = 32ft/sec2 a = 32 ππ£ = 32 ππ‘ ππ£ =
32ππ‘
v = 32t + c1 ππ = 32π‘ + π1 ππ‘ ππ =
32π‘ + π1 ππ‘
S = 16t2 + c1 + c2 when t = 0, v = 0 c1 = 0 v = 32t when t = 0 , s = 0 c2 = 0 s = 16t2 400 16
π‘=
π‘=
20 4
t = 5 sec v = vt *since it is a free falling body, its velocity is ( - ) vt = -32t vt = -32(5) vt = -160 ft/sec
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
12
EXERCISE 10.1
PRODUCT OF SINES AND COSINES
1. Κ sin 5π₯ sin π₯ ππ₯ =
5. Κ cos 3π₯ β 2π cos π₯ + π ππ₯ 1 = Κ[cos π’ + π£ + cosβ‘ (π’ β π£)]ππ₯ 2
2 sin π’ sin π£ ππ₯
πππ‘ π’ = 3π₯ β 2π
= [cos π’ β π₯ β cos(π’ + π£)]ππ₯ π’ = 5π₯
π£ =π₯+π
π£=π₯
1 = Κ[cos 5π₯ β π₯ β cosβ‘ (5π₯ + π₯)]ππ₯ 2 1 = Κ[cos 4π₯ β cos 6π₯]ππ₯ 2
= =
πππ ππ πππππ β +πͺ π ππ
β
1 sin 6π₯ 6
π’ + π£ = 3π₯ β 2π + π₯ + π = 4π₯ β π
ο·
1 = [Κ cos 4π₯ππ₯ β Κ cos 6π₯ππ₯ 2 1 1 [ sin 4π₯ 2 4
ο·
π’ β π£ = 3π₯ β 2π β π₯ + π = 2π₯ β 3π 1 = Κ[cos 4π₯ β π + cos(2π₯ β 3π)]ππ₯ 2
]+πΆ
πππ cos 4π₯ β π = cos 4π₯πππ π + π ππ4π₯π πππ = βπππ 4π₯ πππ cos 2π₯ β 3π = cos 2π₯πππ 3π + sin 2π₯π ππ 3π = β cos 2π₯
3. Κ sin 9π₯ β 3 cos π₯ + 5 ππ₯ 1 = Κ [sin 9x β 3 + x + 5 + sin 9π₯ β 3 β π₯ β 5 ππ₯ 2
1 = Κ[sin 5π₯ + 2 + sin(3π₯ β 8)]ππ₯ 2 πππ‘ π§ = 5π₯ + 2 ππ§ =5 ππ₯ ππ§ = ππ₯ 5 1
= 2 [β cos π§ =β
1 5
1 2
= Κ(cos 4π₯ β cos 2π₯)ππ₯ 1
1
1
= 2 [β 4 sin 4π₯ β 2 sin 2π₯] + πΆ π π = β π¬π’π§ ππ β π¬π’π§ ππ + πͺ π π
; πππ‘ π€ = 3π₯ β 8 ππ€ ; =3 ππ₯ ππ€ ; = ππ₯ 3
1
β 3 πππ π€] + πΆ
π π πππ ππ + π β πππ ππ β π + πͺ ππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
13
EXERCISE 10.1
7.
PRODUCT OF SINES AND COSINES
π
4 π ππ 8π₯ πππ 3π₯ππ₯
= 2Κ[sin 8π₯ + 3π₯ + π πππ₯ 8π₯ β 3π₯ ππ₯
5 = Κ[cos π’ β π£ β cosβ‘ (π’ + π£)]ππ₯ 2
= 2Κ[π ππ11π₯ + sin 5π₯]ππ₯
πππ‘ π’ = 4π₯ +
πππ‘ π’ = 11π₯ ; πππ‘ π£ = 5π₯
1
ππ’ = 11 ; ππ₯
ππ£ =5 ππ₯
ππ’ = ππ₯ ; 11
ππ£ = ππ₯ 5
1
= 2[β 11 cos 11π₯ β 5 cos 5π₯ ] + πΆ =β
π
9. 5 π ππ 4π₯ + 3 π ππ 2π₯ β 6 ππ₯
π 3
;
π
π
π
π
ο·
π’ β π£ = 4π₯ + 3 β 2π₯ β 6 = 2π₯ + π/2 π π£ = 2π₯ β 6
ο·
π’ + π£ = 4π₯ + 3 + 2π₯ β 6 = 6π₯ + π/6
π π ππ¨π¬ πππ β πππππ + πͺ ππ π
5 π π = Κ[πππ 2π₯ + β πππ 6π₯ + ]ππ₯ 2 2 6 ο·
π
πππ cos 2π₯ + 2 = cos 2π₯πππ = βπ ππ2π₯
ο·
πππ cos 6π₯ +
π π β sin 2π₯ sin 2 2
π 6
π π β π ππ 6π₯ π ππ 6 6 3 1 = πππ 6π₯ β π ππ 6π₯ 2 2 = πππ 6π₯πππ
5 3 1 = Κ[β π ππ 2π₯ β πππ 6π₯ + π ππ 6π₯ ]ππ₯ 2 2 2 5 1
3
2 2
12
= [ πππ 2π₯ β
=
π πππ ππ π
β
π ππ 6π₯ β
π π πππ ππ ππ
1 12
π ππ₯ 6π₯ + πΆ π
β ππ πππ ππ + πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
14
EXERCISE 10.2
POWER OF SINES AND COSINES
1. π ππ3 π₯πππ 4 π₯ππ₯; ππ¦ πΆππ π πΌ = =
=
1 π’5 π’7 β +πΆ 3 5 7
=
1 5 π’ 15
=
π π ππππ ππ β ππππ ππ + πͺ ππ ππ
π ππ4 π₯πππ 4 π₯π πππ₯ππ₯ (1 β πππ 2 π₯)2πππ 4 π₯π πππ₯ππ₯
=
(1 β 2πππ 2 π₯ + πππ 4 π₯)πππ 4 π₯π πππ₯ππ₯
=
(πππ 4 π₯ β 2πππ 6 π₯ + πππ 8 π₯)π πππ₯ππ₯
1
β 21 π’7 + πΆ
5.
π ππ4 π₯πππ 2 π₯ππ₯
Let u = cosx
=
(π ππ2 π₯)2πππ 2 π₯ππ₯
ππ’ = βπ πππ₯ ππ₯
=
1 β πππ 2π₯ 2 1 + πππ 2π₯ ( ) ππ₯ 2 2
-ππ’ = π πππ₯ππ₯ =
= - (π’4 β 2π’6 + π’8 )ππ’ = =
2π’ 6 7
π’5 5
π’9 9
+πΆ
=
π π π ππππ π β ππππ π β ππππ π + πͺ π π π
=
β
β
=
3. π ππ4 3π₯πππ 3 3π₯ππ₯ ; ππ¦ πΆππ π πΌπΌ =
4
2
π ππ 3π₯πππ 3π₯πππ 3π₯ππ₯
1 1 1 1 πππ 2π₯ β ( πππ 2π₯ + πππ 2 2π₯) + ππ₯ 4 2 4 2 2
1 8
1 β 2πππ 2π₯ + πππ 2 2π₯ 1 + πππ 2π₯ ππ₯
1 (1 β 2πππ 2π₯ + πππ 2 π₯ + πππ 2π₯ β 2πππ 2 2π₯ + πππ 3 2π₯)ππ₯ 8
=
1 8
π ππ4 3π₯ 1 β π ππ2 3π₯ πππ 3π₯ππ₯ (π ππ4 3π₯ β π ππ6 3π₯)πππ 3π₯ππ₯
4
(πππ 2 2π₯ 1 + πππ 2π₯ ππ₯ 2
1 (1 β πππ 2π₯ β πππ 2 2π₯ + πππ 3 2π₯)ππ₯ 8
= =
1 4
=
=
=
1 β 2πππ 2π₯ +
1 8
ππ₯ β
πππ 2 2π₯ππ₯ +
πππ 2π₯ππ₯ β
πππ 3 2π₯ππ₯
1
1
1
1
1
2
2
8
2
6
[π₯ β π ππ2π₯ β ( π₯ + π ππ4π₯ + π ππ2π₯ β π ππ3 2π₯]
π πππππ ππππ ππ β β +πͺ ππ ππ ππ
Let u = sin3x ππ’ ππ₯
=
= 3πππ 3π₯ ; ( π’4 β π’6 )
ππ’ 3
= πππ 3π₯ππ₯
ππ’ 3 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
15
EXERCISE 10.2
POWER OF SINES AND COSINES
7. ( π πππ₯ + πππ π₯)2dx =
(π πππ₯ + 2 π πππ₯ πππ π₯ + πππ 2 π₯)ππ₯
=
π πππ₯ππ₯ + 2
=
π πππ₯ππ₯ + 2 π ππ2 πππ π₯ππ₯ + (
1
πππ 2 π₯ππ₯
π ππ2 πππ π₯ππ₯ + 1
1+πππ 2π₯ )ππ₯ 2
Let u = sinx
=
π ππ2 2π₯ π ππ2π₯ ππ₯
=
1 β πππ 2 2π₯ π ππ2π₯ ππ₯
πππ‘ π’ = πππ 2π₯
= -πππ π₯ + 2
π’ ππ’ +
2 3 π’2 3 π
π
1 ππ₯ + 2
1 2
π₯
+2+ π
= -ππππ + π ππππ + π +
π ππ 2π₯ 4
πππππ π
πππ 2π₯ ππ₯ 2
1
βπ π πππππ + ππππ ππ + π π π
= πβ
π₯
ππππ
+π
π ππ7 π₯ πππ 2 π₯ ππ₯
=
π ππ7 π₯ πππ 2 π₯ πππ π₯ ππ₯
=
π ππ7 π₯ 1 β π ππ2 π₯ πππ π₯ππ₯
=
π ππ7 π₯ β π ππ9 π₯ πππ π₯ ππ₯ u=sinx du=cosxdx
=
π’7 β π’9 ππ’
πππ 2 2π₯ ππ₯
1
= 2 β 2 π ππ6π₯ β 5 πππ 5π₯ β πππ π₯ + 2 + 8 π ππ4π₯ + π πππππ πππππ β π β ππ
π’β
+πͺ
π ππ3π₯πππ 2π₯ ππ₯ +
1
ππ’ 2
=
ππ.
= π ππ2 3π₯ + 2π ππ3π₯πππ 2π₯ + πππ 2 2π₯ ππ₯
π₯
β
β1 2
+πΆ
9. (π ππ3π₯ + πππ 2π₯)2 ππ₯
= π ππ2 3π₯ ππ₯ + 2
π’3 3
; π·π’ = β2π ππ2π₯ ππ₯
=
ππ’ = πππ π₯ππ₯ π πππ₯ππ₯ + 2
1 β π’2
=
ππ’ = πππ π₯ ππ₯
=
π ππ3 2π₯ ππ₯
ππ.
πππππ + π +
π
= =
π’8 π’10 β +π 8 10 π π ππππ π β ππ πππππ π + π
π
11. πππ 2 4π₯ ππ₯ 1 + πππ 8π₯ ππ₯ 2
= 1
=2 =
1 + πππ 8π₯ ππ₯
π πππππ + +π π ππ DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
16
EXERCISE 10.3
POWER OF TANGENTS AND SECANTS
1. π‘ππ2 2π₯π ππ 4 2π₯ππ₯
5. ____ 2 π₯ππ₯ β πππ . π¦ = 3 π‘ππ
= π‘ππ2 2π₯π ππ 2 2π₯π ππ 2 2π₯ππ₯
πΉπππ π‘ππ πππ π πππ π‘πππ:
= π‘ππ2 2π₯(1 + π‘ππ2 2π₯)π ππ 2 2π₯ππ₯
ππ¦ π₯ π₯ π₯ = π‘ππ2 π ππ 2 β π ππ 2 + 1 ππ₯ 2 2 2
1
= (π‘ππ2 2π₯ + π‘ππ4 2π₯)π ππ 2 2π₯ππ₯
ππ’
=
1 π’3 2 3
=
+
ππππ ππ π
+
2
π₯ π₯ = π‘ππ2 (π‘ππ2 ) 2 2
(π’2 + π’4 )ππ’ π’5 5
π₯
β 2π‘ππ + π₯ + π
π₯ π₯ = π‘ππ2 (π ππ 2 β 1) 2 2
= (π’2 + π’4 )( 2 ) =
2
π₯ π₯ π₯ = π‘ππ2 π ππ 2 β π‘ππ2 2 2 2
ππ’ = π ππ 2 2π₯ππ₯ 2
1 2
3π₯
π₯ π₯ π₯ = π‘ππ2 π ππ 2 β (π ππ 2 β 1) 2 2 2
ππ’ = 2π ππ 2 2π₯ ππ₯
πππ‘ π’ = π‘ππ2π₯ ;
2
π₯ 2
ππ¦ = π‘ππ4 dx
+π
= πππππππππ, πππ πππππππ ππππ ππ "ππππ "
ππππ ππ
+π
ππ
7. (π πππ₯ + π‘ππ π₯)2 ππ₯ 3.
π‘πππ₯ π ππ 6 π₯ππ₯ ; πΆπ΄ππΈ πΌ
= (π ππ 2 π₯ + 2π πππ₯ π‘ππ π₯ + π‘ππ2 π₯) ππ₯
1
= π‘πππ₯ + 2π πππ₯ + π‘ππ2 π₯ ππ₯
= π‘ππ2π₯ π ππ 4 π₯π ππ 2 π₯ππ₯ 1
= π‘πππ₯ + 2π πππ₯ + (π ππ 2 π₯ β 1) ππ₯
1
= π‘πππ₯ + 2π πππ₯ + π‘πππ₯ β π₯ + π
= π‘ππ2 π₯(1 + π‘ππ2 π₯)2 π ππ 2 π₯ππ₯ = π‘ππ2 π₯(1 + 2π‘ππ2 π₯ + π‘ππ4 π₯)π ππ 2 π₯ππ₯ 1 2
5 2
= πππππ + πππππ β π + π
9 2
2
= (π‘ππ π₯ + 2π‘ππ π₯ + π‘ππ π₯)π ππ π₯ππ₯ πππ‘ π’ = π‘πππ₯ ; 1
ππ’ = π ππ 2 π₯ ; ππ’ = π ππ 2 π₯ ππ₯ ππ₯
5
9
= (π’2 π₯ + 2π’2 π₯ + π’2 π₯)ππ’ 3
7
4π’ 2 + 7
11
=
2π’ 2 3
=
πππππ π πππππ π + π π
π
+
2π’2 11 π
+π ππ
π πππ π π + ππ
+π DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
17
EXERCISE 10.3
POWER OF TANGENTS AND SECANTS
π ππ 3π₯
9. (π‘ππ 3π₯ )4 ππ₯ π ππ 4 3π₯ π‘ππ 4 3π₯
=
1
ππ₯
= π‘ππ3 π₯π ππ β2 π₯ππ₯
= π ππ 4 3π₯π‘ππβ4 3π₯ππ₯
3
= π‘ππ2 π₯π‘πππ₯π ππ β2 π₯π πππ₯ππ₯
= π ππ 2 3π₯π ππ 2 3π₯π‘ππβ4 3π₯ππ₯
3
= (π ππ 2 π₯ β 1)π‘πππ₯π ππ β2 π₯π πππ₯ππ₯
= π ππ 2 3π₯(1 + π‘ππ2 3π₯)π‘ππβ4 3π₯ππ₯ = (π‘ππβ4 3π₯ + π‘ππβ2 3π₯)π ππ 2 3π₯ππ₯ ππ’ πππ‘ π’ = π‘ππ 3π₯ ; = 3π ππ 2 3π₯ ππ₯ ππ’ = π ππ 2 3π₯ππ₯ 3
1
=
β
π’ β2 3
1 π‘ππ β3 3π₯ 3 β3
= β
β
ππ’ = π πππ₯π‘πππ₯ππ₯ 1
3
3
=
+π
2π’ 2 3
1
β 2π’β2 + π π
π‘ππ β1 3π₯ 3
ππππ ππ πππππ β π + π
ππ’ = π πππ₯π‘πππ₯ ππ₯
πππ‘ π’ = π ππ π₯ ;
1
1 π’ β3 β3
3
= (π ππ 2 π₯ β π ππ β2 π₯)π‘πππ₯π πππ₯ππ₯
= (π’2 β π’β2 )ππ’
=3 (π’β4 + π’β2 ) ππ’ =3
π‘ππ 3 π₯ ππ₯ π πππ₯
11.
+π
=
πππππ π β π
π π
+π
ππππ π
π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
18
EXERCISE 10.4
POWER OF COTANGENTS AND COSECANTS
1. πππ‘ 4 π₯ππ π 4 π₯ππ₯
= πππ‘ 4 π₯(1 + πππ‘ 2 π₯)ππ π 2 π₯ππ₯
1
= πππ‘ 2 3π₯ ππ π 2 3π₯ ππ π 2 3π₯ ππ₯
= (πππ‘ 4 π₯ + πππ‘ 6 π₯)ππ π 2 π₯ππ₯ πππ‘ π’ = πππ‘ π₯ ;
1
= πππ‘ 2 3π₯ 1 + πππ‘ 2 3π₯ ππ π 2 3π₯ ππ₯
ππ’ = βππ π 2 π₯ ; ππ’ = βππ π 2 π₯ππ₯ ππ₯
= β (π’4 + π’6 )ππ’ =-
π’5 5
= -
3.
