AP® Exam Practice Questions for Chapter 7...AP® Exam Practice Questions for Chapter 7 1 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, c opied or duplicated,

AP® Exam Practice Questions for Chapter 7 1

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

AP® Exam Practice Questions for Chapter 7

1.

( ) ( ) ( )

3

1 31 3 3 3

00

13

1 1 1 13 3 3 3

3 1

10

3

x

x x

e ex

x

A e e dx ex e e e

==

=

= − = − = ⋅ − − − =

So, the answer is A.

2.

( )( )( )

3 2

3 2

2

7 12 4 2 4

7 10 0

7 10 0

5 2 0

0, 2, 5

x x x x

x x x

x x x

x x x

x

− + + = +

− + =

− + =

− − =

=

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )

2 53 2 3 2

0 2

2 53 2 3 2

0 2

2 54 3 2 4 3 2

0 2

7 71 14 3 4 3

56 625 875 563 4 3 3

25312

7 12 4 2 4 2 4 7 12 4

7 10 7 10

5 5

4 20 125 4 20

A x x x x dx x x x x dx

x x x dx x x x dx

x x x x x x

= − + + − + + + − − + +

= − + + − + −

= − + + − + −

= − + + − + − − − + −

=

So, the answer is C.

3. ( )

[ ]

( )

( )

4

0

4

0

4 cos 4 sin

4 sin 4 cos

2 24 4 0 4 1

2 2

4 2 4

4 2 1

A x x dx

x x

π

π

= −

= +

= + − +

= −

= −


4. 4 sin 2

1sin

2

6

x

x

x π

=

=

=

( )

[ ]

( )

6

0

6

0

2 4 sin

2 4 cos

2 3 0 43

2 3 43

A x dx

x x

π

π

π

π

= −

= +

= + − +

= + −

So, the answer is B.

−1 1 2 3

−1

1

2

3

y

x

y = e3x

y = e

−1 1 2 3 4 5 6

4

6

8

10

12

14

16

x

y

(0, 4)

(2, 8)(5, 14)

y = x3 − 7x2 + 12x + 4

y = 2x + 4

2 AP® Exam Practice Questions for Chapter 7


5.

Write both equations in terms of y and find the points of intersection.

4 4y x x y= − + = −

( )( )

2

2

4 4

3 4 0

4 1 0

1, 4

y y y

y y

y y

y

− = −

− − =

− + =

= −

( ) ( )( )

( ) ( )

4 2

1

4 2

1

43 2

1

313 2

64 313 3 2

1256

4 4

3 4

4

24 16 4

A y y y dy

y y dy

y y y

−

−

−

= − − −

= − + +

= − + +

= − + + − + −

=

So, the answer is D.

6. ( )ln sec

1sec tan

sec

tan

y x

dy x xdx x

x

=

= ⋅

=

4 2

0

4 2

0

4

0

1 tan

sec

sec

s x dx

x dx

x dx

π

π

π

= +

=

=


7.

( )

0.5

0.5 0.5

4

0.5 4 2

x

x x

y edy e edx

=

= =

( )4 420.5

1 11 2 1 4x xs e dx e dx= + = +


8. 3 2

1 2

2

3y x

dy xdx

=

=

( )

( )

( )

8 21 2

3

8

3

83 2

3

1

1

21

3

227 8

338

3

s x dx

x dx

x

= +

= +

= +

= −

=


9.

( )

5 4

1

5 22

1

2

1 36

1 6

6

s x dx

x dx

dy xdx

= +

= +

=

2

3

6

2

dy x dx

y x C

=

= +

Use ( )1, 9 to find C.

( )39 2 1

9 2

7

C

CC

= +

= +=

The equation for the curve is 32 7.y x= +


10.

1 21

02 sin

2.184

V x dx− = −

≈

So, the answer is C.

−2−4 1 6

−2

−1

(0, 4)

(5, −1)

x

y

−1 2 3

−1

1

3

x

y



11. (a)

Find the point of intersection of the graphs.

