417
13-1 Free Response Homework 1. Answer:
3. Answer:
5. Answer:
7. Answer:
2x3 − x2 + 3x+C6x2 − 2x + 3( )⌠
⌡⎮ dx
= 6 x3
3 − 2 x2
2 + 3x+ c
= 2x3 − x2 + 3x+ c
3x23 +C
2x3
⌠
⌡
⎮⎮
dx = 2x−13⌠
⌡⎮ dx
= 2 x23
23+ c
= 3x23 + c
23 x
6 + 54 x4 +C
x3 4x2 + 5( ) ∫ dx
= 4x5 + 5x3( ) ∫ dx
= 4 x6
6+ 5 x
4
4+ c
=23x6 +
54x4 + c
23 x
32 −12x12 +C
x − 6x
⎛⎝⎜
⎞⎠⎟
⌠
⌡
⎮⎮⎮
dx
= x12 − 6x
−12⎛⎝⎜
⎞⎠⎟
⌠
⌡
⎮⎮ dx
= x32
32− 6 x
12
12+ c
= 23 x
32 −12x12 + c
418
9. Answer:
11. Answer:
13.
14 x
4 + x3 + 32 x2 + x+C
x +1( )3 ∫ dx
= x3 + 3x2 + 3x+1( ) ∫ dx
= x4
4 + 3 x3
3 + 3 x2
2 + x+ c
= x4
4 + x3 + 32 x
2 + x+ c
23 x
32 + 65 x52 −12x
12 +C
x + 3 x32 − 6x
⎛⎝⎜
⎞⎠⎟
⌠
⌡
⎮⎮⎮
dx
= x12 + 3 x
32 − 6x−12⎛
⎝⎜⎞⎠⎟
⌠
⌡
⎮⎮ dx
= x32
32+ 3 x
52
52− 6 x
12
12+ c
= 23 x
32 + 65 x
52 −12x12 + c
f x( ) = x3 − 3x2 + 3x+ 2
f x( ) = 3x2 − 6x+ 3( )dx⌠⌡
= x3 − 3x2 + 3x+ cf 0( ) = 2 = 0( )3 − 3 0( )2 + 3 0( )+ cc = 2
419
15.
17.
13-1 Multiple Choice Homework 1. Answer: D
f x( ) = 32 x2 −103 x
32 − 2x+ 323f x( ) = x − 2( ) 3 x +1( )⌠
⌡ dx
= 3x−−5 x − 2( )⌠⌡ dx
= 32 x2 − 5 x
32
32− 2x+ c
f 4( ) =1= 32 1( )2 − 5 1( )32
32− 2 1( )+ c
c = 323
x t( ) = 112 t4 − 13t
3 + 2t2 + 2t + 4
v t( ) = t 2 − 2t + 4( )∫ dt
=13t 3 − t 2 + 4t + c1
v 0( ) = 2 = 13t 3 − t 2 + 4t + c1
c1 = 2
x t( ) = 13t
3 − t2 + 4t + 2⎛⎝⎜
⎞⎠⎟
⌠
⌡⎮ dt
= 112 t4 − 13t
3 + 2t2 + 2t + c2
x 0( ) = 4 = 112 t4 − 13t
3 + 2t2 + 2t + c2c2 = 4
1x2 dx⌠
⌡⎮= x−1
−1+ c
420
3. Answer: A
13-2 Free Response Homework
1. Answer:
3. Answer:
5. Answer:
x 3x dx∫ = 3 x3
2 dx∫ = 3 x5
2
52+ c
120 5x+ 3( )4 +C
u = 5x+ 3; du = 5dx5x+ 3( )3⌠
⌡⎮ dx
= 15 5x+ 3( )3⌠
⌡⎮ 5dx
= 15 u3∫ du
= 15u4
4 + c
= 120u
4 + c
17 x
7 + 12 x4 + x+C
1+ 2x3 + x6( )⌠
⌡⎮ dx = x+ 2 x
4
4 + x7
7 + c
16 2x
2 + 3( )32 +Cu = 2x2 + 3; du = 4xdxx 2x2 + 3( )⌠
⌡⎮⎮ dx = 1
2 2x2 + 3( )⌠
⌡⎮⎮ 2xdx
= 12 u
12⌠⌡⎮ du = 1
2u
32
32+ c = 1
3 2x2 + 3( )32 + c
421
7. Answer:
9. Answer:
13-2 Multiple Choice Homework 1. Answer: C