+
π’7 7
5
ππ’ π π₯ = β3 ππ π 2 3π₯ ππ₯ ππ₯
c β
πππ‘ 5 4π₯ππ₯
= πππ‘ 3 4π₯πππ‘ 2 4π₯ππ₯ 3
1
= πππ‘ 2 3π₯ + πππ‘ 2 3π₯ ππ π 2 3π₯ ππ₯ πππ‘ π’ = πππ‘ 3π₯
+c
ππππ π ππππ π + + π π
πππ 3π₯ ππ π 4 3π₯ ππ₯
5.
2
= πππ‘ 4π₯(ππ π 4π₯ β 1)ππ₯
ππ’ = ππ π 2 3π₯ ππ₯ 3
=β
1 3
=β
1 3
1
5
π’2 + π’2 ππ’ 3
π’2 3 2
7
+
π’2 7 2
+π
π π π π = β ππππ ππ β ππππ ππ + π π ππ
= (πππ‘ 3 4π₯ππ π 2 4π₯ β πππ‘ 3 4π₯)ππ₯ = [πππ‘ 3 4π₯ππ π 2 4π₯ β (ππ π 2 4π₯ β 1)πππ‘4π₯]ππ₯ =
πππ‘ 3 4π₯ππ π2 4π₯ππ₯ β πππ‘4π₯ππ π2 4π₯ππ₯ β πππ‘4π₯ππ₯
πππ‘ π’ = πππ‘ 4π₯ ; 1
ππ’ = ππ π 2 4π₯ππ₯ β4
1
1
=β 4 π’3 ππ’ β 4 π’ππ’ + 4 ππβ‘(πππ 4π₯) 1 π’4 ππ₯ 4
=β 4
= β
β
π’2 2
1 4
+ ππ πππ 4π₯ + π
ππππ ππ ππππ ππ + ππ π
π
+ π ππ πππππ + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
19
EXERCISE 10.4
POWER OF COTANGENTS AND COSECANTS
7.
πππ 5 2π₯ππ₯ π ππ 8 2π₯
πππ 5 2π₯ππ₯ π ππ 5 2π₯
=
πππ‘ 5 2π₯ ππ π 3 2π₯ ππ₯ \
=
πππ‘ 4 2π₯ ππ π 2 2π₯ ππ π 2π₯ πππ‘ 2π₯ ππ₯
=
ππ π 2 2π₯ β 1
=
2
1 π ππ 3 2π₯
ππ₯
ππ π 2 2π₯ ππ π 2π₯ πππ‘ 2π₯ ππ₯
=
ππ π 4 2π₯ β 2 ππ π2 2π₯ + 1 ππ π 2 2π₯ ππ π 2π₯ πππ‘ 2π₯ ππ₯
=
ππ π 6 2π₯ β 2 ππ π 4 2π₯ + ππ π 2 2π₯ ππ π 2π₯ πππ‘ 2π₯ ππ₯
πππ‘ π’ = ππ π 2π₯
=β
1 2
π’6 β 2π’4 + π’2 ππ’
=β
1 π’7 2 7
β
= β
ππ₯
=
πππ‘ β6 π₯ ππ π 2 π₯ ππ π 2 π₯ ππ₯
=
πππ‘ β6 π₯ 1 + πππ‘ 2 π₯ ππ π 2 π₯ ππ₯
=
πππ‘ β6 π₯ + πππ‘ β4 π₯ ππ π 2 π₯ ππ₯
πππ‘: π’ = πππ‘ π₯ ππ’ = β ππ π 2 π₯ ππ₯ ππ₯
= β1
ππ’ = ππ π 2π₯ πππ‘ 2π₯ ππ₯ 2
2π’ 5 5
ππ π 4 π₯ πππ‘ 6 π₯
βππ’ = ππ π 2 π₯ππ₯
ππ’ π(π₯) = β2 ππ π 2π₯ πππ‘ 2π₯ ππ₯ ππ₯ β
9.
+
π’3 3
π’β6 + π’β4 ππ₯
= β1 β =
+π
ππππ ππ ππππ ππ ππππ ππ + β +π ππ π π
πππ‘ β5 π₯ 5
π’β5 π’β3 β +π 5 3
+
πππ‘ β3 π₯ 3
+π
ππππ π ππππ π = + +π π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
20
EXERCISE 10.5
1.
TRIGONOMETRIC SUBSTITUTIONS
π₯ 2 ππ₯
3.
4βπ₯ 2
ππ₯
π₯ = π’ ; π = 2 ; 22 β π₯ 2
π’ = 3π₯
π’ = π π ππ π π₯ π ππ π = 2 π₯ = 2π ππ π ππ₯ = 2πππ πππ
π=2
= =
π’ = ππ‘πππ 3π₯ = 2π‘πππ 2 3π₯ π₯ = π‘πππ ; π‘πππ = 3 2
π₯ 2 ππ₯ 4 β π₯2
2 ππ₯ = π ππ 2 πππ 3
4 π ππ π 2πππ πππ 2πππ π π ππ2 πππ
=4
1 β πππ 2πππ 2 ππ β
π₯ 2
= πππππππ
1 2
ππ₯
=
π₯ 9π₯ 2 + 4 2 π ππ 2 πππ 3
=
2π ππππππ π ]2 + C
π π βπ π π
9π₯ 2 + 4
2π πππ =
1 π₯ π ππ2π + πΆ ; π = π ππβ( ) 2 2
= 2[ππππ ππ β
9π₯ 2 + 4 2
π πππ =
=4
=2
;
π₯ 9π₯ 2 +4
π β ππ +πͺ π
2 π‘πππ2π πππ 3
π πππππ 2π‘πππ
= 1 = 2
1 πππ π π πππ πππ π
ππ
=
1 2
1 ππ π πππ
=
1 2
ππ ππππ
=
1 [-ππβ‘|ππ ππ 2
= β
π ππ π
+ πππ‘π|] + πΆ
πππ + π π β +πͺ ππ ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
21
EXERCISE 10.5
5.
π₯ 2 ππ₯ 3 9βπ₯ 2 2
7.
π₯ 2 ππ₯
=
=
TRIGONOMETRIC SUBSTITUTIONS
9β
9 β π₯2
3 ππ₯ = πππ πππ 2
π₯ = 3π πππ ππ₯ = 3πππ π
=
= =
2π₯
π = π ππβ1 ( )
(3π πππ)2 3 πππ π 9 β (3π πππ)2 3 πππ π 9π ππ2 π 1 β π ππ2 π
3
= =
π ππ2 π (1 β π ππ2 π)
=
π ππ2 π ππ πππ 2 π π‘ππ2 πππ β
3 π₯ = π πππ 2 2π₯ = π πππ 3
π’ = ππ πππ
=
ππ₯
π’ = ππ πππ ; 2π₯ = 3π πππ
π₯ 2 ππ₯
π’=π₯ ; π=3
=
π₯2
π = 3 ; π’ = 2π₯
π₯2 3
9 β π₯2
9β4π₯ 2
3 2
3πππ π ( πππ πππ ) 3 2
( π πππ )2 3πππ π (3πππ πππ ) 3 2
2( π πππ )2 9πππ 2 πππ 2(
9 ) 4π ππ 2 π
π ππ 2 πππ
= π‘πππ β π =
π
π β π¨πππππ + π π π β ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
22
EXERCISE 10.5
ππ₯ π₯ 2 +4 2
9.
TRIGONOMETRIC SUBSTITUTIONS
π₯ = 2 π‘πππ ;
ππ₯
11.
; π€ππππ: π’ = π₯ , π = 2
π₯ π₯ 2 β9
π = 3 ;π’ = π₯
π₯ π‘πππ = 2
π’ = ππ πππ π₯ = 3π πππ ; ππ₯ = 3π ππππ‘πππππ π₯ π₯ π πππ = ; π = π΄πππ ππ 3 3
π₯
π·π₯ = 2 π ππ 2 πππ ; π = ππππ‘ππ 2
x
π₯2 + 4
x π₯2 β 9
2 π₯2 + 4 2
π πππ =
π₯2 + 4
2π πππ =
3
2
4 π ππ 2 π = π₯ 2 + 4 π‘πππ =
2
2 π ππ πππ 4 π ππ 2 π 2
=
2 π ππ 2 πππ = 16 π ππ 4 π
= 1 8
πππ 2 πππ
=
1 8
1 + πππ 2π ππ 2
=
1 8
=
1 ππ + 2
1 1 π 2
=8 =
ππ 1 = 2 8 π ππ π 8
πππ 2π ππ 2
ππ π ππ 2 π
= =
π₯2 β 9 ; 3π‘πππ = 3 ππ₯
π₯ π₯2 β 9
=
π₯2 β 9
3π ππππ‘πππππ 3π πππ(3π‘πππ)
ππ 3
=
1 π 3
=
π π π¨πππππ + π π π
1
+ 4 (2)π ππππππ π + π
π π π½+ ππππ½ππππ½ + π ππ ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
23
EXERCISE 10.5
TRIGONOMETRIC SUBSTITUTIONS
ππ₯
3
π₯ 2 β 16 2 ππ. ( ) π₯3
ππ.
π’ = π₯; π = 4
5 β 12π₯ + 4π₯ 2 = 2π₯ β 9 β 4
π’ = π ππβ ; π₯ = 4π ππβ ; π ππβ =
π₯ 4
π’ = ππ ππβ ; 2π₯ β 3 = 2π ππβ 2π₯ β 3 = 2π ππβ ; 2π₯ = 2π ππβ + 3
π₯ 3 = 64 sec 3 β ; ππ₯ = 4π ππβ π‘ππβ πβ π₯ 2 β 16 ; 4π‘ππβ = 4
π₯ 2 β 16
=4 =4
2ππ₯ = 2π ππβ π‘ππβ ππ₯ = π ππβ π‘ππβ π ππβ =
3
=
5 β 12π₯ + 4π₯ 2
π = 2 ; π’ = 2π₯ β 3
π₯ β = ππππ ππβ 4
π‘ππβ =
2π₯ β 3
( 4π‘ππβ (4π ππβ π‘ππβ πβ )) (64sec3 )
2π₯ β 3 2
β = ππππ ππ
tan4 β πβ secβ‘^2β
π‘ππβ =
(sec 2 β1)^2πβ sec 2 β
2π‘ππβ =
2π₯ β 3 2
2π₯ β 3 2
2π₯ β 3
=4
sec 4 β2 sec 2 β + 1 sec 2 β
=
(π ππβ π‘ππβ ) 2π ππβ 2π‘ππβ
=4
sec 4 β β 2 sec 2 β + 1 sec 2 β
=
1 4
π ππβ π‘ππβ π ππβ π‘ππβ
=4
sec 2 β β 2 + 1/ sec 2 β πβ
=
1 4
πβ
1
= 4(π‘ππβ β 2β + 2 β + π ππβ πππ β =
1 4
= β ; β = ππππ ππ
π π ππ β ππ ππ β ππ β π ππ«ππ¬ππ + +π π ππ
=
2
β4 2
β4
2π₯β3 2
π ππ β π ππππππ + π π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
24
EXERCISE 10.6
π.
π₯2
ADDITIONAL STANDARD FORMULAS
ππ₯ + 25
36 β 9π₯ 2 ππ₯
π.
Let: π’ = π₯
Let: π = 6
π=5
π’ = 3π₯
ππ’ = ππ₯
ππ’ = ππ₯ 3
=
π π π¨πππππ + π π π
π₯ππ₯
π.
1 β π₯4
=
1 3π₯ 1 3π₯ 36 β 9π₯ 2 + 8π΄πππ ππ +π 3 2 3 6 1 3π₯ 2
1
=
Let: π’ = π₯ 2
π₯
36 β 9π₯ 2 + 3 8π΄πππ ππ 2 + π
=3
π π ππ β πππ + ππ¨πππππ + π π π
π=1 ππ’ = ππ₯ 2
π.
1 π₯2 = π΄πππ ππ + π 2 1
Let: π = 5
π π
= π¨πππππππ + π
16π₯ 2 + 25ππ₯
π’ = 4π₯ ππ’ = ππ₯ 4
π.
ππ₯ 49 β 25π₯ 2
Let: π = 7
=
=
1 4π₯ 4
2
16π₯ 2 + 25 +
1 52 4
2
ππ 4π₯ + 16π₯ 2 + 25 + π
π ππ ππππ + ππ + ππ ππ + ππππ + ππ + π ππ π
π’ = 5π₯ ππ’ 2
= ππ₯
=
π ππ β π ππ +π ππ ππ + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
25
EXERCISE 10.7
INTEGRANDS INVOLVING QUADRATIC EQUATIONS
π.
ππ₯ π₯ 2 β3π₯+2
1.
πππππππ‘πππ π‘ππ π ππ’πππ
2π₯ 2
ππ₯
=
1 2
π₯β2
2
π₯ β 3π₯ = β2 π₯ 2 β 3π₯ = 3 2
2
3 π₯β 2
2
π₯β
9 9 = β2 + 4 4
=
(π₯ β
=
π’2 1
β
2
1 2
= ππ
ππ’ 1 π’βπ = ππ +π 2 βπ 2π π’+π
ππ
1 2
1 4
1 2
=
ππ’ + π2
= ππππ‘ππ
3 π’=π₯β 2 π=
π’2
1 1 π= , π’=π₯β 2 2
1 β 4 3 2 ) 2
=
1
+4
1 π’ = ππππ‘ππ + π 2 π
1 4
ππ₯
=
ππ₯ β 2π₯ + 1
3 1 2 2 3 1 π₯β + 2 2
π₯β β
+π
πβπ +π πβπ
1 2
π₯β 1 2
+π
=
π ππππππππ β π + π π
π.
3 β 2π₯ β π₯ 2
=
4β π₯+1
2
π’ = π₯ + 1, π = 2 π2
= =
π₯+1 2
=
β
π’2
π’ π2 π’ 2 2 = π β π’ + ππππ ππ + π 2 π 4
3 β 2π₯ β π₯ 2 + 2 ππππ ππ
π₯+1 2
+π
π+π π+π π β ππ β ππ + πππππππ +π π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
26
EXERCISE 10.7 π.
π₯2
INTEGRANDS INVOLVING QUADRATIC EQUATIONS
ππ₯ β 8π₯ + 7
2π₯β3ππ₯ 4π₯ 2 β1
11.
2π₯ππ₯ 4π₯ 2 β1
Completing the square
=
π₯ 2 β 8π₯ = β7
=2
π₯ 2 β 8π₯ + 16 = β7 + 16 π₯β4
2
=9
π₯β4
2
β9=0
π₯ππ₯ 4π₯ 2 β1
β3
ππ₯ 4π₯ 2 β1
πππ‘ π’ = 4π₯ 2 β 1 ; ππ’ 8
=2
ππ₯ (π₯ β 4)2 + 9
=
3ππ₯ 4π₯ 2 β1
β
=
π’
1
β 3[2 ππ
π ππ|πππ π
ππ’ = π₯ππ₯ 8 4π₯ 2 β1 4π₯ 2 +1 π
β π| β π ππ
+ π] ππβπ ππ+π
+π
π = 3 ;π’ = π₯ β4 = =
π’2
1 π’βπ ππ +π 2π π’+π 1
= 6 ππ =
ππ’ β π2
π₯β4β3 π₯β4+3
+π
=
π πβπ ππ +π π πβπ
(2π₯+7)ππ₯ π₯ 2 +2π₯+5
13.
=
2π₯+2 +5ππ₯ π₯ 2 +2π₯+5 2π₯+2 π₯ 2 +2π₯+5
+5
ππ₯ (π₯+1)2 +4
πππ‘ π’ = π₯ 2 + 2π₯ + 5 ; ππ’ = (2π₯π‘2)ππ₯ =
3+2π₯ ππ₯ π₯ 2 +9
9. = =3
2π₯ππ₯ π₯ 2 +9
ππ₯ + π₯ 2 +9
1
+ 2 π΄πππ‘ππ
π₯+1 2
+π π
=
3ππ₯ + π₯ 2 +9
ππ’ π’
2
β‘ππ|ππ + ππ + π| + π π¨πππππ
π+π + π
π
π₯ππ₯ π₯ 2 +9
πππ‘ π’ = π₯ 2 + 9 ; ππ’ = 2π₯ππ₯ 1
π₯
= 33 π΄πππ‘ππ 3 + 2
ππ’ 2
π’
π
= π¨πππππ π + ππ ππ + π + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
27
EXERCISE 10.7 (π₯β3)ππ₯
15. = =
INTEGRANDS INVOLVING QUADRATIC EQUATIONS 19.
4π₯βπ₯ 2 π₯β2 β1ππ₯
=2
4π₯βπ₯ 2 π₯β2 ππ₯ 4π₯βπ₯ 2
πππ‘ π’ = ππ’ =
4π₯βπ₯ 2
4π₯ β
π₯2
β2(πβ2)ππ₯ 2 4π₯βπ₯ 2
+
17 2
ππ₯ π₯ 2 β4π₯+20
π₯ 2 β 4π₯ + 20 = π₯ β 2
4 β 2π₯
; ππ’ =
2 4π₯ β π₯ 2
ππ₯ = 2[
; 4π₯ β π₯ 2 = 4 β (2 β π₯)2
ππ’ π’
+
17 2
2
πβπ + π
+ 16
ππ₯ ] π₯β2 2 +16
= 2[ππ π₯ 2 β 4π₯ + 20 +
17 1 π₯β2 ( )Arctan 4 2 4
= π ππ ππ β ππ + ππ +
4β(2βπ₯)2
= β ππ β ππ β π¨πππππ
ππ πβπ Arctan + π π
+ π] π
π
π₯+3 ππ₯
17.