3 2

3 2

3 2

2 3

10 1 6

9 6

3

2

3

2

1.3103707

x

x

x

x

x

= +

=

=

=

≈

( ) ( )1.31037073 2 3 2

010 1 6 10 1 6

k

kx dx x dx − + = − +

or

3 2

3 2

3 2

2 3

1 6

1 6

1

6

1

6

y x

y xy x

yx

= +

− =− =

− =

2 3 2 33 210 1 6

1 1

1 12

6 6

ky ydy dy+− − =

(b)

( ) ( )

3 2

1 2 1 2

22 1 2

1 6

36 9

2

1 1 9 1 81

y x

y x x

y x x

= +

′ = =

′+ = + = +

( ) ( )

( )

( )

1

0

1 1 2

0

13 2

0

3 2

1 81

11 81 81

81

1 21 81

81 3

282 1 6.103

243

s x dx

x dx

x

= +

= −

= +

= − ≈

(c)

( ) ( )( )

3 2 3 2

23 2

10 1 6 3 10 1 6

3 9 6

A w h

x x

x

= ⋅

= − + ⋅ − +

= −

( )1.3103707 23 2

03 9 6 143.289V x dx= − ≈

Notes: Use your calculator to find the point of intersection of the two graphs (no work needed for this).

In this intermediate step, round the x-coordinate of the intersection point to more than three decimal places to use in upcoming integrals.

Notes: Be sure to write down the appropriate definite integral before numerically approximating it on your calculator.

Be sure to round the answer to at least three decimal places to receive credit on the exam.



−1 1 2 3 4 5−2

4

6

8

12

(0, 1) k

x

y



12. (a)

The graphs intersect at 0.05564832x ≈ and 1.7915369.x ≈

( )1.7915369

0.05564832ln 2 3

1.471

A x x dx= − −

≈

(b) ( ) ( )( )1.7915369 2 2

0.05564832ln 3 2 3 3

18.783

V x x dxπ= + − − +

≈

(c) Rewrite the equations in terms of y.

ln

1 32 3

2 2

yy x x e

y x x y

= =

= − = +

The graphs intersect at 2.888703y ≈ −

and 0.58307388.y ≈

( )2

0.58307388 2

2.888703

3

2yy e dyπ

−

+ −

Notes: Use your calculator to find the points of intersection of the two graphs (no work needed for this).

In this intermediate step, round the coordinates of the intersection points to more than three decimal places to use in upcoming integrals.



Notes: To use the disk method (washer method), set up this integral in terms of y (using horizontal elements in R).

Setting up the volume integral in terms of x (using vertical elements in R) requires the shell method.

−1−2−3 2 3 4−1

−2

−3

1

2

3

y

x



13. (a)

Find the points of intersection of the graphs

2

2

4

4

1 1

1

1

0 1

0.724492, 1.2207441

y x x y

y y

y y

y yy

= − = ± +

+ =

+ =

= − −≈ −

( )( )

1.2207441 2

0.724492

1.22074413 2 3

0.724492

2 13 3

1

1

1.377

A y y dy

y y

−

−

= + −

= + −

≈

(b) ( ) ( )21.2207441 22

0.7244922 2 1

11.501

V y y dyπ−

= − − − + ≈

(c) ( )( )1.2207441 2

0.7244922 1 1 y y y dyπ

−+ + − or

( ) ( )

( ) ( )

2 20.5248886

0

21.4902161 22

0.524886

1 1

1

x x dx

x x dx

π

π

+ − − +

+ + −

Notes: It is strategic to set up this area integral in terms of y. Setting up the area integral in terms of x requires the sum of multiple definite integrals.

Use your calculator to find the points of intersection of the two graphs (no work needed for this).

In this intermediate step, round the coordinates of the intersection points to more than three decimal places to use in upcoming integrals.

