2e AP Exam Prep Solutions Ch 06 WEB

AP ? Exam Practice Questions for Chapter 6

1

AP? Exam Practice Questions for Chapter 6

y

1.

y=e

3

2

1

y = e 3x

?1

1

2

x

3

?1

e3 x = e

3x = 1

x =

1

3

A =

3x

?0 (e ? e ) dx

13

13

1 ?

?

= ?ex ? e3 x ?

3 ?0

?

(

= e?

) (

1

3

? 13 e ? 0 ?

1

3

)

1

3

=

So, the answer is A.

2.

y = x 3 ? 7x 2 + 12x + 4

y

16

14

12

10

8

6

4

y = 2x + 4

(5, 14)

(2, 8)

(0, 4)

?1

1

2

3

4

5

6

x

x3 ? 7 x 2 + 12 x + 4 = 2 x + 4

x3 ? 7 x 2 + 10 x = 0

x( x 2 ? 7 x + 10) = 0

x( x ? 5)( x ? 2) = 0

x = 0, 2, 5

A =

=

?0 ??( x

2

?0 ( x

2

3

? 7 x 2 + 12 x + 4) ? ( 2 x + 4)?? dx +

3

? 7 x 2 + 10 x) dx +

?2 (? x

5

2

3

= ?? 14 x 4 ? 73 x3 + 5 x 2 ?? + ??? 14 x 4 +

0

(

= 4?

=

56

3

) (

+ 20 + ? 625

+

4

875

3

?2 ??(2 x + 4) ? ( x

5

3

? 7 x 2 + 12 x + 4)?? dx

+ 7 x 2 ? 10 x) dx

7 x3

3

5

? 5 x 2 ??

2

) (

? 125 ? ? 4 +

56

3

)

? 20

253

12

So, the answer is C.

? 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

2

3. A =

¦Ð 4

?0 (4 cos x ? 4 sin x) dx

6.

y = ln (sec x)

dy

1

=

? sec x tan x

sec x

dx

¦Ð 4

= [4 sin x + 4 cos x] 0

= tan x

? ? 2?

? 2 ??

= ?4??

?? + 4??

??? ? ??0 + 4(1)??

2

?

? 2 ???

?? ?

?0

=

?0

=

?0

= 4 2 ?4

(

)

= 4

2 ?1

So, the answer is A.

1

2

A =

7.

6

= [2 x + 4 cos x] 0

?¦Ð

?

= ? + 2 3 ? ? ( 0 + 4)

?3

?

3

+ 2 3 ?4

8.

(0, 4)

?2

?1

1

?2

6

?1

=

?1

?3

=

?3

x

2

( 27 ? 8)

3

38

=

3

y

3

y = ?1, 4

4

2

= ??? 13 y 3 +

(

125

6

So, the answer is D.

1

?1

2

x

3

?1

4

+ 4 y??

?1

) ( 13 + 23 ? 4)

= ? 64

+ 24 + 16 ?

3

=

? 4 y )?? dy

+ 3 y + 4) dy

3 y2

2

1 + x dx

=

9.

2

??1 ??(4 ? y) ? ( y

2

So, the answer is B.

? 4)( y + 1) = 0

4

8

1 + ( x1 2 ) dx

8

y2 ? 3y ? 4 = 0

??1 (? y

1 + 4e x dx

3 2?

?2

= ? (1 + x) ?

3

?

?3

(5, ? 1)

y2 ? 4 y = 4 ? y

=

8

s =

y = ?x + 4 ? x = 4 ? y

A =

2

2 32

x

3

y =

Write both equations in terms of y and find the points

of intersection.

(y

4

1 + ( 2e0.5 x ) dx

dy

= x1 2

dx

y

?4

4

s =

So, the answer is A.

So, the answer is B.

5.

sec x dx

= 2e0.5 x

?0 (2 ? 4 sin x) dx

¦Ð

¦Ð 4

sec 2 x dx

y = 4e0.5 x

¦Ð 6

¦Ð 6

=

¦Ð 4

dy

= 0.5( 4e0.5 x )

dx

¦Ð

x =

1 + tan 2 x dx

So, the answer is B.

4. 4 sin x = 2

sin x =

¦Ð 4

s =

V =

1

?0 ??2 ? sin

?1

2

x?? dx

¡Ö 2.184

So, the answer is C.

? 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

3

y

10. (a)

12

8

Reminders: Use your calculator to find the point of

intersection of the two graphs (no work needed for

this).

6

4

k

(0, 1)

?1

1

?2

2

3

4

x

5

Find the point of intersection of the graphs.

In this intermediate step, round the x-coordinate of

the intersection point to more than three decimal

places to use in upcoming integrals.

10 = 1 + 6 x3 2

9 = 6 x3 2

3

= x3 2

2

?3?

? ?

?2?

23

= x

x ¡Ö 1.3103707

32

?0 ??10 ? (1 + 6 x )?? dx = ?k

k

1.3103707

?10 ? (1 + 6 x3 2 )? dx

?

?

or

y = 1 + 6 x3 2

y ? 1 = 6 x3 2

y ?1

= x3 2

6

? y ? 1?

x = ?

?

? 6 ?

y ? 1?

?

?

? 6 ?

10 ?

?1

23

23

dy = 2 ?

1+ 6 k 3 2

0

? y ? 1?

?

?

? 6 ?

23

dy

y = 1 + 6 x3 2

(b)

? 3?

y¡ä = 6? ? x1 2 = 9 x1 2

? 2?

1 + ( y¡ä) = 1 + (9 x

2

s =

1

?0

12

)

2

= 1 + 81x

1 + 81x dx

=

1 1

(1 ? 81x)1 2 (81) dx

81 ?0

=

1 ?2

(1 + 81x)3 2 ??

81 ?? 3

?0

=

2

(823 2 ? 1) ¡Ö 6.103

243

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.

1

(c) A = w ? h

= ??10 ? (1 + 6 x 3 2 )?? ? 3 ??10 ? (1 + 6 x 3 2 )??

= 3(9 ? 6 x 3 2 )

V = 3?

1.3103707

0

2

(9 ? 6 x 3 2 )

2

dx ¡Ö 143.289

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.

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

4

AP ? Exam Practice Questions for Chapter 6

y

11. (a)

3

2

1

?3 ?2 ?1

?1

2

3

x

4

Reminders: Use your calculator to find the points of

intersection of the two graphs (no work needed for

this).

?2

?3

In this intermediate step, round the coordinates of

the intersection points to more than three decimal

places to use in upcoming integrals.

Rewrite the equations in terms of y.

y = ln x ? x = e

y

y = 2x ? 3 ? x =

1

3

y +

2

2

The graphs intersect at y ¡Ö ? 2.888703

and y ¡Ö 0.58307388.

A =

0.58307388

?? 2.888703

?? 1

3?

y?

?? 2 y + 2 ? ? e ? dy

?

?

?

?

0.58307388

3

?1

?

= ? y2 + y ? e y ?

2

?4

? ? 2.888703

¡Ö ( ? 0.832) ? ( ? 2.303)

= 1.471

(b) V = ¦Ð

? 0.05564832 ( ??ln( x) + 3??

1.7915369

2

? ??( 2 x ? 3) + 3??

2

) dx

¡Ö 18.783

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.

(c) ¦Ð

0.58307388 ??

??2.888703

??

???

2

2?

y + 3?

y

? ? (e ) ? dy

2 ?

??

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.

? 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

5

y

12. (a)

4

3

2

1

?4 ?3 ?2 ?1

2

3

x

4

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.

?2

?3

?4

Find the points of intersection of the graphs

y = x2 ? 1 ? x = ¡À

y +1

Reminders: Use your calculator to find the points of

intersection of the two graphs (no work needed for this).

y + 1 = y2

y + 1 = y4

0 = y4 ? y ? 1

y ¡Ö ? 0.724492, 1.2207441

A =

??0.724492 (

)

1.2207441

Be sure to write down the appropriate definite integral

before numerically approximating it on your calculator.

y + 1 ? y 2 dy

32

= ? 23 ( y + 1) ?

?

1.2207441

1 y3 ?

3 ? ? 0.724492

Be sure to round the answer to at least three decimal

places to receive credit on the exam.

¡Ö 1.377

(b) V = ¦Ð

?

2

? ?0.724492 ??(2 ? y )

1.2207441

2

In this intermediate step, round the coordinates of the

intersection points to more than three decimal places to

use in upcoming integrals.

(

)

2

y + 1 ?? dy

?

? 2 ?

¡Ö 11.501

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.

(c) 2¦Ð

¦Ð

? ?0.724492 ( y + 1)(

)

1.2207441

0.5248886

?0

(

?

??

+¦Ð

y + 1 ? y 2 dy or

) ? (?

x +1

2

?

? 0.524886 ??(

1.4902161

)

2

x + 1 ?? dx

?

) ? ( x ) ?? dx

x +1

2

2

2?

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.

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

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download