8π₯βπ₯ 2 π₯β4 +7ππ₯
=
8π₯βπ₯ 2 π₯β4 ππ₯ 8π₯βπ₯ 2
πππ‘ π’ = ππ’ = =
2π₯+4ππ₯ π₯ 2 β4π₯+20
2(2π₯+4+17)ππ₯ π₯ 2 β4π₯+20
ππ’
= ππ’ β
=
=
πππ‘ π’ = π₯ 2 β 4π₯ + 20 ; ππ’ = (2π₯ β 4)ππ₯
ππ₯
β
(4π₯+9)ππ₯ π₯ 2 β4π₯+20
+7
ππ₯ 8π₯βπ₯ 2
8π₯ β π₯ 2 ; ππ’ =
β2(π₯ β 4)ππ₯ 2 8π₯
β π₯2
8 β 2π₯ 2 8π₯ β π₯ 2
ππ₯
; 8π₯ β π₯ 2
16 β (4 β π₯)2
=β ππ’ + 7
ππ’ 16β(4βπ₯)2
= - ππ β ππ + ππ¨πππππ
πβπ + π
π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
28
EXERCISE 10.8
ALGEBRAIC SUBSTITUTION
ππ₯
1.
5
π§=
3
π₯
5
=
π§ 2 ππ§ π§3 β π§2
=3
3
9π₯ 6 β10π₯ 4 30 π
+π
π
π
ππ (πππ β ππππ ) = +π ππ
ππ§ π§β1
=3
7
3π₯ 6 π₯ 4 = β +π 10 3
2 π₯βπ₯ 3
π’ = π§β1 ππ’ = ππ§ 5.
ππ’ π’
=3
ππ₯ π₯+2
= 3 ππ |π§ β 1 | + π
π§ = π₯+2 π§4 = π₯ + 2 π₯ = 2 β π§4 ππ₯ = β4π§ 3 ππ§
π
= π ππ | π β π | + π
1
1
π₯+2 2
4
= 3 ππ π’ + π
= β4
π§ 3 ππ§ π§3 β π§2
= β4
π§ππ§ π§β1
1
(π₯ 3 βπ₯ 4 ππ₯
3.
3 4β
π’ = π§β1
1
4π₯ 2
π§=
12
ππ’ = ππ§
π₯
π§ = π’+1
ππ₯ = 12π§11 ππ§ 5
=3
(π§ 4 βπ§ 3 ) π§ 11 ππ§ π§8
= 3 (π§ 9 β π§ 8 ) ππ₯ = 3[ π§ 9 ππ§ β
π§ 8 ππ§]
= β4
π’ + 1 ππ’ π’
= β4[
π’ ππ’ + π’
ππ’ ] π’
= β4[π’ + ππ π’ + π = βπ[π β π + ππ π β π + π]
10
9
π§ π§ = 3[ β + π] 10 9 =
3π§10 π§ 9 β +π 10 3 DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
29
EXERCISE 10.8
7.
ALGEBRAIC SUBSTITUTION
1
4 + π₯ππ₯ ;
9. π₯ π₯ + 4 3 ππ₯ π§ = π₯+4
π§ =(4+ π₯)1/2
12
π₯ (π§ 2 β 4)2= π₯ππ₯ = 4π§ 3 β 16π§ ππ§
=
π§ 4π§ 3 β 16π§ ππ§
=
4π§ 4 β 16π§ 2 ππ§
=4
π§5 π§3 β 16 +πΆ 5 3
=
4 (4 + 5
= 4+ π₯
= 4+ π₯
= 4+ π₯
π₯)5/2β
4 4+ π₯ 5
3 2
3 2
3 2
=
4 15
=
π π+ π π
4+ π₯
16 (4 + 3
; π§3 = π₯ + 4
π₯ = π§ 3 β 4 ; ππ₯ = 3π§ 3 ππ§
π§ 2 β 4 = π₯π§ 4 β 8π§ 2 + 16 = π₯ π§=
1 3
π§ 3 β 4 π§ 3π§ 2 ππ§
=
π§ 6 ππ§ β 4
=3
π§ 3 ππ§
7
4 3π§ 3 = β 3π§ 3 + π 7
3 π₯+4 = 7
π₯)3/2+C
7 3
β3 π₯+4
4 3
+π
4
16 β +πΆ 3
12 4 + π₯ β 80 +πΆ 15
=
3 π₯+4 3 7
π₯+4β7 +π 1 3
π₯ π₯ + 4 ππ₯ =
π π+π
π π
π
πβπ
+π
48 + 12 π₯ β 80 +πΆ 15 3 2
π π
12 + 3 π₯ β 20 + πΆ π πβπ +πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
30
EXERCISE 10.8
11.
4β 2π₯+1 1β2π₯
ALGEBRAIC SUBSTITUTION
ππ₯
π§ = 2π₯ + 1 ; π§ 2 = 2π₯ + 1 π§2 β 1 2π₯ = π§ 2 β 1 ; π₯ = ; ππ₯ = π§ππ§ 2 =
= = = =
ππ.
x 5 4 + x 3 dx
π§=
4 + π₯3π§2 = 4 + π₯3 ; π₯ =
ππ₯ =
1 4 β π§2 3
=β
4 β π§ π§ππ§ 1β2
π§ 2 β1 2
ππ§ β 2
=π§β2
2π§ππ§ β π§2 β 2
= π§ β 2 ππ π§ 2 β 2 +
ππ§ 2 π§ β2 1 ππ 2
= ππ + π β π ππ ππ β π +
π π
2π₯π‘1β 2 2π₯+1+ 2
ππ
+π
ππππ β π ππ + π + π
+π
2 3
4 β π§2
5
(π§)
=
1 3
=
1 2π§ 5 8π§ 3 β +π 3 5 3
=
2π§ 5 8π§ 3 β +π 15 9
=
β2π§ππ§ 3 4 β π§2
2 3
β8π§ 2 + 2π§ 4 ππ§ 3
=
2π§ β 1 ππ§ π§2 β 2
3 4 β π§2
4 β π§ 2 π§ β2π§ππ§ 3
=
4π§ β 2ππ§ π§2 β 2
β2π§ππ§
2π§ππ§
3
=
4π§ β 2ππ§ 1+ β π§2 β 2 1β
6
2π§ 4 ππ§ β 8π§ 2 ππ§
4 + π₯ 3 β 40 4 β π₯ 3 +π 45
=
4 + π₯ 3 6 4 + π₯ 3 β 40 +π 45
=
4 + π₯ 3 24 + 6π₯ 3 β 40 +π 45
= =
4 β π§2
π₯ 5 4 + π₯ 3 ππ₯
=
4π§ β π§ 2 ππ§ 2 β π§2
2 3
3
4+π₯ 3 β16+6π₯ 3 45
+π
π π + ππ πππ β π +π ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
31
EXERCISE 10.8
ALGEBRAIC SUBSTITUTION
3
1
15. π₯ 3 (4 + π₯ 2 )2 ππ₯
17. Κ
3+1 =2 2
π₯ = π‘ππ π’ ; ππ₯ = π ππ 2 π’ ππ’
π§2 = 4 + π₯2
= Κ πππ‘ 3 π’ ππ π π’ ππ’
X = 4 β π§2
= Κ πππ‘ π’ ππ π π’ ππ π 2 ( π’ β 1)ππ’ 1
1
dx = 2 4 β π§ 2 β2 (-2zdz)
= π π’ππ . π = ππ π π’ πππ ππ = β πππ‘ π’ ππ π π’ ππ’
= βΚ π 2 β 1 ππ
π§ππ§
1 (4βπ§ 2 )2
= Κ1ππ β Κπ 2 ππ
= (4 β π§ 2 )(π§ 3 )(βπ§ππ§) = Κπ β = =
ππ₯
π‘πππ π₯ 2 + 1 = π‘ππ2 π’ + 1 = π ππ π’ & π’ = π‘ππβ1 π₯
Z = 4 + π₯2
=-
π₯ 4 π₯ 2 +1
β4π§ 4 + π§ 6 ππ§
π 3 3
= ππ π π’ β
4π§ 5 π§ 7 β + +πΆ 5 7
=
=
28π§ 5 + 5π§ 7 +πΆ 35
=
β28( 4+π₯ 2 )5 +5( 4+π₯ 2 )7 +C 35
ππ π 3 π’ 3
+πΆ
ππ + π πππ β π +πͺ πππ
5
= =
=
4 + π₯ 2 (β28 + 5(4 + π₯ 2 ) +πΆ 35 4+π₯ 2
5
(β28+20+5π₯ 2 ) 35
π + ππ
π
+πΆ
(πππ β π)
ππ
+πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
32
EXERCISE 10.8
ππ. ( π₯=
ALGEBRAIC SUBSTITUTION
ππ₯ π₯ 2 (81 +
π₯4
)
1 1 ; ππ₯ = β 2 ππ§ π§ π§
1 π§2
βππ§ π§2
81 +
πππ‘ π₯ = 1 π§
=
1
3 4
1 π₯
ππ§
β π₯3
β π§2
1 π§4
ππ§
β π§2
1/π§ 4
1
(π§ 2 β1) 3 π§ π§4
=
πππ‘ π’ = 81π§ 4 + 1 ; ππ’ = 324π§ 3 ππ§
β1 81
π§=
1
81π§ 4 + 1
1 324
1 , π§2
π§ 2 β1 3 π§3
π§3
=
=
(π₯βπ₯ 3 )1/3 ππ₯ π₯4
=
1 π4
βππ§ π§2 81π§ 4 +1 π§6
=
21.
(β
ππ§ ) π§2
1
π§ 2 β 1 3 π§ππ§
=β
ππ’ πππ‘ π’ = π§ 2 β 1
3
π’4 81π§ 4 + 1
1 4
+π
=β
1 2
1
π’3 ππ’ 4
π ππ + ππ =β +π ππ ππ
=
1 π’3 β 2 ( 4 )+c 3
=β
3 2 π§ β1 8 3
= β8
1 π₯2
β1
4 3
4 3
+π +π
4
=
3 1-x2 3 β 8 x2 +c
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
33
EXERCISE 10.9
INTEGRATION OF RATIONAL FUNCTIONS OF SINES AND COSINES
π·π₯ 1+πππ π₯
1.
ππ₯ = π πππ₯ + πππ π₯ + 3
π.
2ππ§ 1+π§ 2 2π§ 1βπ§ 2 + +3 1+π§ 2 1+π§ 2
2 π·π§
=
1+π§ 2 1βπ§ 2 1+π§ 2
1+
2ππ§
=
1+π§ 2 1+π§ 2 + 1βπ§ 2 1+π§ 2
=
2ππ§ 1 2
π§+2
π§+2 2 π’ 2 = ππππ‘ππ + π = ππππ‘ππ + π 7 7 π π 2
= ππ₯ 4+2 π πππ₯
2 ππ§
=
=
1+ π§ 2 2π§ 1+π§ 2
1 2π§+1 ππππ‘ππ 7 7
π π
4+2
2ππ§
=
1+π§ 2
ππ§
=
=
1 2 2
+
3
= 2
π’2
3 2
ππππ‘ππ
π π
π§+2 3 2
ππ’ 2 π’ = ππππ‘ππ + π 2 +π π π
4
1
2
π
ππππππ
ππππ π + π π
+ π
1+π§ 2
π.
π πππ₯ππ₯ = 2ππ§ =2 1 β π§2
= π§+
+ πΆ
4+4π§ 2 + 4π§ 1+π§ 2
2 ππ§ 1 3 ; π€ππππ: π’ = π§ + ; π = 4π§ 2 + 4π§ + 4 2 2
=2
2
2ππ§
=
4π§ 4+ 1+π§ 2
=
7 4
1
π = πππ + π π
3.
+
ππ’ 7 1 βΆ π€ππππ π = , π’ = π§ + π’2 + π2 2 2
=2
ππ§ = π§ + π
1+2π§β2π§ 2 +3+3π§ 2 1+π§ 2
2ππ§ = 4 + 2π§ + 2π§ 2
=
2ππ§ 2
=
1+π§ 2
1+2π§β2π§ 2 +3 1+π§ 2
2 ππ§
=
2ππ§
=
1+π§ 2
+ π=
1 3
ππππ‘ππ
π
ππππππ
π πππ π + π π
2π§+1 3
+ π
2
= 2π ππ
π+π’ πβπ’
1 + π§ 2 2ππ§ . 1 β π§2 1 + π§2 π2
ππ’ π€ππππ π = 1, π’ = π§ β π’2
+ π = ππ
1+π§ 1βπ§
+πΆ π
π + πππ π π π+π = ππ + π = ππ π + π ππ πβπ π β πππ π
+π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
34
EXERCISE 10.10
INTEGRATION BY PARTS
1. π₯πππ π₯ππ₯ =
π₯πππ π₯ππ₯
; πππ‘ ππ£ = πππ₯ππ₯ π£ = π πππ₯
= π₯π πππ₯ β
π’=π₯ ππ’ = ππ₯
π πππ₯ππ₯
= πππππ + ππππ + πͺ
3. π βπ₯ πππ 2π₯ππ₯ ; π’ = πππ 2π₯
ππ£ = π βπ₯ ππ₯ ; π’ = π ππ2π₯
;
π£ = -π βπ₯
ππ’ = π ππ2π₯ππ₯ ;
; ππ£ = π βπ₯ ππ₯
; ππ’ = 2πππ 2π₯ππ₯ ; π£ = -π βπ₯
= -π π₯ πππ 2π₯ β 2π βπ₯ π ππ2π₯ππ₯ = -π π₯ πππ 2π₯ β 2 π βπ₯ π ππ2π₯ππ₯ = -π βπ₯ πππ 2π₯ β 2[-π βπ₯ π ππ2π₯ β
-π βπ₯ 2πππ 2π₯ππ₯
= -π βπ₯ πππ 2π₯ + 2π βπ₯ π ππ2π₯ β 4 π βπ₯ πππ 2π₯ππ₯ πππ 4 π βπ₯ πππ 2π₯ππ₯ π‘π πππ‘π π ππππ =
2π βπ₯ π ππ 2π₯βπ βπ₯ πππ 2π₯ 5
=
πβπ π
+πΆ
πππππ β πππππ + πͺ
5. ππππ‘ππ2π₯ππ₯ ; ππ£ = ππ₯ ; π’ = ππππ‘ππ2π₯ 2ππ₯
π£ = π₯ ; ππ’ = 1+π₯ 2 2π₯ππ₯ 1+4π₯ 2
= π₯ππππ‘ππ2π₯ β = π₯ππππ‘ππ2π₯ β 2 1
ππ₯ 1+4π₯ 2 ππ’ π’
= π₯ππππ‘ππ2π₯ β 4 π π
= πππππππππ β ππ π + πππ + πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
35
EXERCISE 10.10
INTEGRATION BY PARTS
7. π ππ 3 π₯ππ₯ ; ππ£ = π ππ 2 π₯ππ₯ ; π’ = π πππ₯ π£ = π‘πππ₯
ππ’ = π πππ₯π‘πππ₯
π ππ 2 π₯π πππ₯ππ₯
=
= π πππ₯π‘πππ₯ β
π ππ 2 π₯ β 1 π πππ₯ππ₯
= π πππ₯π‘πππ₯ + ππ π πππ₯π‘πππ₯ β π ππ 3 π₯ππ₯ ; πππ
π ππ 3 π₯ππ₯ ππ πππ‘π π ππππ
2 π ππ 3 π₯ππ₯ = π πππ₯π‘πππ₯ + ππ π πππ₯ + π‘πππ₯ π π
π ππ 3 π₯ππ₯ =
ππππππππ + ππ ππππ + ππππ + πͺ
9. π₯πππ 2 2π₯ππ₯ ; ππ£ = π2π₯ππ₯ π£= 1 π₯ 2
=π₯ =
π₯2 2
1 π₯ 2
;
1
+ 8 π ππ4π₯
1
1
ππ’ = π₯ 1
+ 8 π ππ4π₯ β (2 π₯ + 8 π ππ4π₯)ππ₯
1 8
1 4
+ π₯π ππ4π₯ β π₯ 2 +
π
π’ =π₯
π
1 πππ 4π₯ 32
+πΆ
π
= π ππ + π πππππ + ππ πππππ + πͺ
11.
π₯ππππ πππ₯ππ₯ 1βπ₯ 2
;
ππ£ =
π₯ππ₯
;
1βπ₯ 2
π£ = - 1 β π₯2
;
π’ = ππππ πππ₯ ππ’ =
= - 1 β π₯ 2 ππππ πππ₯ β (- 1 β π₯ 2 )(
ππ₯ 1βπ₯ 2
ππ₯ 1βπ₯ 2
)
= - 1 β π₯ 2 ππππ πππ₯ + ππ₯ = π β π β ππ πππππππ + πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
36
EXERCISE 10.10
INTEGRATION BY PARTS
13. π πππ₯ππ 1 + π πππ₯ ππ₯ ;
π’ = ππ 1 + π πππ₯ ππ’ =
1 1+π πππ₯
;
ππ£ = π πππ₯ππ₯
πππ π₯ππ₯ ;
π£ = πππ π₯
πππ 2 π₯ππ₯ 1+π πππ₯
= -πππ ππ 1 + π πππ₯ +
1βπ ππ 2 π₯ππ₯ 1+π πππ₯
= -πππ ππβ‘|1 + π πππ₯ | + = -πππ ππ 1 + π πππ₯ +
1 β π πππ₯ ππ₯
= -πππ ππ 1 + π πππ₯ + π₯ + πππ π₯ + πΆ = β πππ ππ π + ππππ + π + ππππ + πͺ
15.
