AP® Exam Practice Questions for Chapter 4

AP ? Exam Practice Questions for Chapter 4

1

AP? Exam Practice Questions for Chapter 4

1. Use the Trapezoidal Rule to find an approximation of

4

?0 g ( x) dx

4

?0 g ( x) dx.

¡Ö

4?0

?0 + 2(3) + 2(7) + 2(12) + 2(18) + 2( 25) + 2(33) + 2( 42) + 52??

2(8) ?

=

1

(332) = 83

4

So, the answer is B.

2.

5

?0 f ( x) dx

=

3

5

5. g ( ?1) =

?0 f ( x) dx + ?3 f ( x) dx

1

(3 + 5)3 + 2(? 2)

2

= 8

4

= ? ? f (t ) dt

=

?1

0

= ? ? f (t ) dt ?

?1

So, the answer is C.

3.

4

? ( x ? 5)2

+9

= ? 12 ( ? 2) ?

1

2

3

4

?0 f (t ) dt ? ?3 f (t ) dt

(4)(? 2)

? 12 ( 2)

= 4

dx

So, the answer is C.

Let u = x ? 5, du = dx, and a = 3.

6.

1

u

?1

?

4? 2

dx = 4 ? arctan + C ?

u + a2

a

a

?

?

x ?5

?1

?

= 4? arctan

+ C?

3

?3

?

4

x?5

= tan ?1

+C

3

3

So, the answer is D.

7. f ¡ä( x) =

=

x3 + 6

a = 1, b = 5

b

f (b ) = f ( a ) +

?a f ¡ä( x) dx

f (5) = f (1) +

?1

5

x3 + 6 dx

¡Ö 2 + 24.672

= 26.672

So, the answer is D.

So, the answer is A.

3¦Ð

1

?0

3¦Ð

?0

2

= 0.4

b ¡Ö 2.214

=

8. Average value =

[? cos x]¦Ðb

cos b = ? 0.6

1 2 3

(4t ? 4t ) dt

2 ?0

2

1

= ??t 4 ? 2t 2 ??

0

2

1

= (16 ? 8)

2

= 4 units sec

2

= 0.4

1 + cos b = 0.4

4. v(t ) = 4t 3 ? 4t , 0 ¡Ü t ¡Ü 2

?0 v(t ) dt

¦Ð

?b sin x dx

? cos ¦Ð + cos b = 0.4

So, the answer is A.

1

2?0

?1

?4 f (t ) dt

2

2 3¦Ð

? 2x

?

+ sin x ? dx =

? 2

x

¦Ð ?0

+

1

3

?

?

3¦Ð

2

?ln x 2 + 1 ? cos x?

?0

3¦Ð ?

2

=

2

3¦Ð

2

? 2x

?

+ sin x ? dx

? 2

x

+

1

?

?

? 9¦Ð 2

?

+ 1 ? ( ?1)?

?ln

4

??

??

¡Ö 0.879

So, the answer is B.

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2

AP ? Exam Practice Questions for Chapter 4

9. (a)

12

?0

C ¡ä(t ) dt = C (12) ? C (0)

= 40 ? 65

= ? 25¡ãC

The total temperature lost from t = 0 to t = 12 is 25¡ãC.

(b)

1

12

12

?0

Note: Could simply explain as ¡°change in

temperature from time

C (t ) represents the average temperature for

0 ¡Ü t ¡Ü 12.

1

12

12

?0 C (t ) dt

=

1 ? 3

C (t ) dt +

12 ???0

+

=

5

7

?3 C (t ) dt + ?5 C (t ) dt

?

?7 C (t ) dt + ?8 C (t ) dt ??

8

12

1 ?3 ? 0

5?3

(65 + 57) +

(57 + 50)

12 ?? 2

2

7 ?5

(50 + 46)

2

8?7

+

(46 + 44)

2

12 ? 8

+

(44 + 40)??

2

?

+

¡Ö 49.917¡ãC

So, the average temperature of the coffee is

about 49.917¡ãC.

(c) C ¡ä( 4) ¡Ö

C (5) ? C (3)

5?3

50 ? 57

2

= ? 3.5

=

1 pt: difference quotient with units

So, when t = 4, the temperature of the coffee is

changing about ? 3.5¡ãC per minute.

(d) C (t ) =

? C¡ä(t ) dt

=

? ? 2 cos 0.5t dt

? 1 ?

= ? 2?

? ? cos(0.5t )(0.5) dt

? 0.5 ?

= ? 4(sin 0.5t ) + K

= ? 4 sin 0.5t + K

Because C(t ) is continuous at C = 12, use C (12) = 40

to find K.