12 1+ x
4( )12 +Cu = 1+ x4; du = 4x3dx
x3
1+ x4
⌠
⌡
⎮⎮⎮
dx = 14
4x3
1+ x4
⌠
⌡
⎮⎮⎮
dx
= 14 u
−12⌠⌡⎮ du = 1
4u
12
12+ c = 1
2 1+ x4 + c
130 x2 + 4( )15 − 47 x2 + 4( )14 +1613 x2 + 4( )13 + cu = x2 + 4; du = 2xdx; x2 = u − 4
x5 x2 + 4( )12⎛
⎝⎜
⎞
⎠⎟
⌠
⌡
⎮⎮⎮⎮
dx
= 12 x4 x2 + 4( )12⎛
⎝⎜
⎞
⎠⎟
⌠
⌡
⎮⎮⎮⎮
2xdx
= 12 u − 4( )2u12⌠
⌡⎮⎮ du
= 12 u2 − 8u −16( )u12⌠
⌡⎮⎮ du
= 12 u14 − 8u13 −16u12( )⌠
⌡⎮⎮ du
= 12u15
15 − 8u14
14 −16u13
13 + c
= 130 x2 + 4( )15
− 47 x2 + 4( )14
+1613 x2 + 4( )13
+C
xx2 − 4
dx⌠⌡⎮
= 12
2xx2 − 4
dx⌠⌡⎮
= 12
duu
⌠⌡⎮
= 12
lnu + c
422
3. Answer: D
5. Answer: E
13-3 Free Response Homework
1.
3.
5.
7.
60x 1+ x dx∫ = 60 u2 −1( )u du∫ = 60 u3 − u( ) du∫ = 60 u4
4− u
2
2⎛⎝⎜
⎞⎠⎟+ c
x x2 −1( )4 dx∫ = 122x x2 −1( )4 dx∫ = 1
2u4 du∫ = 1
2u5
5+ c = 1
10x2 −1( )5 + c
15 sin x5 +C
u = x5; du = 5x4dx
x4 cos x5( )⌠
⌡⎮ dx = 1
4 cos x5( )⌠
⌡⎮ 4x4dx = 1
4 cos u( )⌠⌡⎮ du = 1
4 sinu + c
13 tan 3x−1( )+C
u = 3x−1; du = 3dx
sec2 3x−1( )( )⌠
⌡⎮ dx = 1
3 sec2 3x−1( )( )⌠
⌡⎮ 3dx = 1
3 sec2u( )⌠
⌡⎮ du = 1
3 tanu + c
15 tan
5 x+C
u = tan x; du = sec2 x dx
tan4 xsec2 x( )⌠
⌡⎮ dx = u4∫ du = 1
5u5 + c
16 e
6x +C
u = 6x; du = 6dx
e6x( )⌠
⌡⎮⎮ dx = 1
6 e6x( )⌠
⌡⎮⎮ 6dx = 1
6 eu( )⌠
⌡⎮⎮ du = 1
6 eu + c
423
9.
11.
13-3 Multiple Choice Homework 1. Answer: A
3. Answer: C
5. Answer: D
13-4 Free Response Homework 1. Answer: 24
14 ln x
2 +1( )( )2 +Cu = ln x2 +1( ); du = 2x
x2 +1dx
x ln x2 +1( )x2 +1
⌠
⌡
⎮⎮⎮
dx = 12 ln x2 +1( ) 2x
x2 +1⌠
⌡⎮⎮ dx = 1
2 u∫ du =12u2
2 + c = 14 u
2 + c
− 23cot32 x+C
u = cot x; du = −csc2 xdx
cot x csc2 x( )⌠
⌡⎮ dx = − cot x −csc2 x( )⌠
⌡⎮ dx = − u
12⌠
⌡⎮ du = − 2
3u3
2 + c
x3 + 2 + 1x2 +1
⎛⎝⎜
⎞⎠⎟ dx⌠
⌡⎮= x3( ) dx∫ + 2( ) dx∫ + 1
x2 +1⎛⎝⎜
⎞⎠⎟ dx⌠
⌡⎮= x4
4+ 2x + tan−1 x + c
x − 2x −1
dx⌠⌡⎮
= 1+ −1x −1
⎛⎝⎜
⎞⎠⎟ dx⌠
⌡⎮= x − ln x −1( ) + c
u = cos x; du = − sin xdx
6sin xcos2 x dx∫ = −6 u2 du∫ = −2u3 + c
x2 + 5( ) dx0
3
∫ = x3
3 + 5x⎡
⎣⎢
⎤
⎦⎥
0
3
= 33
3 + 5 3( )⎛
