13-1 Free Response Homework 1. Answer: 2x3 − x2 + 3x+C

6x2 − 2x + 3( )⌠⌡⎮ dx

= 6 x

3

3 − 2 x2

2 + 3x+ c

= 2x3 − x2 + 3x+ c

3. Answer: 3x

23 +C 2x3

⌠

⌡

⎮⎮

dx = 2x−13⌠

⌡⎮ dx

= 2 x23

23+ c

= 3x23 + c

5. Answer: 23 x

6 + 54 x4 +C

x3 4x2 + 5( ) ∫ dx

= 4x5 + 5x3( ) ∫ dx

= 4 x6

6+ 5 x

4

4+ c

=23x6 +

54x4 + c

7. Answer: 23 x

32 −12x12 +C

x − 6

x⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮⎮

dx

= x12 − 6x

−12⎛⎝⎜

⎞⎠⎟

⌠

⌡

⎮⎮ dx

= x32

32− 6 x

12

12+ c

= 23 x

32 −12x12 + c

9. Answer: 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

11. Answer: 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

13. 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

15. f x( ) = 32 x2 −103 x

32 − 2x+ 323 f 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

17. 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

13-1 Multiple Choice Homework 1. Answer: D

1x2 dx⌠

⌡⎮= x−1

−1+ c

3. Answer: A

x 3x dx∫ = 3 x3

2 dx∫ = 3 x5

2

52+ c

13-2 Free Response Homework

1. Answer: 120 5x+ 3( )4 +C

u = 5x+ 3;  du = 5dx

5x+ 3( )3⌠

⌡⎮ dx

= 15 5x+ 3( )3⌠

⌡⎮ 5dx

= 15 u3∫ du

= 15u4

4 + c

= 120u

4 + c

3. Answer: 17 x

7 + 12 x4 + x+C

1+ 2x3 + x6( )⌠

⌡⎮ dx = x+ 2 x

4

4 + x7

7 + c

5. Answer: 16 2x

2 + 3( )32 +C

u = 2x2 + 3;  du = 4xdx

x 2x2 + 3( )⌠

⌡⎮⎮ dx = 1

2 2x2 + 3( )⌠

⌡⎮⎮ 2xdx

= 12 u

12⌠⌡⎮ du = 1

2u

32

32+ c = 1

3 2x2 + 3( )32 + c

7. Answer: 12 1+ x

4( )12 +C

u = 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

9. Answer: 130 x2 + 4( )15 − 47 x2 + 4( )14 +1613 x2 + 4( )13 + c

u = 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

13-2 Multiple Choice Homework 1. Answer: C

xx2 − 4

dx⌠⌡⎮

= 12

2xx2 − 4

dx⌠⌡⎮

= 12

duu

⌠⌡⎮

= 12

lnu + c

3. Answer: D

60x 1+ x dx∫ = 60 u2 −1( )u du∫ = 60 u3 − u( ) du∫ = 60 u4

4− u

2

2⎛⎝⎜

⎞⎠⎟+ c

5. Answer: E x x2 −1( )4 dx∫ = 1

22x x2 −1( )4 dx∫ = 1

2u4 du∫ = 1

2u5

5+ c = 1

10x2 −1( )5 + c

13-3 Free Response Homework

1. 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

3. 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

5. 15 tan

5 x+C

u = tan x;  du = sec2 x dx

tan4 xsec2 x( )⌠

⌡⎮ dx = u4∫ du = 1

5u5 + c

7. 16 e

6x +C

u = 6x;  du = 6dx

e6x( )⌠

⌡⎮⎮ dx = 1

6 e6x( )⌠

⌡⎮⎮ 6dx = 1

6 eu( )⌠

⌡⎮⎮ du = 1

6 eu + c

9. 14 ln x

2 +1( )( )2 +C u = 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

11. − 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

13-3 Multiple Choice Homework 1. Answer: A

x3 + 2 + 1x2 +1

⎛⎝⎜

⎞⎠⎟ dx⌠

⌡⎮= x3( ) dx∫ + 2( ) dx∫ + 1

x2 +1⎛⎝⎜

⎞⎠⎟ dx⌠

⌡⎮= x4

4+ 2x + tan−1 x + c

3. Answer: C x − 2x −1

dx⌠⌡⎮

= 1+ −1x −1

⎛⎝⎜

⎞⎠⎟ dx⌠

⌡⎮= x − ln x −1( ) + c

5. Answer: D u = cos x;  du = − sin xdx

6sin xcos2 x dx∫ = −6 u2 du∫ = −2u3 + c

13-4 Free Response Homework 1. Answer: 24

x2 + 5( ) dx0

3

∫ = x3

3 + 5x⎡

⎣⎢

⎤

⎦⎥

0

3

= 33

3 + 5 3( )⎛

⎝⎜⎞

⎠⎟− 03

3 + 5 0( )⎛

⎝⎜⎞

⎠⎟= 24

3. Answer: 0 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