Be sure to write down the appropriate definite integral before numerically approximating it on your calculator.


Notes: To use the disk method (washer method), set up this integral in terms of y.

Setting up this volume integral in terms of x, via the shell method, requires the sum of multiple definite integrals.

Be sure to write down the appropriate definite integral before numerically approximating it on your calculator.


Notes: It is strategic to set up the shell method here in order to set up a single integral.

Setting up this volume integral in terms of x, via the disk/washer method, requires the sum of multiple definite integrals.

−1−2−3−4 2 3 4

−2

−3

−4

1

2

3

4

x

y



14. (a)

(b) 22

2

2 2

2

23 2

2

22

2 2

1 13

2 2

1 13

6 4

19 25

3 3

44

3

y yA dy

y y dy

y y y

−

−

−

− = + −

= − +

= − +

= − −

=

15. (a) ( )22 2 22

00 02

2V x dx x dx xππ π π = = = =

(b) ( )

( )

2 22 2

0

2 4

0

25

0

2

4

14

5

162

5

V y dy

y dy

y y

π

π

π

π

= −

= −

= −

=

(c) ( ) ( )

( )

22 2

0

22

0

2 2

2 4

x dx

x dx

π

π

+ −

= + −

(d) ( )

( )2 22 2

00

1 1 3 3

2 2 2 4

3 23 3 3

4 8 8 2

A bh x x x

V x dx x

= = =

= = = =

3 pts: sketch of correct region (with boundary curves labeled)

Notes: It is strategic to set up this area integral in terms of y. Setting up this area integral in terms of x requires the sum of multiple definite integrals.

The answer does not need to be simplified.

Notes: To use the disk method (washer method), set up this integral in terms of y.

Setting up the volume integral in terms of x requires the shell method.

2 pts: integral

Notes: To use the disk method (washer method), set up this integral in terms of x.

Setting up the volume integral in terms of y requires the shell method.

y

x

y = 2x + 2x = y2 + 2

y = 2

y = −2

12

−2 1 3 4−1

−3

3



16.

(a) ( )2 3

0

24

0

8

18

4

16 4

12

A x dx

x x

= −

= − = −=

(b) ( )( )

2 3

0

2 4

0

22 5

0

2 8

2 8

12 4

5

96

5

V x x dx

x x dx

x x

π

π

π

π

= −

= −

= −

=

or

( )283

0

8 2 3

0

85 3

0

3

5

96

5

V y dy

y dy

y

π

π

π

π

=

=

=

=

(c) 23

6

2 2 8

xA xπ π = =

2 6

0

27

0

8

1

8 7

128

5616

7

V x dx

x

π

π

π

π

=

=

=

=

Notes: To use disks, set up this integral terms of y.

Setting up the volume integral in terms of x requires the shell method.

1 3

2

4

6

R

S

y

x

(2, 8)

(0, 0)



17.

(a) ( )

( )

2 2

0

2 2

0

23 2

0

2

2

1

3

84

320

3

A x x dx

x x dx

x x

= − −

= +

= +

= +

=

(b) ( ) [ ]( )

( ) ( )( )

2 2 22

0

2 2 4 2

0

2 4 2

0

25 3 2

0

4 4 2

16 8 16 16 4

4 16

1 48

5 3

32 3232

3 5

736

15

V x x dx

x x x x dx

x x x dx

x x x

π

π

π

π

π

π

= − − − − − −

= + + − − +

= + +

= + +

= + +

=

(c) ( ) ( ) ( ) ( )

( )

2 2 2 2

2 2

0

2 2

0

2 2 4 4 2 0 4 0

1 2

8 2 5 1 4

P

x dx

x dx

= − + − − + − + − −

+ +

= + + +

−2−3−4 3 4−1

−2

−3

−4

2

3

4

T

x

y

AP® Exam Practice Questions for Chapter 7...AP® Exam Practice Questions for Chapter 7 1 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, c opied or duplicated,

Documents