⎝⎜⎞
⎠⎟− 03
3 + 5 0( )⎛
⎝⎜⎞
⎠⎟= 24
424
3. Answer: 0
5. Answer:
7. Answer: 0.070
9. Answer:
11. Answer:
u =1− x2; du = 2x dxu −1( ) = 0, u 1( ) = 0
x 1− x2( ) dx−1
1
∫ = 12 2x 1− x2( ) dx−1
1
∫ = 12 u
12 dx0
0
∫ = 0
−3313
x+ 5( ) x2 − 3( ) dx−2
2
∫ = x3 + 5x2 − 3x−15( ) dx−2
2
∫ = x4
4 + 5 x3
3 − 3 x2
2 −15x⎡
⎣⎢
⎤
⎦⎥−2
2
= −3313
u = cosx; du = −sin x dx
u π2
⎛⎝⎜
⎞⎠⎟= 0, u π
6⎛⎝⎜
⎞⎠⎟= 32
cos5 xsin x dxπ6
π2∫ = − u5 du3
2
0
∫ = − u6
6⎡
⎣⎢
⎤
⎦⎥
32
0
= .070
0.54912
2dx2x+ 5−1
2⌠
⌡⎮⎮ = 12 ln 2x+ 5
⎤⎦⎥−1
2
= 12 ln9− ln3⎡⎣ ⎤⎦ =
12 ln3= 0.549
0.693u = ln x; du = 1x dx
u 2( ) = ln2, u 4( ) = ln4 dxx lnx2
4⌠
⌡⎮⎮ = duuln2
ln4⌠
⌡⎮⎮ = lnu⎡⎣ ⎤⎦ln2
ln4 = lnln4 − lnln2 = 0.693
425
13-4 Multiple Choice Homework 1. Answer: C
3. Answer: A
5. Answer: D
13-5 Free Response Homework
1. Answer:
3. Answer: 4
1x+ 2x⎛
⎝⎜⎞⎠⎟ dx
2
6⌠⌡⎮
= ln x + x2⎡⎣ ⎤⎦2
6= ln6 + 36( )− ln2 + 4( ) = ln6 − ln2 + 32 = ln 6
2+ 32
u = tan x; du = sec2 x dx
u π4
⎛⎝⎜
⎞⎠⎟=1, u π
3⎛⎝⎜
⎞⎠⎟= 3
sec2 xtan x
dxπ
4
π3⌠
⌡⎮= du
u1
3⌠⌡⎮
= lnu[ ]13 = ln 3 − ln1
f x( ) + 3( ) dx2
4
∫ = f x( )dx2
4
∫ + 3 dx2
4
∫ = 6 + 3x]24 = 6 + 12 − 6( ) = 12
83
A = x2 +1( ) dx−1
1⌠⌡ = x3
3 + x⎡
⎣⎢
⎤
⎦⎥−1
1
= 13
3 +1⎡
⎣⎢
⎤
⎦⎥ −
−1( )33 + −1( )
⎡
⎣
⎢⎢
⎤
⎦
⎥⎥= 8
3
426
5. Answer: 9
7. Answer:
9a. Answer:
9b. Answer:
A = cosx dx0
2π∫ = cosxdx
0
π2∫ − cosxdxπ
2
3π2∫ + cosxdx3π
2
2π∫
= sin x⎡⎣ ⎤⎦0π2 − sin x⎡⎣ ⎤⎦π 2
3π2 + sin x⎡⎣ ⎤⎦3π 2
2π
= sinπ2 − sin0⎡⎣⎢
⎤⎦⎥0
π2− sin 3π2 − sinπ2⎡⎣⎢
⎤⎦⎥π 2
3π2+ sin2π − sin 3π2⎡⎣⎢
⎤⎦⎥3π 2
2π
= 1− 0( )− −21−1( )+ 0− −1( )( )= 4
A = x 9 − 4x2 dx−32
32⌠⌡⎮
= 2 x 9− 4x2 dx0
32∫ = − 12 9− 4x2 −4xdx( )0
32∫ = − 12 u du9
0∫
= − 12u32
32
⎡
⎣
⎢⎢
⎤
⎦
⎥⎥9
0
= 0− − 13932⎛
⎝⎜⎞⎠⎟
⎛
⎝⎜⎞
⎠⎟= 9
4 2A = − y= 3sinx 1− cos x
−π
0∫ dx = 3 1− cos x sin xdx( )−π
0∫ = 3 u du
2
0∫
= 3u3
2
32
⎡
⎣
⎢⎢
⎤
⎦
⎥⎥2
0
= 2 0− 23
2⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟= 4 2
− 32m
displacement = 3t − 5( )0
3
∫ dt = 32 t
2 − 5t⎡⎣⎢
⎤⎦⎥0
3
= 13.5−15( )− 0( ) = −1.5
416 m
427
11. Answer: 52 feet
13-5 Multiple Choice Homework 1. Answer: C
3. Answer: D
5. Answer: B
distance= 3t − 50
3
∫ dt = − 3t − 5( )0
53∫ dt + 3t − 5( )53
3
∫ dt
= − 32 t
2 − 5t⎡⎣⎢
⎤⎦⎥0
53+ 32 t
2 − 5t⎡⎣⎢
⎤⎦⎥53
3
= 416
distance= 4t +1( )1
5
∫ dt = 2t2 + t⎡⎣ ⎤⎦15= 50+ 5( )− 2+1( ) = 52
4x3 + 6x− 1x
⎛⎝⎜
⎞⎠⎟dx
1
2⌠
⌡
⎮⎮⎮
= x4 + 3x2 − ln x⎡⎣ ⎤⎦12
= 24 + 3 2( )2 − ln2⎡⎣⎢
⎤⎦⎥− 14 + 3 1( )2 − ln1⎡⎣⎢
⎤⎦⎥= 24 − ln2 units2
a t( ) = 12t 2
v t( ) = 12t 2 dt∫= 4t 3 + c
v 0( ) = 6 = 4 0( )3 + cc = 6
x t( ) = 4t 3 + 6( )0
2∫ dt = t 4 + 6t⎡⎣ ⎤⎦0
2= 16 +12 = 28
x1+ x2
dx0
3⌠⌡⎮
== 12
2x1+ x2
dx1
4⌠⌡⎮
= 12
u1
2
⎡
⎣⎢⎢
⎤
⎦⎥⎥1
4
= 2 −1= 1
428
429
Integrals Practice Test Part 1: CALCULATOR REQUIRED Multiple Choice (3 pts. each) 1. Answer: B
2. Answer: B
3. Answer: C
4. Answer: E
5. Answer: B
6. Answer: E
x t( ) = 2 + 3t 2 + 6t( )dt0
1
∫ = 2 + t 3 + 3t 2⎡⎣ ⎤⎦01= 6
f x( ) dx−5
5
∫ = f x( ) dx−5
2
∫ − f x( ) dx5
2
∫ = −17 − −4( ) = −13
x2
ex3 dx⌠
⌡⎮= − 1
3−3x2
ex3 dx⌠
⌡⎮= − 1
3eu du∫ = − 1
3eu + c = − 1
3e−x
3+ c
sin t dt0
x
∫ = −cost[ ]0x = −cos x − −cos0( ) = 1− cos x
x2 +1x
dx1
e⌠⌡⎮
= x + 1x
⎛⎝⎜
⎞⎠⎟ dx
1
e⌠⌡⎮
= x2
2+ ln x⎡
⎣⎢
⎤
⎦⎥
1
e
= e2
2+1⎛
⎝⎜⎞⎠⎟− 1
2− 0⎛
⎝⎜⎞⎠⎟ =
e2
2+ 1
2
x 4 − x2 dx∫ = − 12
4 − x2 −2xdx( )∫
= − 124 − x2( )3232
+ c
= − 134 − x2( )32 + c
430
7. Answer: D
8. Answer: D
x7 + k( ) dx−2
2
∫ = 16
x8
8+ kx
⎡
⎣⎢
⎤
⎦⎥−2
2
= 16
28
8+ 2k⎛
⎝⎜⎞⎠⎟−
−2( )8
8− 2k
⎛
⎝⎜⎞
⎠⎟= 16
4k = 16k = 4
sin 2x + 3( )∫ dx = 12sin 2x + 3( )∫ 2dx( )
= 12sinu∫ du
= 12
−cosu( ) + c
= − 12cos 2x + 3( ) +C
431
Integrals Practice Test Part 2: NO CALCULATOR ALLOWED Free Response
1. Answer:
2. Answer:
3. Answer:
Area = 12
A = 1x21
2⌠⌡⎮
dx = −1x
⎡⎣⎢
⎤⎦⎥1
2
= −12− −1( ) = 1
2
Area = 2 ln5
A = x1+ x2−2
2⌠⌡⎮
dx = 12
2x1+ x20
2⌠⌡⎮
dx = ln 1+ x2( )⎡⎣ ⎤⎦02= 2 ln5 − 0( ) = 2ln5
Displacement = t2 − t( ) dt0
5
∫
Total Distance Traveled = − t2 − t( ) dt0
1
∫ + t2 − t( ) dt1
5
∫