? 4 sin ??0.5(12)?? + K = 40

K ¡Ö 38.8823

Reminder: Use more than three decimal places

when representing the constant of integration

(avoid premature rounding in this intermediate

step) so that the final answer can be rounded to

three decimal places. Perhaps store the value of

the constant in your calculator for use in the

subsequent computation.

So, C (15) = ? 4 sin (0.5 ? 15) + 38.8823

¡Ö 35.130¡ãC.

Reminder: Round each answer to at least three decimal

places to receive credit on the exam.

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AP ? Exam Practice Questions for Chapter 4

10. (a)

6

?0 M (t ) dt

=

6

?0

¦Ð

3

¦Ðt

dt

12

6

¦Ð ? 6 ?¦Ð ?

= 2 ? sin

? ?dt

0

12 ? 12 ?

6

sin

6

¦Ðt ?

?

= 2 ?? cos ?

12

?

?0

= 2 ??0 ? ( ?1)??

= 2

So, 2 inches of snow will melt during the 6 hour

period.

(b) S (3) ? M (3)

¦Ð?

?¦Ð

= ??0.006(9) ? 0.12(3) + 0.87?? ? ? sin ?

4?

?6

¡Ö 0.194 in. h

(c) I (t ) =

=

? ??s(t ) ? M (t )?? dt

?

? ??(0.006t

2

? 0.12t + 0.87) ?

¦Ð

6

= 0.002t 3 ? 0.06t 2 + 0.87t + 2 cos

sin

¦Ðt

12

¦Ðt ?

dt

12 ??

+C

I (0) = 40 = 2 + C ? C = 38

I (t ) = 0.002t 3 ? 0.06t 2 + 0.87t + 2 cos

¦Ðt

+ 38

12

(d) I ¡ä(t ) = 0 ? S (t ) = M (t )

t ¡Ö 4.2406 hours

(using a graphing utility)

I ( 4.2406) ¡Ö 41.6519 inches

Reminders:

Round each answer to at least three decimal places to receive

credit.

No work is needed for these computations (after setting each

up); use your calculator to compute/evaluate/solve.

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4

AP ? Exam Practice Questions for Chapter 4

11. (a) v(t ) = sin

¦Ðt

2 pts: answers with reason

4

The particle is moving to the right on the t-intervals

(0, 4) and (8, 9) because v(t ) > 0 on these intervals.

(b)

4

? 0 sin

¦Ðt

4

¦Ðt

8

?4 sin

dt ?

4

dt +

9

?8 sin

¦Ðt

4

dt

Reminder: Be sure to explicitly identify each function

by name. Referring to

as ¡°it,¡± ¡°the function,¡± or

¡°the graph¡± may not receive credit on the exam.

1 pt: definite integrals

¦Ð

¦Ðt

cos

4

4

a(3) = v¡ä(3)

(c) v¡ä(t ) =

¦Ð (3)

cos

4

4

¡Ö ? 0.555

¦Ð

=

Reminder:

Because a(3) < 0 and v(3) > 0, the acceleration

and velocity are in opposite directions. This

means that the particle is slowing down.

(d) s(t ) =

=

=

does not need to be evaluated or

simplified. Leaving the answer as

or

is sufficient.

? v(t ) dt

? sin

4

¦Ð

= ?

¦Ðt

dt

4

? sin

¦Ðt ?¦Ð ?

4

¦Ðt

¦Ð

cos

Reminder: The answer does not need to be evaluated

or simplified.

? ? dt

4 ?4?

4

+C

Use s(0) = ? 4 to find C.

?4 = ?

?4 +

4

¦Ð

4

¦Ð

cos 0 + C

= C

So, s (t ) = ?

4

¦Ð

cos

¦Ðt

4

Therefore, s (3) = ?

4

¦Ð

+

4 ? 4¦Ð

cos

¦Ð

.

3¦Ð

4 ? 4¦Ð

+

4

¦Ð

¡Ö ? 1.83.

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AP ? Exam Practice Questions for Chapter 4

12. F ( x ) =

5

x

?3 f (t ) dt

0

(a) F (0) =

?3 f (t ) dt

3

= ? ? f (t ) dt

0

= ? ? 14 ¦Ð ( 2) +

?

5

= ?¦Ð ?

2

2

1

2

(2 + 3)(1)??

F ¡ä(0) = f (0) = 3

F ( 4) =

=

4

?3 f (t ) dt

1

2

(?1)(1)

= ? 0.5

(b) The graph of F ( x) does not have a minimum value

because F ