Page 1
AP® Exam Practice Questions for Chapter 6 1
© 2017 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 6
1.
( )
( ) ( )
3
1 3 3
0
1 33
0
13
1 1 13 3 3
13
3 1
1
3
0
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.
−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
Page 2
2 AP® Exam Practice Questions for Chapter 6
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
= −
= +
= + − +
= −
= −
So, the answer is A.
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.
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
= +
=
=
So, the answer is B.
7.
( )
0.5
0.5
0.5
4
0.5 4
2
x
x
x
y edy edx
e
=
=
=
( )4 20.5
1
4
1
1 2
1 4
x
x
s e dx
e dx
= +
= +
So, the answer is A.
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
= +
= +
= +
= −
=
So, the answer is B.
9.
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
Page 3
AP® Exam Practice Questions for Chapter 6 3
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
10. (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
= +
=
=
=
≈
( ) ( )1.313 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 0
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= − ≈
−1 1 2 3 4 5−2
4
6
8
12
(0, 1) k
x
y
Reminders: 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.
Reminders: 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.
Reminders: 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.
Page 4
4 AP® Exam Practice Questions for Chapter 6
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
11. (a)
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 ≈
( ) ( )
0.58307388
2.888703
0.583073882
2.888703
1 3
2 2
1 3
4 2
0.832 2.303
1.471
y
y
A y e dy
y y e
−
−
= + −
= + −
≈ − − −
=
(b) ( ) ( )( )1.7915369 2 2
0.05564832ln 3 2 3 3
18.783
V x x dxπ= + − − +
≈
(c) ( )2
0.58307388 2
2.888703
3
2yy e dyπ
−
+ −
−1−2−3 2 3 4−1
−2
−3
1
2
3
y
x
Reminders: 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.
Reminders: 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.
Notes: To use disks (washers), 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.
Page 5
AP® Exam Practice Questions for Chapter 6 5
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12. (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 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
+ − − +
+ + −
−1−2−3−4 2 3 4
−2
−3
−4
1
2
3
4
x
y
Note: Because R is horizontally simple, it is strategic to set up this area integral in terms of y (using horizontal elements in R). Setting up the area integral in terms of x (using vertical elements in R) requires the sum of multiple definite integrals because R is not vertically simple.
Reminders: 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.
Be sure to round the answer to at least three decimal places to receive credit on the exam.
Notes: To use disk (washers), set up this integral in terms of y (using horizontal elements in R).
Setting up this volume integral in terms of x (using vertical elements in R), via the shell method, requires the sum of multiple definite integrals because R is not vertically simple.
Reminders: 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.
Notes: Because R is horizontally simple, it is strategic to set up the shell method here (using horizontal elements in R) in order to set up a single integral.
Setting up this volume integral in terms of x (using vertical elements in R), via the disk/washer method, requires the sum of multiple definite integrals because R is not vertically simple.
Page 6
6 AP® Exam Practice Questions for Chapter 6
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
13. (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
−
−
−
− = + −
= − +
= − +
= − −
=
14. (a) ( )22 2 22
00 02
2V x dx x dx xππ π π = = = =
(b) ( ) ( )2 222 2 4
0 0
25
0
2 4
1 164 2
5 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
= = =
= = = =
y
x
y = 2x + 2x = y2 + 2
y = 2
y = −2
12
−2 1 3 4−1
−3
3
3 pts: sketch of correct region (with boundary curves labeled)
Note: Because the region is horizontally simple, it is strategic to set up this area integral in terms of y (using horizontal elements of R). Setting up this area integral in terms of x (using vertical elements in R) requires the sum of multiple definite integrals because R is not vertically simple.
Reminder: The answer does not need to be simplified.
Notes: To use disks (washers), 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.
2 pts: integral
Notes: To use disks (washers), set up this integral in terms of x (using vertical elements in R).
Setting up the volume integral in terms of y (using horizontal elements in R) requires the shell method.
Page 7
AP® Exam Practice Questions for Chapter 6 7
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.
(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
( )28
0
8 2 3
0
85 3
0
3
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
=
=
=
=
1 3
2
4
6
R
S
y
x
(2, 8)
(0, 0)
Notes: To use disks, set up this integral 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.
Page 8
8 AP® Exam Practice Questions for Chapter 6
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.
(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