5. Answer: −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

7. Answer: 0.070 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

9. Answer: 0.549 12

2dx2x+ 5−1

2⌠

⌡⎮⎮ = 12 ln 2x+ 5

⎤⎦⎥−1

2

= 12 ln9− ln3⎡⎣ ⎤⎦ =

12 ln3= 0.549

11. Answer: 0.693

u = ln x;  du = 1x  dx

u 2( ) = ln2, u 4( ) = ln4

dxx lnx2

4⌠

⌡⎮⎮ = duuln2

ln4⌠

⌡⎮⎮ = lnu⎡⎣ ⎤⎦ln2

ln4 = lnln4 − lnln2 = 0.693

13-4 Multiple Choice Homework 1. Answer: C

1x+ 2x⎛

⎝⎜⎞⎠⎟ dx

2

6⌠⌡⎮

= ln x + x2⎡⎣ ⎤⎦2

6= ln6 + 36( )− ln2 + 4( ) = ln6 − ln2 + 32 = ln 6

2+ 32

3. Answer: A 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

5. Answer: D

f x( ) + 3( ) dx2

4

∫ = f x( )dx2

4

∫ + 3 dx2

4

∫ = 6 + 3x]24 = 6 + 12 − 6( ) = 12

13-5 Free Response Homework

1. Answer: 83

A = x2 +1( ) dx−1

1⌠⌡ = x3

3 + x⎡

⎣⎢

⎤

⎦⎥−1

1

= 13

3 +1⎡

⎣⎢

⎤

⎦⎥ −

−1( )33 + −1( )

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥= 8

3

3. Answer: 4

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

5. Answer: 9

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

7. Answer: 4 2 A = − 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

9a. Answer: − 32m

displacement = 3t − 5( )0

3

∫ dt = 32 t

2 − 5t⎡⎣⎢

⎤⎦⎥0

3

= 13.5−15( )− 0( ) = −1.5

9b. Answer: 416 m

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

11. Answer: 52 feet distance= 4t +1( )

1

5

∫ dt = 2t2 + t⎡⎣ ⎤⎦15= 50+ 5( )− 2+1( ) = 52

13-5 Multiple Choice Homework 1. Answer: C

4x3 + 6x− 1x

⎛⎝⎜

⎞⎠⎟dx

1

2⌠

⌡

⎮⎮⎮

= x4 + 3x2 − ln x⎡⎣ ⎤⎦12

= 24 + 3 2( )2 − ln2⎡⎣⎢

⎤⎦⎥− 14 + 3 1( )2 − ln1⎡⎣⎢

⎤⎦⎥= 24 − ln2 units2

3. Answer: D 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

5. Answer: B

x1+ x2

dx0

3⌠⌡⎮

== 12

2x1+ x2

dx1

4⌠⌡⎮

= 12

u1

2

⎡

⎣⎢⎢

⎤

⎦⎥⎥1

4

= 2 −1= 1

Integrals Practice Test Part 1: CALCULATOR REQUIRED Multiple Choice (3 pts. each) 1. Answer: B x t( ) = 2 + 3t 2 + 6t( )dt

0

1

∫ = 2 + t 3 + 3t 2⎡⎣ ⎤⎦01= 6

2. Answer: B

f x( ) dx−5

5

∫ = f x( ) dx−5

2

∫ − f x( ) dx5

2

∫ = −17 − −4( ) = −13

3. Answer: C x2

ex3 dx⌠

⌡⎮= − 1

3−3x2

ex3 dx⌠

⌡⎮= − 1

3eu du∫ = − 1

3eu + c = − 1

3e−x

3+ c

4. Answer: E

sin t dt0

x

∫ = −cost[ ]0x = −cos x − −cos0( ) = 1− cos x

5. Answer: B

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

6. Answer: E

x 4 − x2 dx∫ = − 12

4 − x2 −2xdx( )∫

= − 124 − x2( )3232

+ c

= − 134 − x2( )32 + c

7. 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

8. Answer: D

sin 2x + 3( )∫ dx = 12sin 2x + 3( )∫ 2dx( )

= 12sinu∫ du

= 12

−cosu( ) + c

= − 12cos 2x + 3( ) +C

Integrals Practice Test Part 2: NO CALCULATOR ALLOWED Free Response

1. Answer: Area = 12

A = 1x21

2⌠⌡⎮

 dx = −1x

⎡⎣⎢

⎤⎦⎥1

2

= −12− −1( ) = 1

2

2. Answer: Area = 2 ln5

A = x1+ x2−2

2⌠⌡⎮

 dx = 12

2x1+ x20

2⌠⌡⎮

 dx = ln 1+ x2( )⎡⎣ ⎤⎦02= 2 ln5 − 0( ) = 2ln5

3. Answer: Displacement = t2 − t( ) dt0

5

∫ Total Distance Traveled = − t2 − t( ) dt0

1

∫ + t2 − t( ) dt1

5

∫