π π₯ π₯ππ₯ (π₯+1)2
;
π’ = ππ₯ π₯ ππ’ = π π₯ π₯ + 1 ππ₯
=-
ππ₯π₯ π₯+1
+ π π₯ ππ₯ =
ππ£ = (π₯+1)2 ππ₯
;
π£ =-
π₯3
=
1 ππππ ππ 3
=
1 3 π₯ ππππ πππ₯ 3
+ (3 πππ π β
=
1 3 π₯ ππππ πππ₯ 3
+
=
1 3 π₯ ππππ ππ 3
ππππ ππππ; 1
=3
1
1 3
1βπ₯ 2
3β 1βπ₯ 2 9 1βπ₯ 2 2+π₯ 2
π₯3 1βπ₯ 2
ππ₯ 1βπ₯ 2
;
ππ£ = π₯ 2 ππ₯ ; π£ =
π₯3 3
ππ₯
1
+
1 π₯+1
ππ +πͺ π+π
17. π₯ 2 ππππ πππ₯ππ₯; π’ = ππππ πππ₯ ; ππ’ = β3
1
;
9
πππ 3 π )+ 9
πΆ
+πΆ +πΆ
ππ₯ ; π = 1 ; π£ = π₯ ; π₯ = π ππ π ; ππ₯ = πππ π ππ ;
1 β π₯ 2 = πππ π
π ππ 3 π (πππ πππ) πππ π
1
= 3 π ππ2 π π πππππ π π
= (βππππ +
ππππ π )+ π
πͺ DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
37
EXERCISE 10.11
INTEGRATION OF RATIONAL FUNCTIONS
12π₯+18 π₯+2 π₯+4 (π₯β1)
1.
12π₯ + 18 π΄ π΅ πΆ = + + π₯ + 2 π₯ + 4 (π₯ β 1) (π₯ + 2) (π₯ + 4) (π₯ β 1) 12π₯ + 18 = π΄ π₯ + 4 π₯ β 1 + π΅ π₯ + 2 π₯ β 1 + πΆ π₯ + 2 (π₯ + 4) 12π₯ + 18 = π΄(π₯ 2 + 3π₯ β 4) + π΅ π₯ 2 + π₯ β 2 + πΆ(π₯ 2 + 6π₯ + 8) 12π₯ + 18 = π΄π₯ 2 + 3π΄π₯ β 4π΄ + π΅π₯ 2 + π΅ β 2π΅ + πΆπ₯ 2 + 6πΆπ₯ + 8πΆ π΄π₯ 2 + π΅π₯ 2 + πΆπ₯ 2 = 0 3π΄π₯ + π΅π₯ + 6πΆπ₯ = 12π₯ 4π΄ + π΅ + 8πΆ = 18 π΄=1 π΅ = β3 πΆ=2
=
ππ₯ + (π₯+2)
β3ππ₯ + (π₯+4)
2ππ₯ (π₯β1)
= ππ π + π β π ππ π + π + πππ π β π β‘
3.
1=
ππ₯ π₯β1 (π₯β4) π΄ π΅ + (π₯ β 1) (π₯ β 4)
1 = π΄ π₯ β 4 + π΅(π₯ β 1) 1 = π΄π₯ β 4π΄ + π΅π₯ β π΅
=
β1 3 (π₯ β1)
1
ππ₯ +
1 3 (π₯ β4)
1
π΄+π΅ =0 β4π΄ β π΅ = 1 π΄ = βπ΅ 1 π΅= 3
ππ₯
= β 3 ππ π₯ β 1 + 3 ππ π₯ β 4 + πΆ
=
π ππ πβπ π ππ πβπ
+C
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
38
EXERCISE 10.11
INTEGRATION OF RATIONAL FUNCTIONS
6π₯ 2 +23π₯β9 ππ₯ (π₯ 3 +2π₯ 2 β3π₯)
5.
6π₯ 2 + 23π₯ β 9 ππ₯ π₯ π₯ + 3 (π₯ β 2) 6π₯ 2 + 23π₯ β 9 =
π΄ π΅ πΆ + + π₯ (π₯ + 3) (π₯ β 1)
6π₯ 2 + 23π₯ β 9 = π΄ π₯ + 3 π₯ β 1 + π΅ π₯ π₯ β 1 + πΆ π₯ (π₯ + 3) 6π₯ 2 + 23 β 9 = π΄ π₯ 2 + 2π β 3 + π΅ π₯ 2 β π₯ + πΆ(π₯ 2 + 3π₯) π΄+π΅+πΆ =6 2π΄ β π΅ + 3πΆ = 23 β3π΄ + 0π΅ + 0πΆ = β9 π΄=3 π΅ = β2 πΆ=5 =3
ππ₯ π₯
β2
ππ₯ + (π₯+3)
5
ππ₯ (π₯β1)
= πππ π β πππ π + π + πππ π β π + πͺ π₯ 3 +5π₯ 2 +9π₯+7 ππ₯ π₯ 2 +5π₯+4
7.
π₯ 3 + 5π₯ 2 + 9π₯ + 7 ππ₯ π₯ + 4 (π₯ + 1)
By division of polynomials, 5π₯ + 7 π΄ π΅ = + π₯ + 4 (π₯ + 1) (π₯ + 4) (π₯ + 1) 5π₯ + 7 = π΄ π₯ + 1 + π΅(π₯ + 4)
= =
π₯ππ₯ + ππ π
+
ππ ππ π
13 ππ₯ 3
(π₯ + 4) π+π +
+ π ππ π
ππ π₯ = 4, 13 π΄= 3 ππ π₯ = β1 2 π΅= 3
2 ππ₯ 3
(π₯ + 1) π+π +πͺ
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39
EXERCISE 10.11
9.
INTEGRATION OF RATIONAL FUNCTIONS
2π₯+1 π₯β2 (π₯β3)2
2π₯ + 1 =
π΄ π΅ πΆ + + (π₯ β 2) (π₯ β 3) (π₯ β 3)2
2π₯ + 1 = π΄ π₯ β 3
2
+π΅ π₯β3 π₯β2 +πΆ π₯β2
2π₯ + 1 = π΄ π₯ 2 β 6π₯ + 9 + π΅ π₯ 2 β 5π₯ + 6 + πΆ π₯ β 2 π΄+π΅ =0 β6π΄ β 5π΅ + πΆ = 2 9π΄ + 6π΅ β 2πΆ = 1 π΄=5 π΅ = β5 πΆ=7 =
5ππ₯ + π₯β2
β5ππ₯ + π₯β3
7ππ₯ π₯β3
2
7
= 5ππ π₯ β 2 β 5ππ π₯ β 3 + π₯β3 = πππ
πβπ π + πβπ (π β π)
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
40
EXERCISE 10.11
11.
INTEGRATION OF RATIONAL FUNCTIONS
2π₯β5 ππ₯ π₯(π₯β1)
2π₯ β 5 π΄ π΅ πΆ π· = + + + π₯(π₯ β 1) π₯ (π₯ β 1) (π₯ β 1)2 (π₯ β 13 2π₯ β 5 = π΄ π₯ β 1
3
+ ππ₯ π₯ β 1
2
+ ππ₯ π₯ β 1 + π·π₯
2π₯ β 5 = π΄π₯ 3 β 3π΄π₯ 2 + 3π΄π₯ β π΄ + π΅π₯ 3 β 2π΅π₯ 2 + π΅π₯ +C π₯ 2 β πΆπ₯ + π·π₯ 2π₯ β 5 = π΄π₯ 3 = 3π΄π₯ 2 β 2π΅π₯ 2 + π΅π₯ + πΆπ₯ 2 β πΆπ₯ + π·π₯ 2π₯ β 5 = π΄π₯ 3 + π΅π₯ 2 β 3π΄π₯ 2 β 2π΅π₯ 2 + πΆπ₯ 2 + 3π΄π₯ + π΅π₯ β πΆπ₯ + π·π₯ β π΄ 2π₯ β 5 = π΄ + π΅ π₯ 3 + β3π΄ β 2π΅ + π π₯ 2 + 3π΄ + π΅ β πΆ + π· π₯ β π΄ π΄+π΅ =0 β3π΄ β 2π΅ + πΆ = 0 3π΄ + π΅ β πΆ + π· = 2 βπ΄ = β5 π΄=5 π΅ = β5 πΆ=5 π· = β3 =
5ππ₯ + π₯
=5
ππ₯ β5 π₯
β5ππ₯ + (π₯ β 1)
5ππ₯ + (π₯ β 1)2
ππ₯ +5 (π₯ β 1) 5
β3ππ₯ (π₯ β 1)3
ππ₯ β3 (π₯ β 1)2
ππ₯ (π₯ β 1)3
3
= 5ππ π₯ β 5ππ π₯ β 1 β (π₯β1) + 2(π₯β1)2 + πΆ = πππ
π π π β + +πͺ π β π (π β π) π(π β π)π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
41
EXERCISE 10.11
INTEGRATION OF RATIONAL FUNCTIONS
3π₯ 2 +17π₯+32 π₯ 3 +8π₯ 2 +16π₯
13.
3π₯ 2 + 17π₯ + 32 π₯(π₯ + 4)2 3π₯ 2 + 17π₯ + 32 π΄ π΅ πΆ = + + 2 π₯(π₯ + 4) π₯ (π₯ + 4) (π₯ + 4)2 π΄+π΅ =3 8π΄ + 4π΅ + πΆ = 17 16π΄ = 32 π΄=2 π΅=1 πΆ=3 2ππ₯ + π₯
=
ππ₯ + (π₯ + 4)
β3ππ₯ (π₯ + 4)2 π
= πππ π + ππ π + π + π+π 2π₯+1 3π₯β1 (π₯ 2 +2π₯+2)
15.
2π₯ + 1 π΄ π΅ 2π₯ + 2 + πΆ = + 2 2 3π₯ β 1 (π₯ + 2π₯ + 2) (3π₯ β 1) π₯ + 2π₯ + 2 2π₯ + 1 = π΄ π₯ 2 + 2π₯ + 2 + π΅ 2π₯ + 2 3π₯ β 1 + πΆ 3π₯ β 1 2π₯ + 1 = π΄ π₯ 2 + 2π₯ + 2 + π΅(6π₯ 2 + 4π₯ + 2) + πΆ 3π₯ β 1 π΄+π΅ =0 2π΄ + 4π΅ + 3πΆ = 2 2π΄ + 2π΅ β πΆ = 1 5 π΄=β 2 5 π΅= 2 πΆ = β1 =β
5 2 5
ππ₯ 5 + (3π₯ β 1) 2
(2π₯ + 2) ππ₯ β π₯ 2 + 2π₯ + 2
π₯2
ππ₯ + 2π₯ + 2
5
= β 2 ππ 3π₯ β 1 + 2 ππ π₯ 2 + 2π₯ + 2 β ππ π₯ 2 + 2π₯ + 2 =
π ππ + ππ + π ππ β ππ ππ + ππ + π π ππ β π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
42
EXERCISE 10.11
17.
INTEGRATION OF RATIONAL FUNCTIONS
5π₯ 2 βπ₯+17 ππ₯ π₯+2 (π₯ 2 +9)
5π₯ 2 β π₯ + 17 π΄ π΅ 2π₯ + πΆ = + 2 π₯ + 2 (π₯ + 9) π₯ + 2 π₯2 + 9 5π₯ 2 β π₯ + 17 = π΄ π₯ 2 + 9 + 2π΅π₯ + πΆ (π₯ + 2) 5π₯ 2 β π₯ + 17 = π΄π₯ 2 + 9π΄ + 2π΅π₯ 2 + 4π΅π₯ + πΆπ₯ + 2πΆ 5π₯ 2 β π₯ + 17 = π΄π₯ 2 + 2π΅π₯ 2 + 4π΅π₯ + πΆπ₯ + 9π΄ + 2πΆ 5π₯ 2 β π₯ + 17 = π΄ + 2π΅ π₯ 2 + 4π΅ + πΆ π₯ + 9π΄ + 2πΆ π₯ 2 = π΄ + 2π΅ = 5 π₯ = 4π΅ + πΆ = β1 π = 9π΄ + 2πΆ = 17 π΄ + 2π΅ = 5 β 2 4π΅ + πΆ = β1
= β2π΄ β 4π΅ = β10 =
4π΅ + πΆ = β1 β2π΄ + πΆ = β11
β2π΄ + πΆ = β1 β 2 9π΄ + 2πΆ = 17
= 4π΄ β 2πΆ = 22 =
9π΄ + 2πΆ = 17 13π΄ = 39
A=3 9(3)+2C=17
4B-5=-1
27+2C=17
4B=-1+5
2C=17-27
4B=4
2C=-10
B=1
C=-5 =
=3
3 1 2π₯ β 5 + ππ₯ π₯+2 π₯2 + 9 ππ₯ π₯+2
+2
π₯ππ₯ β π₯ 2 +9
5
ππ₯ π₯ 2 +9 π
= πππ π + π + ππ ππ + π β π π¨πππππ
π + π
πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
43
EXERCISE 10.11
19.
INTEGRATION OF RATIONAL FUNCTIONS
4π₯ 2 +21π₯+54 π₯ 2 +6π₯+13
4 β 3π₯ β 2 π₯ 2 + 6π₯ + 13 π΄ 2π₯ + 6 + π΅ π₯ 2 + 6π₯ + 13 π΄ 2π₯ + 6 + π΅ = 3π₯ β 2 2π΄ + π΅ = 3 π΅ = β11 π΄=
3 2
3 2π₯ + 6 ππ₯ ππ₯ =Κ4β[ Κ 2 + (β11Κ 2 + 6π₯ + 13)] 2 π₯ + 6π₯ + 13 π₯ = β11Κ
π₯2
ππ₯ + 6π₯ + 9 + 13 β 9
ππ₯ π₯ + 3 2 + 13 β 9
= β11Κ 1 2
= β11( ππππ‘ππ = 4π₯ β
π₯+3 ) 2
3 ππβ‘| π₯ 2 2 π
2
+ 6π₯ + 13| β
= ππ β π ππβ‘| ππ + ππ + ππ| +
11 π₯+3 ππππ‘ππ 2 2 ππ π+π ππππππ π + π
πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
44
EXERCISE 10.11
21.
INTEGRATION OF RATIONAL FUNCTIONS
π₯ 3 +7π₯ 2 +25π₯+35 π₯ 2 +5π₯+6
π₯+2+
9π₯ + 23 ππ₯ + 5π₯ + 6
π₯2
9π₯ + 23 π΄ π΅ = + π₯ + 3 (π₯ + 2) π₯ + 3 π₯ + 2 9π₯ + 23 = π΄ π₯ + 2 + π΅(π₯ + 3)
x=-3 9(-3)+23= A(-3+2)+B(-3+3) -27+23=A(-1)+B(0) -4=-A A=4 If x=-2 9(-2)+23= A(-2+2)+B(-2+3) -18+23=A(0)+B 5=B B=5 β2 5 + ππ₯ π₯+3 π₯+2
=
π₯+2+
=
π₯ππ₯ + 2 ππ₯ β 4
ππ₯ π₯+3
+5
ππ₯ π₯+2
ππ = + ππ β πππ π + π + πππ π + π + π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
45
EXERCISE 10.11
23.
INTEGRATION OF RATIONAL FUNCTIONS
π₯ 2 βπ₯β8 (2π₯β3)(π₯ 2 +2π₯+2)
π΄ π΅ 2π₯ β 2 + πΆ + 2 2π₯ β 3 π₯ + 2π₯ + 2
A(π₯ 2 + 2π₯ + 2) + π΅ 2π₯ + 2 2π₯ β 3 + πΆ(2π₯ β 3) A(π₯ 2 + 2π₯ + 2) + π΅ 4π₯ 2 β 2π₯ β 6 + πΆ(2π₯ β 3) A+4B=1 2A-2B+2C=-1 2A-6B-3C=-8 1
A=-2 1
A=2 C=1 ππ₯ 1 + (2π₯ β 3) 2
β1
2π₯ + 2 ππ₯ + π₯ 2 + 2π₯ + 2
1 βππ(2π₯ β 3) + ππβπ₯ 2 + 2π₯ + 2β + 2
ππ₯ π₯ 2 + 2π₯ + 2 ππ₯ π₯ + 1 2 + 12
1
= β 2 β ππ 2π₯ β 3 + ππβπ₯ 2 + 2π₯ + 2β + ππππ‘ππ π₯ + 1 + πΆ =
π ππ + ππ + π ππβ β + ππππππ π + π + π π ππ β π
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46
EXERCISE 10.11
25. Κ
INTEGRATION OF RATIONAL FUNCTIONS
π₯ 5 +2π₯ 3 β3π₯ π₯ 2 +1 3
=Κ
π₯ 5 + 2π₯ 3 β 3π₯ π₯ 6 + 3π₯ 4 + 2π₯ 2 + 1
=
π΄ 2π₯ + π΅ πΆ 2π₯ + π· πΈ 2π₯ + πΉ + 2 + 2 π₯2 + 1 π₯2 + 1 π₯ +1 2 π₯ +1 3
= π΄ 2π₯ π₯ 2 + 1
2
+ π΅ π₯2 + 1
2
3
+ πΆ 2π₯ π₯ 2 + 1 + π· π₯ 2 + 1 + πΈ 2π₯ + πΉ
= π΄ 2π₯ π₯ 4 + 2π₯ 2 + 1 + π΅ π₯ 4 + 2π₯ 2 + 1 + πΆ 2π₯ 3 + 2π₯ + π· π₯ 2 + 1 + πΈ 2π₯ + πΉ = π΄ 2π₯ 5 + 4π₯ 3 + 2π₯ + π΅ π₯ 4 + 2π₯ 2 + 1 + πΆ 2π₯ 3 + 2π₯ + π· π₯ 2 + 1 + πΈ 2π₯ + πΉ 1 2
π₯ 5 : 2π΄ = 1
; π΄=
π₯4: π΅ = 0
; π΅=0
π₯ 3 : 4π΄ + 2πΆ = 2
; πΆ=0
π₯ 2 : 2π΅ + π· = 0
; π·=0
π₯: 2π΄ + 2πΆ + 2πΈ = β3 ; πΈ = 0 π: π΅ + π· + πΉ = 0 =
; πΉ=0
π π ππ ππ + π + π π π +π
π
+πͺ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
47
EXERCISE 10.11
INTEGRATION OF RATIONAL FUNCTIONS
27. π₯ 4 + 2π₯ 3 + 11π₯ 2 + 8π₯ + 16 π₯(π₯ 2 + 4)2 π΄ π΅ 2π₯ + πΆ π· 2π₯ + πΈ [ + + 2 ][(π₯ 2 + 4)2 ] π (π₯ 2 + 4) (π₯ + 4)2
A π₯2 + 4
2
+ π΅ 2π₯ π₯ (π₯ 2 + 4) + πΆ π₯ 2 + 4 (π₯) + π·(2π₯)(π₯) + πΈ(π₯)
A(π₯ 4 + 8π₯ 2 + 16) + π΅ 2π₯ 4 + 8π₯ 2 + πΆ π₯ 3 + 4π₯ + π·2π₯ 2 + πΈπ₯ π₯ 4 : π΄ + 2π΅ = 1
A=1
π₯3: πΆ = 2
B=0
π₯ 2 : 8A+8B+2D=11
C=2
X: 4C + E=8
D = 3/2
C : 16A = 16
E=0
=
ππ₯ 2ππ₯ 3 + 2 + π₯ π₯ +4 2
= πππ₯ + 2
1 2
ππππ‘ππ
2π₯ππ₯ (π₯ 2 + 4)2 π₯ 2
β
3 2 π₯ 2 +4
+πΆ
π π = πππ + ππππππ β +πͺ π π ππ + π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
48
EXERCISE 11.1
SUMMATION NOTATION
π=10
β π = 10
π.
π
12π 3
π.
π
π=1
π=1
π3
π=10
π=1 2
10 10 + 1 4
= 12
π3 β π
=
π=10
= 12
π(π β 1)(π + 1) π=1
π3 β π
=
2
π=1 π=10
=
= 3(100 121 )
π=10 3
π + π=1
= πππππ
=
π π=1
10 2 10+1 2 4
β
10 10+1 2
= ππππ π=10
(12π 2 + 4π )
π.
π=ππ
π.
π=1 π=10
π=1
π
π=10
9π 2 + 6π + 1
=
π=1
10(10 + 1)(2 10 + 1) 10(10 + 1) = 12 +4 6 2
π=1
π2 + 6
=9
= 2 110 21 + 2 110 = ππππ
π
π=π
π=10
π2 + 4
= 12
ππ + π
=9
π+
10(10+1)(2 10 +1) 6
+6
1 10(10+1) 2
+ 10
= ππππ
π. ππ β ππ + ππ β ππ + β― + (ππ β ππ ) π
=
ππ β ππ π=π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
49
EXERCISE 11.1
SUMMATION NOTATION
ππ. π π₯1 βπ₯1 + π π₯2 βπ₯2 + β― + π π₯π βπ₯π π
=
π(ππ ) βππ π=π
ππ. 14 + 24 + 34 + β― + π4 π
ππ
= π=π
ππ. π1 π1+π2 π2+π3 π3 + β― + ππ ππ π
=
ππ ππ π=π
ππ. π’13 + π’23 + π’33 + β― + π’π3 π
πππ
= π=π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
50
EXERCISE 11.2
π.
THE DEFINITE INTEGRAL
2 3π₯ 2 ππ₯ 1
π = 0 ;π = 2 βπ₯ = =
(π₯ β 1)ππ₯
π=0
;
2β0 π
βπ₯ =
1β0 π
2 π
= πππ 2
= πππ 3
4π 2 2 ( ) π2 π
= πππβ‘3
8π 2 π3
πββ
= πππ 24 πββ
= ππππββ 24
π 2 +1 2π+1 6 π3 1 0 0 0 2π 3 +π 2 +2π 2 +π 6π 3
} 1 π
β
1 2π3 + π2 + 2π2 + π π3 6
1 0 π 2 βπ π2
β1
2 3
= β1 =β
π π
5 2π₯ 1
5.
π(π + 1)(2π + 1) 1 6 π3
π π3
]β
1 π π+1 2π+ π2 6
= ππππββ 2
πββ
πββ
1 π
πββ
2π 2 ( ) π π
π π
π₯ 2 β π₯ππ₯
= πππ 2
π=1
ππ =
π2
2π π 3
;
= { πππ 2 [ ( 2 )
2 =0+π π
= πππ π=β
π=1
πββ
ππ = π + πβπ₯
=
1 2π₯ 0
3.
βπ₯ =
+ 3ππ₯
5β1 π
;
ππ = 1 + π
4 π
4
=π
1
= ππππββ 24 =π
4π
4
=ππππ=β (1 + π ) β π + 3π 4 π
=ππππ=β =ππππ=β
4π π
+ 16
16π π2
+ π2 +
+
3π
π(π+1) 2
+ 3π
= ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
51
EXERCISE 11.2
2
π.
THE DEFINITE INTEGRAL
π₯ 3 ππ₯
0
βπ₯ =
2 2π ; ππ = π π 3
= πππ
2π π
= πππ
8π 3 π3
πββ
πββ
2 π 2 π
16 π2 π + 1 πββ π 4 4
2
= πππ
π3
4 2 2 (π (π + 2π + 1) πββ π 4
= πππ
= ππππββ
4π 4 π4
8
+ π3 +
4π 2 π3
=4+0+0 = 4
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
52
EXERCISE 11.3
2
π.
SOME PROPERTIES OF THE DEFINITE INTEGRAL
3π₯ 2 β 2π₯ + 1 ππ₯
3
π.
1
2
3π₯ 3 2π₯ 2 + +π₯ 3 2
=
π’ = π₯2 + 1 ππ’ = 2π₯ππ₯
=8β4+2β1+1β1 =
=5
π₯ππ₯ π₯2 + 1
=
1 2 1 2
3 2
ππ’ π’
ππ 10 β ππ 5
= π. πππ 3
3π₯ 2 +
π. 1
4 ππ₯ π₯2 9.
3
3π₯ 4 + 3 π₯
=
0
=
4
= 27 β 3 β 1 + 4 =
0 ππ¦ β1 β(π₯ 2 +2π₯β1)
β1
ππ¦ β(π₯ + 2π₯ + 1 β 1 β 1)
0
ππ π
= β1
ππ¦ β[ π₯ + 1
0
= β1 7
π.
3
1+
π₯2
ππ₯
π’ = 1 + π₯2 ππ’ = 2π₯ππ₯ 1
=2 =
3
4 1+π₯ 2 3
4
+ 2]
ππ¦ β π₯+1
2
0
ππ¦
β1
2β π₯+1
=
0
2
+2
2
πππ‘ π = 2 ; π’ = (π₯ + 1) = π΄πππ ππ =
π₯+1 2
+π
π π
ππ π
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53
EXERCISE 11.3 π
ππ. 0
SOME PROPERTIES OF THE DEFINITE INTEGRAL 1
π₯ππ₯ π₯2 + π
0
πππ‘ π’ = π₯ 2 + π ; ππ’ = 2π₯ππ₯ ; ππ’ π 2
= 1 2
ππ’ = π₯ππ₯ 2
1
π₯ β 2 β π₯ ππ₯ 0
π’
π
=
2π₯ β π₯ 2 ππ₯
ππ.
π π
ππ’ π’
=
1 π πππ’ 0 2
=
1 π ln π₯ 2 + π 0 2
1 = ln π 2 + π β ππ 0 + π 2
π ; πππ β πππ = ππ π
1 π2 + π 1 π π+1 = ππ = ππ 2 π 2 π 1
= 2 ππ π + 1 = ππ π + 1
π πππ =
π
4
=
π
=8
1 ππππ‘ππ 1 2
=
π π
4
4
0
u= x; du=dx; a=2
=
2πππ π β 2π πππ β 4π ππππππ πππ
π ππ2 ππππ 2 π ππ
0 π
1 π₯ = ππππ‘ππ 2 2
2π πππ = π₯
π
=8 2 ππ₯ 0 π₯ 2 +4
; 2πππ π = 2 β π₯
π₯ = 2π ππ π ; ππ₯ = 4π ππππππ πππ π΄π‘ π₯ = 1, π = π 4 ; π₯ = 0, π = 0
1 2
= ππ π + π
ππ.
2βπ₯ 2 π₯ ; 2 2
cos π =
=2
π 0
4
1 β πππ 2π 2
1 + πππ 2π ππ 2
1 β πππ 2 2π ππ
=π π
1
ππ.
π₯π π₯ ππ₯
0
π’=π₯ ; ππ’ = ππ₯ ; = π₯π π₯ β
ππ£ = π π₯ ππ₯ π£ = ππ₯
1 π₯ π ππ₯ 0
= π₯π π₯ β π π₯
= 1β1+0β1 = 1
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54
EXERCISE 11.3
π 2
ππ.
SOME PROPERTIES OF THE DEFINITE INTEGRAL
π
π ππ π₯πππ₯ππ₯
0
πππ‘ π’ = π πππ₯ ; ππ’ = πππ π₯ππ₯ π’2 ππ’
=
6β1 6β3 6β5 2β1 6+2 6+2β2 6+2β4 6+2β6
=2
π’3 3
=
π ππ 3 π₯ 3
=
π ππ6 π’ πππ 2 π’ ππ’
π
π
=
π
=2
π 2
=
π₯ π₯ π ππ6 πππ 2 ππ₯ 2 2 0 π₯ ππ₯ π’ = ; ππ’ = 2 2 ππ.
2
π π
π 2
ππ πππ
π 4
ππ.
π ππ2 4π₯ πππ 2 2π₯ ππ₯
8
π 2
ππ.
π ππ6 π₯πππ 4 π₯ ππ₯
π
=
2β1 2β1 2+2 2+2β2
=4
1 2
π
=
6β1 6β3 6β5 (4β1)(4β3)( 2 ) (6+4)(6+4β2)(6+4β4)(6+4β6)(6+4β8)
=
ππ πππ
=
π 2
π ππ
2
ππ.
4 β π₯2
3 2
ππ₯ ; πππ‘ π₯ = 2π ππβ
0
ππ₯ = 2πππ β π ππβ π 2
ππ.
2
π ππ7 π₯
=
4 β 2π ππβ
=
ππ = ππ
(2πππ β πβ )
0
π (4β1)(7β3)(7β5) 7(7β2)(7β4)(7β6)
3
2 2
2
=
3
(4 πππ 2 β )2 2πππ β πππ β πβ
0 2
=
8 πππ 3 β 2πππ β πβ
0
=( =
4β1
4β3
4 4β2
π 2
ππ ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
55
EXERCISE 12.1
1. π¦ = 3π₯ 2 ;
AREA UNDER A CURVE
ππππ π₯ = 1 π‘π π₯ = 2
3. π₯π¦ = β1 ; ππππ π₯ = 1 π‘π π₯ = 2 π¦=β
1 π₯
2
π΄=
2
π¦ππ₯
π΄=
1
π΄=
2 3π₯ 2 ππ₯ 1 2 3
π΄= π₯
π΄= 2
π¦ππ₯ 1
1 3
β 1
π΄=
2 1 β ππ₯ 1 π₯ 2
π΄ = [β ππ π₯] 3
π¨ = π ππ. πππππ
1
π΄ = {[β ππ 2] β [β ππ 1]} π΄ = β ππ 2; ππ’π‘ π‘ππππ ππ ππ πππππ‘ππ£π ππππ, πππππ, π¨ = πππ ππ. πππππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
56
EXERCISE 12.1
AREA UNDER A CURVE
5. π¦ = 3πππ₯, π₯ = 2 π‘π π¦ = 4
9. π₯ + π¦ = 3 & π‘ππ πππππππππ‘π ππ₯ππ
π
ππ΄ =
π¦ππ₯
0 4
π΄= 3
πππ₯ππ₯ 2
= 3[π₯πππ₯ β π₯] = 3[4 ππ 4 β 4] β 3[2 ππ 2 β 2] = 3[4 ππ 4 β 4 β 2ππ 2 + 2] = 3[8ππ2 β 2ππ2 β 2] π΄=
= 3[6ππ2 β 2] = 6[3ππ2 β 1]
3 0
3 β π₯ ππ₯
π΄ = 3π₯ β
π¨ = π[πππ β π] ππ. πππππ
π₯3 2
π΄= 3 3 β π¨= 7. π¦ = 9 β π₯ 2 π΄=
3 β3
; π₯ = β3 π‘π π₯ = 3
3 0 3 2 2
π ππ. πππππ π
4 β π₯ 2 ππ₯
π¨ = π ππππππ πππππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
57
EXERCISE 12.1
AREA UNDER A CURVE
11. π¦ 2 = 4π₯, π₯ = 1 πππ π₯ = 4
ππ. π₯π¦ = 1, π¦ = π₯, π₯ = 2, π¦ = 0
4
π΄=
4π₯ππ₯ 1 4
π΄= π΄=
1
4π₯ 2 ππ₯ 1
π₯π¦ = 1; π¦ = π₯
8 3 π₯4 3
π₯(π₯) = 1
8(4)3/2 8(1)3/2 π΄= β 3 3 π΄=
64 3
β
8 3
ππ π¨= ππ. πππππ π
π₯=1 ; π¦=1 2
π΄1 = 1
; (1,1)
1 ππ₯ π₯
= (ππ π₯) = ππ 2 β ππ 1 π΄1 = ππ 2 π π. π’πππ‘π π΄2 = =
1 ππ 2
1 1 1 2
π΄2 =
1 π π. π’πππ‘π 2
π΄π‘ = π΄1 + π΄2 π π¨ = (ππ π + )ππ. πππππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
58
EXERCISE 12.2
AREA BETWEEN TWO CURVES
5. y = x 2 ; y = 2 β x 2
1. π¦ = π₯ 2 ; π¦ = 2π₯ + 3 π¦ = 2π₯ + 3 π₯ 2 = 2π₯ + 3 π₯ 2 β 2π₯ β 3 = 0 π₯β3 π₯+1 = 0 π₯ = 3, π₯ = β1 π΄=
3 β1
ππ¦ = 2π₯ ; (0,0) ππ₯ π₯ = 0 ,π¦ = 0 π2 π¦ = 2 (ππππππ£π π’ππ€πππ) ππ₯ 2
2π₯ + 3 β π₯ 2 ππ₯
= [π₯ 2 + 3π₯ β
π₯3 ] 3 3 -1
= 32 + 3(3) β
(3)3
πππππ‘ ππ πππ‘πππ πππ‘πππ: β (β1)2 + 3(β1) β
3
(β1)3 3
5
= 9+3 π¨=
3. π₯ 2 = π¦ β 1
(2π₯ + 2)(π₯ β 1) 2π₯ + 2 = 0π₯ β 1 = 0 2 2π₯ = β π₯ = 1 2
; π₯ =π¦β3
π₯ = β1 π¦ = 1
Y1=Y2 π¦β3 2 =π¦β1 π¦ 2 β 6π¦ + 9 = π¦ β 1 π¦β5 π¦β2 =0 π¦ = 5 ,π¦ = 2 π₯ =5β3=2 2
π₯2 = 2 β π₯2 π₯2 β 2 + π₯2 = 0
ππ ππ. πππππ π
π΄=
y1= y2
ππ΄ = [π1 β π2]ππ₯ 1 β1 1
=
π₯ + 3 β (π₯ 2 + 1) ππ₯
2 β1
=
π₯2 π₯3 + 2π₯ β 2 3
π₯ + 2 β π₯ 2 ππ₯
=
22 2
+ 2(2) β
=
10 3
+6 =A=
7
23 3
= 2π₯ β 2 3
2
2π₯ 3 3
β =
β1 2 2 π π
+ 2(β1) β
(β1)3 3
=
2
2
= 2 β 3 β [β2 + 3]
=2β +2β
-1
27 6
(2 β 2π₯ 2 ) ππ₯
β1
β1
=
(2 β π₯ 2 β π₯ 2 ) ππ₯
ππ΄ =
2 3
=
12β4 3
π ππ. πππππ π
ππ. πππππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
59
EXERCISE 12.2
AREA BETWEEN TWO CURVES
7. π¦ = π πππ₯ ; π₯ = πππ π₯ ; π₯ = x 0 90 180 270 360
y 0 1 0 -1 0
π΄2 =
x 0 90 180 270 360
π 2 π 4
πππ π₯ =
π 2
11. π¦ = π₯ 3 , π¦ = 8, π₯ = 0 ππ¦ = 3π₯ 2 ππ₯
y 1 0 -1 0 1
, 0 = 3π₯ 2
π¦ = 0 ,π₯ = 0 π2 π¦ = 6π₯(ππππππ£π π’ππ€πππ) ππ₯ 2 πππππ‘ ππ πππ‘πππ πππ‘πππ:
π πππ₯ππ₯ = [-πππ π₯]
y1= y2
π 4
π 2
= [-πππ ] β [-πππ ] = π΄1 =
π 4
π 2 π 4
2 2
π₯3 = 8
π π πππ π₯ππ₯ = π πππ₯ = π πππ₯ β π ππ 2 4
π₯3 β 8 = 0 π₯3 = 8 π₯=
2 = 1β 2
3
8
π₯=2
π¨π β π¨π = π β π ππ. πππππ
π€πππ π₯ = 2 π¦ = 8 , (2,8) π€πππ π₯ = β2
9. π₯ 2 = 4π¦ , π¦ = π¦=
π₯2
8 π₯ 2 +4
4
π₯ 2 π₯ 2 + 4 = 32 2 8 π₯2 π΄= β ππ₯ 2 4 β2 π₯ + 4 π΄ = 4.95
π¦ = β2
3
, π¦ = β8
(-2,-8) ππ΄ = [π1 β π2]ππ₯ 2
ππ΄ =
(8 β π₯ 3 ) ππ₯
0
= ππ β π = ππ ππ. πππππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
60
EXERCISE 12.2
AREA BETWEEN TWO CURVES
13. π¦ = 2π₯ + 1 , π¦ = 7 β π₯ , π₯ = 8
π
8
π΄=
15. π¦ = πππ₯ 3 , π¦ = πππ₯; π₯ = π
2π₯ + 1 β 7 β π₯ ππ₯
π΄= 1
2
π
8
=
2π₯ + 1 β 7 + π₯ ππ₯
=
8 2
3π₯ β 6 ππ₯
=
3π₯ 2 = β 6π₯ 2 =
3(8)2 2
π πππ₯ 3 1
π’ = πππ₯ 3
8 2
β 6(8) β
[(πππ₯ 3 ) β (πππ₯)]ππ₯
1
2
=
(π¦2 β π¦1 )ππ₯
ππ’ = 3(2)2 2
β 6(2)
= ππ₯ ; ππ’ = π
1
= ππ ππ πππππ
πππ₯
; π£ = π₯ ; π’ = πππ₯ ; ππ£ = ππ₯
3π₯ 2 ππ£ π₯3
= π₯πππ₯ 3 β
π 1
β
= π₯πππ₯ 3 β
3π₯ 2 π₯( 3 ) π₯ 3π₯
= π₯πππ₯ 3 β 3π₯
π 1
ππ₯ π₯
π β [π₯πππ₯ β 1
;
π£=π₯
ππ₯ π π₯( )] π₯ 1
β [π₯πππ₯ β π₯] π1
= π ππ. πππππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
61
EXERCISE 12.2
AREA BETWEEN TWO CURVES
17. π¦ 2 = 2ππ₯ , π¦ 2 = 4ππ₯ β π2
π€πππ π₯ = β4π
π¦ 2 = 2ππ₯π¦ 2 = 4ππ₯ β π2
π₯ = (β4π)2
π₯= ππ₯ ππ¦
π¦2 2π
;x=
π¦ 2 +π 2 4π
2π¦
=
16π2 2π
= 8π
= 2π
ππ₯ π¦ = ππ¦ π
π΄
π
ππ΄ =
0 = 0 ; (0,0)
π
[ βπ
π¦ 2 + π2 π¦ 2 β ]ππ¦ 4π 2π
π2 π₯ 1 = ππππ π‘π π‘ππ πππππ‘ ππ¦ 2 π
=
π π¦ 2 +π 2 βπ¦ 2 ( 4π )ππ§ βπ
πππππ‘ ππ πππ‘πππ πππ‘πππ:
=
π¦3 π2 π¦ 2π¦ 3 + β 12π 4π 12π
=
π3 π2 π 2π3 (βπ)3 π2 (βπ) 2(βπ)3 + β β + β 12π 4π 12π 12π 4π 12π
=
π3 β 2π3 + π3 β 2π3 π3 + π3 + 12π 4π
=
β2π 3 12π
X1 = X2 π¦ 2 π¦ 2 + π2 = 2π 4π 4ππ¦ 2 = 2ππ¦ 2 + 2π3 4ππ¦ 2 β 2ππ¦ 2 β 2π3 = 0 2ππ¦ 2 β 2π3 = 0 2ππ¦ 2 = 2π3 2π3 π¦2 = 2π 2 π¦ = π2 π¦ = π2 π¦ = Β±π π2
+
2π 3 4π
=
a -a
β2π 3 +6π 3 12π
4π 3
= 12π A=
a2 sq. 3
units
π
X1 = X2=2π = 2 π€πππ π₯ = 4π π₯= =
(4π)2 2π
16π2 2π
= 8π DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
62
EXERCISE 12.2
AREA BETWEEN TWO CURVES
ππ. π¦ 2 = π₯ + 1 ; π¦ = 1 β π₯
ππ. π¦ 2 = 4π₯ ; π¦ = 4π₯ β 4
π£1= π¦ 2 β 1; π¦π₯ = 1 ππ₯ = 2π¦ ; π₯2 = 1 β π¦ ππ¦ π₯ = 0; π¦ = 0 π2 π₯ = 2 (ππππππ£π π‘π π‘ππ πππππ‘) ππ¦ 2 πππππ‘ ππ πππ‘πππ πππ‘πππ 1 π₯1= π¦2 ; π¦ 2 β 1 = π¦ π¦2 + π¦ β 2 = 0 (π¦ β 1)(π¦ + 2) π¦β1=0 π¦+2=0
4π₯ = π¦ 2 2π₯ = π¦ + 4
y=1
π₯=
ππ₯ 1 = 2π¦ ππ¦ 4 1 0 = 2π¦ 4 0=0 0,0
y=-2
π£=0
π¦2 π¦+4 π₯= 4 2
π2π₯ ππ¦ 2
π¦=3
π€πππ π₯ = 1, π¦ = 2
= (concave to the right)
πππππ‘ ππ πππ‘πππ πππ‘πππ π¦2 π¦ + 4 = 4 2
π€πππ π₯ = 2, π¦ = 5 π€πππ π¦ = 1, π₯ = 0 π€πππ π¦ = 2, π₯ = 3 π€πππ π¦ = 3, π₯ = 8 π‘πππ;
2π¦ 2 β 4π¦ + 4(4) 2π¦ 2 β 4π¦ β 16 = 0
ππ΄ = π2 β π1 ππ¦ 1
1
1 β π¦ β π¦ 2 β 1 ππ¦
ππ΄ = β2
2π¦ β 8 π¦ + 2 2π¦ β 8 = 0π¦ + 2 = 0 π¦ = 4; π₯ = 4(1, 2)
β2
π΄ = 1 β π¦ β π¦2 + 1 π΄ = 2βπ¦β
1 β2
(4, 4)
π¦ 2 1β2
ππ΄ = (π₯2 β π₯1 )ππ¦
3 1
π΄ = 2π¦ β
π΄ = 2(1) β
π¦2 π¦ β 2 3
(1)2 (1)3 β 2 3
1
1
π΄= β2 β π΄ = 2(β2) β
(β2)2 (β2)3 β 2 3
4 π¦+4 β2 2
β
π¦2 4
ππ¦
π¨ = π ππ. πππππ
8
π΄=2β2β3+4+2β3 π¨=
π ππ. πππππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
63
EXERCISE 12.2
AREA BETWEEN TWO CURVES
23. π¦ 2 = π₯ + 4 , π₯ β 2π¦ + 1 = 0
3
π΄=
2π¦ β 1 β π¦ 2 β 4 ππ¦
β1 3
3 + 2π¦ β π¦ 2 ππ¦
= β1
= 3π¦ + π¦ 2 β =
3 π¦3 3 β1
ππ π
25. π¦ = π 2π₯ , π¦ = π , π₯ = 2
2
π΄=
π 2π₯ β π π₯ ππ₯
0
= =
π 2π₯ β ππ₯ 2 π4 2
2 0 1
β π2 β 2 + 1
= π π ππ β π
π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
64
EXERCISE 12.4
VOLUME OF A SOLID OF REVOLUTION
1. π¦ = π₯ 2 β 2π₯ , π₯ β ππ₯ππ , ππππ’π‘ π‘ππ π₯ β ππ₯ππ ππ¦ = 2π₯ β 2 , πππ’ππ‘π π‘π π§πππ ππ₯ 0 = 2π₯ β 2 ; π¦ = 12 β 2(1) π₯=1
;
π¦ = β1
π2π¦ =2 ππ₯ 2
π¦ = π₯ 2 β 2π₯
π£ 1, β1 x y
0 0
1 -1
2 0
3 3
1 -1
ππ£ = ππ¦ 2 ππ₯
dx
ππ£ = π π₯ 2 β 2π₯ 2 ππ₯
2
y (1,-1)
-2
Κππ£ = πΚ π₯ 4 β 4π₯ 3 + 4π₯ 2 ππ₯ π£=π =π
π₯5 5
β
1 2 5
4π₯ 4 4
5
+
4π₯ 3 3
β 24 +
4 2 3
=π
32 32 β 16 + 5 3
=π
96 β 240 + 160 15
=π
16 15
π½=
3
β 0
πππ ππππππ ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
65
EXERCISE 12.4
VOLUME OF A SOLID OF REVOLUTION
π. π₯ + π¦ = 5 ; π¦ = 0 ; π₯ = 0 ; ππππ’π‘ π¦ = 0
π. π₯ + π¦ = 6 ; π¦ = 3 ; π₯ = 0 ; ππππ’π‘ π¦ β ππ₯ππ
π€πππ π₯ = 0 ; π¦ = 5
π₯ = (6 β π¦)
π€πππ π¦ = 0 ;
π₯=5
ππ£ = ππ₯ 2 ππ¦
ππ£ = ππ¦ 2 ππ₯
; ππ’π‘ π¦ = 5 β π₯
ππ£ = π 6 β π¦ ππ¦
π¦2 = 5 β π₯
2
ππ£ = π 36 β 12π¦ + π¦ 2 ππ¦
ππ£ = π 5 β π₯ 2 ππ₯ π£
5
ππ£ = π 0
π£ 0
25 β 10π₯ + π₯ 2 ππ₯
π=π
10π₯ 2 π₯ 3 + 2 3
25 5 β 5 5
π = π 125 β 125 + π½=
2
125 3
0
π=
+
1 5 3
3
36 β 12π¦ + π¦ 2 ππ¦
π¦2 π¦3 π = π 36π¦ β 12 + 2 3
0
π = π 25π₯ β
3
ππ£ =
β 0
36 3 β 6 3
2
+
π = π 36 3 β 6 9 +
β0
1 3 3
2
β 0
1 3 27
π = π[ 36 3 β 6 9 + 9]
ππππ ππππππ π
π = π(9)(12 β 6 + 1) π = π(9)(7)
y
π½ = πππ ππππππ
π₯=0
π¦π₯ = 0 5
(0,6)
3 2 1 ππ₯
π₯+π¦ =6
5 y
π¦=3
3 3
5
x π¦=0
ππ¦ 1 0
(6,0) π₯ 1
3
5
π₯
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
66
EXERCISE 12.4
VOLUME OF A SOLID OF REVOLUTION
π. π₯π¦ = 4, π₯ = 2, π¦ = 4; ππππ’π‘ π¦ = 4
9. π¦ 2 = 4ππ₯, π₯ = π; ππππ’π‘ π₯ = π
π = ππ 2 π
2
π = ππ π π = π(4 β π¦)2 ππ₯
π = π(π β π₯)2 π π£
π = π 4β π£
4 π₯ 2
ππ£ = π 0
0
ππ₯ 32 16 (16 β + 2 ) ππ₯ π₯ π₯
2π
ππ£ =
2 0
β2π 2π
π= π
π(π β
(π2 β
β2π
16 π = π 16 2 β 32ππ2 β β0 2
π = π π2 π¦ β
π = 8π 4 β 4 ππ 2 β 1
π = π2 2π β
π½ = ππ π β π ππ π ππ. πππππ
π¦2 2 ) ππ¦ 4π
π¦2 π¦4 + ) ππ¦ 2 16π2
π¦3 π¦5 + 6 16 5 π2
2π3 2π5 β2π3 β2π5 + β π2 β2π β + 6 16(5)π2 16 16(5)π2 2
1
π = 4π3 π 1 β 3 + 5 π½=
ππππ π ππ. πππππ ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
67
EXERCISE 12.4
VOLUME OF A SOLID OF REVOLUTION
ππ. π¦ = π ππ π₯, π₯ = 0, π¦ = 1; ππππ’π‘ π¦ = 1
π = ππ 2 π π = π(1 β π¦)2 ππ₯ π 2
π£
π£= 0
π (1 β π ππ π₯ )2 ππ₯
0 π 2
π= π
0
1 β 2π πππ₯ + π ππ2 π₯ ππ₯
π = π[π₯ + 2 πππ π₯ +
π₯ π ππ2π₯ β ] 2 4
π= π
3π₯ π₯ π ππ 2π₯ + 2 πππ π₯ + β 2 2 4
π= π
3π₯ π ππ 2π₯ + 2 πππ π₯ β 2 4
π= π
3π + 0 β 4(0) β 0 + 2 + 0 4
π= π½=
3π 2 4
β 2π
π ππ β π ππ. πππππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
68
EXERCISE 12.5
THE WASHER METHOD 9
1. π¦ = π₯ 2 , π₯ = 3, π¦ = 0; ππππ’π‘ π‘ππ π¦ β ππ₯ππ
π=π
32 β π₯ 2 ππ¦
0 9
π=π
9 β π¦ ππ¦ 0
π = π 9π¦ β
π¦2 9 2 0
π = π 9(9) β π½=
(9)2 9 2 0
πππ πͺπΌπ©π°πͺ πΌπ΅π°π»πΊ π
3. π¦ 2 = 4ππ₯, π₯ = π; ππππ’π‘ π‘ππ π¦ β ππ₯ππ x 0 a
π¦ 2 = 4ππ₯
y 0 2a
dy x 2π
X=a
π2 β π₯ 2 ππ¦
π= π β2π 2π
π¦2 = π (π β 4π β2π 2π
= π
2
π2 β
β2π
2
)ππ¦
π¦4 ππ¦ 16π2
π¦5 2π = π π π¦β 80π2 β2π 2
= π (2π3 β
32π5 32π5 3 ) β (β2π + ) 80π2 80π2
= π (2π3 β
2π 3 )β 5
π½=
(2π3 +
2π 3 ) 5
πππ ππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
69
EXERCISE 12.5
THE WASHER METHOD
5. π₯ 2 +π¦ 2 = π2 , π₯ = π π π = 4π 0 π2 β π¦ 2 + π ππ¦ π = 4π π2 π¦ β π = 4π π = 4π π½=
π¦3 3
+ ππ¦
π3 π3 β 3 β 2π 3 β ππ 3
ππ
a
(-a,0)
o
(a,0)
x=b
ππ ππ π
7. π₯ 2 + π¦ 2 = 25 , π₯ + π¦ = 5 ; π¦ = 0 π= π π½=
5 0
25 β π₯ 2 β 5 β π₯
2
ππ₯
ππππ ππ. πππππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
70
EXERCISE 12.5
THE WASHER METHOD
9. π¦ 2 = 4π₯, π₯ 2 = 4π¦; ππππ’π‘ π‘ππ π₯ β ππ₯ππ
π¦2 = π¦1
2
π₯2 4π₯ = 4 3 64π₯ = π₯ π₯ = 4, π¦ = 4: πππΌ (4,4) 4
π=π
4π₯
2
β
0 4
π₯2 4
2
ππ₯
π₯4 ππ₯ 16 0 π₯5 4 π = π 2π₯ 2 β 80 0 (4)5 π = π 2(4)2 + 80 πππ π½= πͺπΌπ©π°πͺ πΌπ΅π°π»πΊ π π=π
11. π¦ 2 = 8π₯, π = 2π₯; ππππ’π‘ π¦ = 4
4π₯ β
π¦2 = π¦1 8π₯ = 2π₯
2
8π₯ = 4π₯ 2 π₯ = 2, π¦ = 4: πππΌ (2,4) 4π₯ 3 2 π = π 4π₯ 2 β 3 0 4(2)3 π = π 4(2) + 3 2
π½=
πππ πͺπΌπ©π°πͺ πΌπ΅π°π»πΊ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
71
EXERCISE 12.6
THE CYLINDRICAL SHELL METHOD
π. 4π¦ = π₯ 3 , π¦ = 0, π₯ = 2, ; ππππ’π‘ π₯ = 2
V = 2Ο
2 π₯π¦ππ₯ 0
V = 2Ο
2 0
V = 2Ο
2 π₯2 [ 0 2
V = 2Ο V = 2Ο V = 2Ο V=
2βπ₯
π₯4 4
β
β
(2)4 4
β
π₯3 4
dx
π₯4 ]ππ₯ 4
π₯5 20
2 0
(2)5 20
3 5
ππ cubic units π
2 0
3. π₯ = 4π¦ β π¦ 2 , π¦ = π₯ , ππππ’π‘ π¦ = 0
V = 2Ο
3 π₯π¦ππ¦ 0
V = 2Ο
3 0
V = 2Ο
3 0
V = 2Ο
4π¦ β π¦ 2 β π¦ π¦ππ¦ 4π¦ 2 β π¦ 3 β π¦ 2 ππ¦
4 2 π¦ 3
V = 2Ο π¦ 3 β
π¦4 4
V = 2Ο (3)3 β V=
πππ πππππ π
1 4 π¦ 4
β
1
β 3 π¦3
3 0
(3)4 4
3 0
πππππ
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72
3 0
EXERCISE 12.6
THE CYLINDRICAL SHELL METHOD
π
5. π¦ = π πππ₯, π¦ = πππ π₯, π₯ = 2 π = 2π π½=
π 2 π 4
π₯ π πππ₯ β πππ π₯ ππ₯
π π + ππ β ππ ππ. πππππ π
7. π₯ = 2 π¦ π = 2π π½=
9 0
π 2
,π₯ = 0 ,π¦ = 0
Y=9
9 β π¦ 2 π¦ ππ¦
πππππ ππ. πππππ π
9.π¦ = πππ₯ , π₯ = π , π¦ = 0 π = 2π
π π₯ 1
πππ₯ ππ₯ (e,1)
π½ = ππ. ππ ππ. πππππ
(1,0) X=e
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73
EXERCISE 12.6
THE CYLINDRICAL SHELL METHOD
11. π¦ 2 = 8π₯ , π₯ = 0 , π¦ = 4 ; about π¦ = 4
4
π = 2π
4βπ¦ 0
=
π 4
4
π¦2 ππ¦ 8
4π¦ 2 β π¦ 3 ππ¦
0
=
π 4π¦ 3 4 3
=
πππ π
β
4 π¦4 4 0
13. ( π₯ β 3 ) 2 + π¦ 2 = 9; ππππ’π‘ π‘ππ π¦ β ππ₯ππ . 3
π = 8π
2 x 9 ο ( x ο 3) dx
0 π = 8π(
β ( 9 β ( x β 3 ) 2 3 27 π πππ₯ β 3 9 )2 + + (π₯ β 3)( 9 β π₯ β 3 3 2 3 2
π = 8π(27π πππ β
2
3 0
27 π ππ β 1) 2
27 π = 8π( )(βπ ππ β 1 + π πππ) 2 π π = 108π( ) 2 π½ = πππ π
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74
EXERCISE 12.6
THE CYLINDRICAL SHELL METHOD
15. π₯ 2 + π¦ 2 = π2 ; ππππ’π‘ π₯ = π π > π π
π= π
πβπ₯ βπ π
π= π
2
β πβπ₯
2
ππ¦
π 2 β ππ₯ + π₯ 2 β π 2 β ππ₯ + π₯ 2 ππ¦
βπ π
π= π
4ππ₯ππ¦ βπ
πππ‘π: π₯ 2 + π¦ 2 = π2 π = 4ππ π = 4ππ
π βπ
π¦ 2 β π2
=π₯=
π¦ 2 β π2 ππ¦
π¦ π2 π π¦ 2 β π2 β ln π¦ + π¦ 2 β π2 + π βπ 2 2
π½ = ππ π ππ π π₯ 2 + π¦ 2 = π2
a
a
a
a
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75
EXERCISE 12.7
π.
VOLUME OF SOLIDS WITH KNOWN CROSS SECTIONS
π₯ 2 + π¦ 2 = 36
π. 9π₯ 2 + 16π¦ 2 = 144
π2 π΄ π₯ = , 2
π = 2π¦
π΄ π₯ = 2π¦ 2 ,
π¦=
36 β π₯ 2
6
π£=
π΄ π₯ ππ₯ β6
1 π΄ π₯ = (2π¦)(π¦) 2 π΄ π₯ = π¦2 8
0
2π₯ 2 ππ₯
π£=
π¦ 2 ππ₯
π=2
6
π=2
β6
π£=
1 π΄ π₯ = ππ 2
6 2(3π₯ β6
β π₯ 2 )ππ₯
8 144β9π₯ 2 0 16
ππ₯
π½ = ππ ππ. πππππ
π = πππ ππ. πππππ
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EXERCISE 12.7
VOLUME OF SOLIDS WITH KNOWN CROSS SECTIONS
π.
π΄ π¦ = (1 β π₯)(2π¦ 2 ) 2
π = 2 (1 β π₯ )2π¦ 2 ππ¦ 0 2
π = 2 (1 β 0
π¦2 2 )π¦ ππ¦ 4
64
π = 15 π½ = π. ππππ ππ. πππππ
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77
EXERCISE 12.8
LENGTH OF AN ARC
3
π. π¦ = π₯ 2 ππππ π₯ = 0 π‘π π₯ = 5
2
2
3
π¦ = π₯2 3 1 ππ¦ = π₯ 2 ππ₯ 2 ππ¦ 3 1 = π₯2 ππ₯ 2 π = 0
π=
=
ππ¦ 1 + ( )2 ππ₯ ππ₯
=
5 0
1+
1+ β 0
2
π=
π₯
0
π¦3 1
π₯3
2
ππ₯
1
π₯3 + π¦3
2
5 0
1
9
9 5
2
3. π‘ππ πππ‘πππ ππ¦ππππ¦πππππ π₯ 3 + π¦ 3 + π3
2 3
2
ππ₯
2
πππ‘π: π3 = π₯3 + π¦3
3 1 ( π₯ 2 )2 dx 2
2
9
9 4
π3
π=
1 + π₯ ππ₯
2
π₯3
0
π = ππ. πππ πππππ
9
π= 0
ππ₯
1
π3 1
π₯3
ππ₯ 2
3π₯ 3 9 π= π 2 0 1 3
π=
3π 2
π=4
3π 2
πΊ = ππ
X=0
x=5
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EXERCISE 12.8
LENGTH OF AN ARC
5. π¦ = π΄πππ πππ π₯ , ππππ π¦ =
π 6
π‘π π¦ =
π 2
π¦ = π΄πππ πππ π₯ ; ππ π πππ¦ = π₯ 1 πππ π¦ππ¦ = ππ₯ π πππ¦
7. πππ ππππ ππ π‘ππ ππ¦πππππ π₯ = π π β π πππ , π¦ = π(1 β πππ π) π₯ = π(π β π πππ) ππ₯ = π(ππ β πππ πππ)
π¦ = π(1 β πππ π) ππ¦ = π(π πππππ)
ππ₯ ππ
ππ¦ ππ
= π(1 β πππ π)
ππ₯ πππ π¦ = ππ¦ π πππ¦
2π
π =
ππ₯ = πππ‘π¦ ππ¦ π= =
π 2 π 6
π 2 π 6
2
+ π2 sin2 π
0
π =π 1+
π2 1 β πππ π
= ππ πππ
ππ₯ 2 ππ¦
ππ¦
2π 0
1 β πππ π
2
+ sin2 π
π = ππ
1 + πππ‘ 2 π¦ ππ¦
πΊ = π. πππππ πππππ
9. πππ πΆπππππππ π = 2 1 β πππ π
π 2
π = 2 1 β πππ π ππ = 2 π πππ ππ ππ = 2π πππ ππ π 2 = 4(1 β πππ π)2 2π
π=
4(1 β πππ π)2 + 4π ππ2 π ππ
0
π=2 π 6
2π 0
(1 β πππ π)2 + π ππ2 π ππ
πΊ = ππ πππππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
79
EXERCISE 12.9
AREA OF A SURFACE OF REVOLUTION
1. π₯ 2 + π¦ 2 = 16 ; ππππ π₯ = 2 π‘π π₯ = 4
3. π¦ 2 = 12π₯ ; ππππ π₯ = 0 π‘π π₯ = 3
4
π = 2π
3
π¦ππ
π = 2π
2
π¦=
π¦ = 12π₯
16 β π₯ 2
ππ¦ 1 = 16 β π₯ 2 ππ₯ 2
1 2
β
ππ¦ 1 = 12π₯ ππ₯ 2
(β2π₯)
2
ππ =
ππ¦ 1+ ππ₯
ππ =
π₯2 1+ ππ₯ 16 β π₯ 2
ππ =
16 β + 16 β π₯ 2
16 β π₯ 2
π = 2π
π₯2
(12)
ππ =
16 β π₯ 2
36 ππ₯ 12π₯
12π₯ + 36
ππ =
12π₯ 2 3π₯ + 9
ππ₯
12π₯
ππ₯
ππ₯
3
12π₯
2 3π₯ + 9
0
ππ₯
2
π = 2π
1+
π = 2π
4 4
ππ =
ππ₯
π₯2
1 2
ππ¦ 6 = ππ₯ 12π₯
ππ¦ π₯ =β ππ₯ 16 β π₯ 2
ππ =
π¦ππ 0
4 16 β π₯ 2
π = 4π ππ₯
3 0
12π₯
ππ₯
3π₯ + 9 ππ₯
πΊ = πππ. πππ ππ. πππππ
4 4ππ₯ 2
πΊ = πππ ππ. πππππ
5. π¦ = π₯ 3 ; ππππ π₯ = 0 π‘π π₯ = 1 ππ¦ = 3π₯ 2 ππ₯ ππ = π = 2π
1 + 9π₯ 4 ππ₯ 1 3 π₯ 0
1 + 9π₯ 4 ππ₯
πΊ = π. ππππ ππ. πππππ
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80
EXERCISE 12.9
AREA OF A SURFACE OF REVOLUTION
7. π₯ = πππ 2π¦ ; ππππ π¦ = 0 π‘π π¦ = π 4
π = 2π
π 4
π₯ππ
0
ππ₯ = β π ππ 2π¦(2) ππ¦ ππ = π = 2π
1 + 4 π ππ2 2π¦ π 4
0
πππ 2π¦ 1 + 4 π ππ2 2π¦ ππ¦
πΊ = π. πππππ ππ. πππππ
9. 4 β π₯ 2 ππππ π₯ = 0 π‘π π₯ = 2 2
π = 2π
π₯ππ 0
ππ¦ = β2π₯ ππ₯ ππ = 1 + 4π₯^2 ππ₯ π = 2π
2 π₯ 0
1 + 4π₯ 2 ππ₯
πΊ = ππ. ππππ ππ. πππππ
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81
EXERCISE 12.9
AREA OF A SURFACE OF REVOLUTION
13. π¦ = ππ₯ ; π₯ = 0 ; π₯ = 1 ; ππππ’π‘ π‘ππ π₯ β ππ₯ππ 1
π = 2π
ππ₯
1 + π2
ππ₯
0 1
π = 2π π
1 + π2 =
π₯ ππ₯ 0
π = 2π π π = 2π π
1 + π2( π₯2/2 )
1 0
1 + π2( Β½ )
πΊ = π π π + ππ
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82
EXERCISE 13.1
1.
FORCE OF FLUID PRESSURE
πΉ = π€π΄π₯ = (62.5ππ/ππ‘ 3 )(96ππ‘ 2 )(4ππ‘) = 24000ππ
π=
πΉ π΄
π=
π€π΄π₯ π΄
3. πΉ = π€π΄π₯ 1 πΉ=π€ 5 3 2
2 2 + ( )(3) 3
π = πππ ππ
2 3
π = π€π₯
5
62.5ππ3 1ππ‘ 2 π=( )(4ππ‘)( ) ππ‘ 144ππ 2 π=
3 5
(625)(4) 144
5 3
π· = π. ππ πππ
5
5. πΉ = 50π€
12ft
πππ π = 3ππ‘ 8ft
π₯
πΉ = π€π΄π₯ 50 =
1 π 3 2
50 =
π2 2
1 π 3
100 = π2 π = ππππ
3 5
h 5
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83
EXERCISE 13.1
FORCE OF FLUID PRESSURE
7. πΉ = π€π΄π₯ = π€[(π)(3)(2)](2) π = πππ π π = 6 = major axis π = 4 = πππππ ππ₯ππ
0
y π₯
b a
A=πππ
x
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
84
EXERCISE 13.2
WORK
1.
π
π€=
π π₯ ππ₯ π
π π₯ = ππ₯ 40 ππ = π π€=
;
π€ππππ π₯ =
1 ππ‘, 2
π π₯ = 40 ππ ; π = 0,
π = 14 β 10 = 4
1 ππ‘ , π = 80 2
4 80π₯ππ₯ 0
π = πππ ππ β ππ
3.
π
π€=
π π₯ ππ₯ π
π π₯ = ππ₯ ; π€ππππ π₯ = π€=
1 πΏ ππ‘, 10
π π₯ = 5 ππ
π = 0, π = πΏ
πΏ 50 π₯ππ₯ 0 πΏ
π = πππ³ ππ β ππ
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85
EXERCISE 13.2
WORK
5. π = πΉπ ππ€ = π€ ππ£ 60 β π₯ ππ€ = ππ 2 π€ 60 β π₯ ππ₯ ππ€ = 9ππ€(60 β π₯)ππ₯ π€
10
ππ€ = 9ππ€ 0
60 β π₯ ππ₯ 0
π€ = 9ππ€ 60π₯ β π₯ 2 π€ = 9ππ€ 60π₯ β
10 0
π₯ 2 10 2 0
π€ = 9ππ€ 600 β 50 π = ππππππ ππ. ππ
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86
EXERCISE 13.2
WORK
9.
π
π€=π€
πππ π
πππππ‘πππππ = π π₯ π€ π₯ π; π€ππππ π = 10 ππ‘, π€ = 2π₯, π = ππ¦ π₯2 + π¦2 = π2 ; π₯ =
π 2 β π¦ 2 ; π€ππππ π = 2
2
π€=π
6βπ¦
10 ππ‘ 2π₯ ππ¦
β2
2
π€ = 20π
6βπ¦
22 β π¦ 2 ππ¦
β2
π = ππππ π ππ β ππ
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87
EXERCISE 13.3
FIRST MOMENT OF A PLANE AREA
1. π¦ 2 = 4π₯, π‘ππ π₯ β ππ₯ππ πππ π₯ = 4 ππ₯ =
5. π¦ 2 = 4π₯ πππ π₯ 2 = 4π¦
1 4 4π₯ππ₯ 2 0
π΄π = ππ ππ¦ =
4 π₯ 0
4π₯ ππ₯
π΄π = ππ. π
π¦2 = 4π₯
4π₯ =
π₯4 16
64π₯ β π₯ 4 = 0
π₯=4
π₯ 64 β π₯ 3 = 0 π₯1 = 0, π₯2 = 4 1
ππ = 2 3. π₯ = 4
π΄π =
4 π
4π₯ β
π₯4 16
ππ¦
ππ π
π
ππ = ππ = ππ = ππ = π΄π =
πππ΄ π 4
π₯ 4 β π₯ ππ¦ π 4 π 4
4π₯ β π₯ 2 ππ¦ π¦2 β
π
π¦4 ππ¦ 16
πππ ππ DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
88
EXERCISE 13.3
FIRST MOMENT OF A PLANE AREA
7. π =
3 0
3βπ¦ [
=
3 0
3βπ¦
27π β9β9 4 27π π= β 18 4 27πβ72 π= 4
9 β π¦ 2 β 3 β π¦ ππ¦
π=
3 + π¦ 3 β π¦ β (3 β π¦)2 ππ¦
3 1 3 (3 β π¦)2 (3 + π¦)2 β (3 β π¦)2 ππ¦ 0 3 3 3 = 0 3 9 β π¦ 2 ππ¦ β 0 π¦ 9 β π¦ 2 ππ¦ β 0 (3 β π¦)2 ππ¦
=
3 9 0 9βπ¦ 2 3
*π΄ =3 πππ π =
β π¦2
ππ¦
π [ππ β π] π
π¦
π πππ = 3
9 β π¦2
3πππ π =
π΄=
π¦
3π πππ = π¦; π = ππππ ππ 3
3πππ πππ = ππ¦ π π¦ = 3; π = 2
π. π₯ = 4π¦ β π¦ 2 , π¦ = π₯
π¦ = 0; π = 0 =3 = 27
π 2
0
π 2
3πππ ππππ πππ
0
= 27
π 2
πππ 2 πππ = 27
0
π
1 + πππ 2π ππ 2
π π ππ2π 2 π 27π + = 27 = 2 4 4 4 0 3
*π΅ = β
π¦ 9 β π¦ 2 ππ¦
0
π’ = 9 β π¦2 ππ’ = β2π¦ππ¦ β = = =
1 2
ππ’ 2
9βπ¦ 2 3 2
= π¦ππ¦ | 30 3 2
1 2 ( )[(9 β 9) β 2 3 1 β27 = β9 3
*πΆ =β
@ π¦ = 3; π’ = 0 π¦ = 0; π’ = 9
3 (3 β 0
3 2
π
ππ¦ =
1 2
ππ¦ =
1 3 2 0
π΄π =
ππ π
(9 β 0) ]
π
π₯π2 β π₯π2 ππ¦ 4π¦ β π¦ 2
2
β π¦ 2 ππ¦
π¦)2 ππ¦
π’ =3βπ¦ ππ’ = βππ¦ = =
(3βπ¦)3 3 |0 3 3 0 3 β 3 3
= β9
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EXERCISE 13.4
CENTROID OF A PLANE AREA
1. π₯ + 2π¦ = 6, π₯ = 0, π¦ = 0 Solving for A ππ΄ = π¦ππ₯ 6 ππ΄ 0
=
6 0
π΄ = [3π₯ β
π₯
3 β 2 ππ₯
π₯2 6 ] 4 0
π΄= 3 6 β
36 4
π¨ = π ππ. πππππ
Solving for π₯
Solving for π¦
π΄π₯ =
6 ππ 0
ππ΄
π΄π₯ =
6 π₯ 0
3β
π₯2 2
ππ₯
π΄π¦ = 2
π΄π₯ =
6 0
3π₯ β
π₯2 2
ππ₯
π΄π¦ =
3π₯ 2 2
π΄π₯ = [
β
π΄π¦ =
6 ππ 0 1
π₯3 6 ] 3 0
ππ΄
6 π₯ π₯ (3 β 2 ) (3 β 2 )ππ₯ 0
1 6 (9 β 2 0 1
3
3π₯ +
π₯2 )ππ₯ 4 π₯3
π΄π¦ = 2 [9π₯ β 2 π₯ 2 + 12 ] 60 1
9π₯ = 18
π¦ = 3 (3)
π = π πππππ
π = π ππππ
Centroid: (2,1)
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EXERCISE 13.4
CENTROID OF A PLANE AREA
3. π¦ = π πππ₯, π¦ = 0 ππππ π₯ = 0 β π
A= π¦ππ₯ =
π 0
π πππ₯ππ₯
= βπππ π₯ A=2 π¦
ππ₯ = =
1
π 0
1
π 0
=2
π₯πππ; π₯π = π₯
=
π 0
π¦ 2 ππ₯
=
π 0
π ππ2 π₯ ππ₯
π’ = π₯ ; ππ£ = π πππ₯
1 π 1βπππ 2π₯ ( ) ππ₯ 2 0 2 1 π₯
= 2 (2 β 2 1 π₯
= 2 (2 β
π ππ 2π₯ 2
π₯π¦ππ΄ π₯π πππ₯ππ₯
ππ’ = ππ₯ ; π£ = βπππ π₯ ππ₯
= βπππ π₯ β βπππ π₯ππ₯
π ππ 2π₯ ) 4
= [βπ₯πππ π₯ + π πππ₯]
π 4
ππ₯ = (2) π₯ =
π 0
ππ¦ =
π π¦ ( )π¦ππ₯ 0 2
=2
=
π¦πππ; π¦π = 2
= βππππ π + π πππ + 0 β π ππ0
π 2
=π π
π¦ = ( 4 )(2) =
Centroid:
π 8
π π , π π
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EXERCISE 13.4
CENTROID OF A PLANE AREA
7. π¦ 2 = π₯ 3 , π¦ = 2π₯ 4
π΄=
3
(2π₯ β π₯ 2 )ππ₯ 0
2 5 π΄ = [π₯ 2 β π₯ 2 ] 5 π΄ = [16 β π΄=
64 ] 5
16 π π. π’πππ‘π 5
4
π΄π₯ =
π¦π₯ππ₯
π΄π¦ =
0 4
π΄π₯ =
3
(2π₯ β π₯ 2 )π₯ππ₯
π΄π¦ =
0 4
π΄π₯ =
5
(2π₯ 2 β π₯ 2 )ππ₯
π΄π¦ =
0
2 2 7 π΄π₯ = [ π₯ 3 β π₯ 2 ] 3 7 π₯ =
π΄π¦ =
7 5 2 2 [ (4)3 β (4)2 16 3 7
π₯=
5 128 257 [ β ] 16 3 7
π₯=
40 π’πππ‘π 21
π¦=
πͺπππππππ :
1 2 1 2 1 2
4
π¦ 2 ππ₯
0 4
5
[(2π₯)2 β π₯ 2 ]ππ₯
0 4
(4π₯ 2 β π₯ 3 ) ππ₯
0
1 4 3 π₯4 4 π₯ β 2 3 4 0
10 π’πππ‘π 3
ππ ππ , ππ π
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EXERCISE 13.4
CENTROID OF A PLANE AREA
9. π₯ 2 + π¦ 2 = 25,
π₯+π¦ =5
25π β 50 4 25 π΄= (π β 2) 4 π΄=
5
ππ¦ = 5
=
25 β π₯ 2 β 5π₯ ππ₯
π₯ 0
π₯ 25 β
π₯ 2 ππ₯
5
β
0
5 x 25 β π₯ 2 β 5π₯ ππ₯
π΄= π΄=
0 5 0
25 β π₯ 2 ππ₯ β 5
π΄βΆ
5 0
A 25 β π₯ 2 ππ₯
5 ππ₯ 0
+
B
25 β 5 π₯ππ₯ 0
C
25 β π₯ 2
5 cos π =
π₯ 5 sin π = π₯ ; π = arcsin 5 π 5 cos π = ππ₯ @π₯ = 5 ; π = 2 π₯ =0; π =0 =
π 2
5 cos π β 5 cos π
0
= 25
π 2
0
1 = 25 2
1 + πππ 2π cos πππ β cos π = 2 2
π 2
0
1 ππ + 2
2
π 2
0
β/2 π +0 4 0 25π = 4 5 π΅ βΆ β 5 0 ππ₯ 5 = β5π₯ 0 = β25
= 25
π₯2
πΆβΆ 2 25 = 2
5 0
25π 25 β 25 + 4 2 25 π΄ = 25π β 2 β΄π΄=
cos 2πππ
π₯2
π’ = 25 β π₯ 2 ππ’ = β2π₯ππ₯ =
1 25βπ₯ 2 2 β2 3
β
2
3
=β
25βπ₯ 2 2
5π₯ππ₯ + 0
3
5
5
5π₯ 2 2
+
π₯3 3
5
5π₯ 2 2
π₯ 2 ππ₯
0
5 0
π₯3 3 0
β + 0 125 125 125 = β β + β β β0+0 3 2 3 3 125 250 375 + 500 =β + = β 2 3 6 125 ππ¦ = 6 2 1 5 ππ₯ = 25 β π₯ 2 β 5π₯ 2 ππ₯ 2 0 5 1 1 5 = 25 β π₯ 2 ππ₯ β 5 β π₯ 2 ππ₯ 2 0 2 0 1 π₯3 π₯3 = 25π₯ β β 1/2 25π₯ + β 5π₯ 2 3 3 =
3
1 125 1 125 125 β β 125 + β 125 2 3 2 3
β 0
125 6 β΄ ππππ£πππ πππ ππππ‘ππππ: ππ₯ =
ππ₯ π₯= = π΄ π₯=
125 6 25 πβ2 4
=
125 4 β 6 25 π β 2
10 3 πβ2
ππ₯ π¦= = π΄ π¦=3
10 πβ2
125 6 25 πβ2 4
πͺπππππππ ππ ππ
ππ ππ , π π βπ π π βπ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
2
93
EXERCISE 13.5
CENTROID OF A SOLID OF REVOLUTION
1. π¦ 2 = π₯ ; π¦ = 3 ; π₯ = 0 ; ππππ’π‘ π‘ππ π¦ β ππ₯ππ
π. π₯ 2 π¦ = 4, π₯ = 1, π₯ = 4, π¦ = 0 ππππ’π‘ π₯ β ππ₯ππ π
3βπ¦
ππ₯π§ =
ππππ£ π
π¦
π¦2 = π₯ ππ₯
4
π¦ 2π π₯π¦ππ₯ = 2 2
= 2π 1 4
=π 1
= 16π ππ₯π§ =
ππ ππ
;
1
=
π₯ π¦ 0 0 9 3
4
π¦ π₯ππ₯ = π 1
1
ππ₯ π₯ β2 = 16π π₯3 β2
4
1
2
4 π₯2
2
16 π₯ 4 π₯ππ₯ = 16π π₯ 4
ππ₯π§ π¦= π
4
π₯ππ₯
π₯ π₯ππ₯ π₯4
4
β8π π₯2
= 1
4 1
15π 2 π
π 2 ππ₯
π=π 9
π = 2π
π₯π¦ππ₯ = 2π 0
ππ ππ = 2π
ππ₯π§ = 381.70
4
π₯ 3 β π¦ ππ₯ 0
9
ππ₯π§ =
π
9
0
3+π¦ 2
4 π₯2
=π 1
π₯
π₯ ππ₯
π₯ 1
ππ₯π§ 381.70 π¦= = π 152.68 π¦ = 2.5 π, π. π, π
ππ₯ = π 1
4
= 16π
π=
21π 4
π¦=
ππ₯π§ π
= 0,
10 , 7
=
4
2
β4
16 ππ₯ = 16π π₯4
π₯ β3 ππ₯ = 16π β3
4
= β 1
4
1
ππ₯ π₯2
16π 3π₯ 3
4 1
15π 2 21π 4
0
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
94
EXERCISE 13.5
π. π₯ 2 = 4π¦ x 0 1 2 4
,
4
4π₯
2
0 4
=π 0
=
π₯2 β 4
y 0 1 2 4
2
ππ₯
π₯4 4π₯ β ππ₯ 16
96π 5 4
ππ₯π§ = 2π 0
=
π¦ 2 = 4π₯ ππππ’π‘ π₯ β ππ₯ππ x 0 1/4 1 4
y 0 ΒΌ 1 4
=π
CENTROID OF A SOLID OF REVOLUTION
4π₯ + 2
π₯2 4
π₯
4π₯ β
π₯2 ππ₯ 4
128π 3
ππ₯π§ π
=π¦=
π = π,
128 π 3 96π 5
ππ ,π π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
95
EXERCISE 13.5
CENTROID OF A SOLID OF REVOLUTION
ππ. π¦ 2 = 4π₯, π¦ = π₯ ππππ’π‘ π₯ = 0
X 0 1/4 1 4
Y 0 1 2 4
X 1 2 3 4
Y 1 2 3 4
4
π = 2π
π( 4π β π)ππ₯ 0
π = 26.80829731 ππ’. π’πππ‘π ππ₯π§ = 2π
π¦π π₯ππ₯ 4
ππ₯π§ = 2π ( 0
4π₯ + π₯ )π₯ 2
4π₯ β π₯ ππ₯
ππ₯π§ = 64π/3 π¦=
ππ₯π§ π
= 2.5
y=(0, 2.5, 0)
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
96
EXERCISE 13.6
MOMENT OF INERTIA OF A PLANE AREA
1. 2π₯ + π¦ = 6 , π₯ = 0 , π¦ = 0 ; ππππ’π‘ π₯ β ππ₯ππ
5. π₯ = 2 π¦ , π₯ = 0, π¦ = 4
dy
x
y
4-y
(4,4)
dx π₯ 0 3
π¦ 6 0
πΌπ₯ =
6 2 π¦ π₯ππ¦ 0
= 6
6 2 6βπ¦ π¦ 0 2
π₯ 0 4
ππ¦
1 6π¦ 2 β π¦ 3 ππ¦ 2 0 1 π¦4 = 2π¦ 3 β 2 4
π¦ 0 4
=
1 = 2 6 2
6 3 β 4
4
πΌπ¦ =
0 4
4
π 2 ππ΄
=
π₯ 2 4 β π¦ ππ₯
0
= ππ
=
3
3. π¦ = π₯ , π₯ = 8 , π¦ = 0 ; π€ππ‘π πππ ππππ‘ π‘π π¦ = 0
4 2 π₯ 0
π°π =
π₯2
4β2
ππ₯
πππ ππ
dy x
πΌπ₯ = =
2 2 π¦ π₯ππ¦ 0 2 5
π¦ ππ¦
0
π°π =
= =
2 2 3 π¦ (π¦ )ππ¦ 0 6 6
π¦ 6
=
2 6
ππ π DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
97
EXERCISE 13.6
MOMENT OF INERTIA OF A PLANE AREA
7. π¦ 2 = 8π₯ , π¦ = 2π₯
9. π¦ = 4π₯ 2 , π¦ = 4π₯ ; π€ππ‘π πππ ππππ‘ π‘π π¦ β ππ₯ππ π¦ = 4π₯ 2 π¦ = 4π₯
X1 X2
(1,4) dy
dx
y (0,0)
π₯ 0 1
π₯ π¦ π₯ π¦ 0 0 0 0 1 2 2 1 2 2 4 2 4
π¦ 0 4
π
πΌπ¦ = 4
πΌπ₯ =
0
π¦ 2 (π₯π β π₯π ) ππ¦
4
πΌπ₯ = πΌπ₯ = π°π =
0
2
π¦ π¦ π¦ 2 ( β ) ππ¦ 2 8
4 π¦3 ( 0 2
β
π₯ π¦ 0 0 1 4
πΌπ¦ = Iy =
π
π₯ 2 (π¦π’ β π¦π ) ππ₯
1 2 π₯ (4π₯ 0
β 4π₯ 2 ) ππ₯
1 5
π¦4 ) ππ¦ 8
ππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
98
EXERCISE 13.6
MOMENT OF INERTIA OF A PLANE AREA
11. π¦ 2 = 8π₯ , π₯ = 0 , π¦ = 4 , with respect to π¦ = 4
13. π¦ = π₯ , π¦ = 2π₯ , π₯ + π¦ = 6, π€ππ‘π πππ ππππ‘ π‘π π₯ = 0 π₯+π¦ = 6
(6 β π₯ β 2π₯) 6 β 3π₯ =
4
πΌπ₯ = =
1 8
4βπ¦
2
0 4
π¦2 ππ¦ 8
16π¦ 2 β 8π¦ 3 + π¦ 4 ππ¦
0
=
1 16π¦3 8 3
=
ππ ππ
β 2π¦ 4 +
4 π¦5 5 0
π₯ π¦ 0 0 1 1 2 2
π¦ = 2π₯
π₯ π¦ 0 0 1 2 2 4
π₯ 0 1 2
π¦ 0 5 4
π
π°π = π°π = π°π =
π
ππ ππ β ππ π π
π π π π
πβπ β
π π
π π
ππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
99
EXERCISE 13.7
MOMENT OF INERTIA OF A SOLID OF REVOLUTION
1. π¦ = 2 π₯ , π¦ = 0 , π₯ = 4 ;about π₯ = 0
3. ππ₯ + ππ¦ = ππ , π₯ = 0 , π¦ = 0 ;about the y-
axis
4
πΌπ¦ = 2π 4
= 4π
π₯ 3 2 π₯ β 0 ππ₯
0
π₯
7
2 ππ₯
=
0 9
2π₯ = 4π 9
= 4π
= 4π =
2 4 9 1024 9
πππππ π
2
π
πΌπ¦ = 2π
4
= 0
9
2
β
2 0 9
9
2
4
2ππ π
0 π
π₯ 3 π β π₯ 4 ππ₯
0
2ππ π π
π
π
π₯ 3 ππ₯ β
0
2ππ π₯4 = π π 4
0
ππ β ππ₯ β 0 ππ₯ π
π₯3
π₯ 4 ππ₯
0 π 0
=
2ππ π5 π5 β π 4 5
=
2ππ π
=
ππ ππ ππ
π₯5 β 5
π 09
π5 20
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
100
EXERCISE 13.7
MOMENT OF INERTIA OF A SOLID OF REVOLUTION
5. 2π₯ + 3π¦ = 6 , π₯ = 0 , π¦ = 0 ; about the x-
9. π₯π¦ = 4 , π¦ = π₯ , π¦ = 1 ; aboutπ¦ = 0
axis
πΌπ₯ =
3
π 2
4
6 β 2π₯ 3
0
β 0 ππ₯ 2
3 16π₯ 4 β192π₯ 3 +864π₯ 2 β1728π₯+1296
=
π 2 0
=
πππ π
81
ππ₯
πΌπ₯ = 2π 2
= 2π
π¦3
1
4 β π¦ ππ¦ π¦
4π¦ 2 β π¦ 4 ππ¦
1
7. π¦ 2 = 3π₯ , π¦ = π₯ ; aboutπ₯ = 0
π¦3 = 2π 4 3 = 2π =
3
πΌπ¦ = 2π
28 3
β
2
π¦5 β 5 1
2 1
31 5
πππ ππ
π₯ 3 π₯ 3 β π₯ ππ₯
0
3
= 2π
π₯
7
3 β π₯ 4 ππ₯
2
0 3
= 2π
3
π₯
7
0
= 2π 54 β =
3 2 ππ₯
β
π₯ 4 ππ₯
0
243 5
πππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
101
EXERCISE 13.7
MOMENT OF INERTIA OF A SOLID OF REVOLUTION
11. π¦ = π₯ 2 , π¦ = 2π₯ ;about the y-axis
2
πΌπ¦ = 2π 2
= 2π
π₯ 3 2π₯ β π₯ 2 ππ₯
0 1
2π₯ 4 β π₯ 5 ππ₯
πΌπ¦ = 2π
0 2
= 2π 2
π₯ 4 ππ₯ β
0
=
64 5
β
2 0
π₯5 = 2π 2 5 = 2π
13. π¦ = π₯ 3 , π₯ = 1 , π¦ = 0 ; about π₯ = β1
2
π₯6 β 6 0
32 3
2
π₯ 5 ππ₯
1
= 2π
π₯+1
3
π₯ 3 β 0 ππ₯
0
π₯ 6 + 3π₯ 5 + 3π₯ 4 + π₯ 3 ππ₯
0
= 2π
π₯7 7
0
=
+
π₯6 2
+
3π₯ 5 5
+
1 π₯4 4 0
ππππ ππ
πππ ππ
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
102
EXERCISE 13.7
MOMENT OF INERTIA OF A SOLID OF REVOLUTION
15. π¦ = 2π₯ , π₯ = 1 , π¦ = 0 ; about π₯ = 2
1
πΌπ¦ = 2π 1
= 4π
2βπ₯
3
2π₯ ππ₯
0
8π₯ β 12π₯ 2 + 6π₯ 3 β π₯ 4
0
= 4π 4π₯ 2 β 4π₯ 3 + =
3π₯ 4 2
β
1 π₯5 5 0
πππ π
DIFFERENTIAL & INTEGRAL CALCULUS | Feliciano & Uy
103
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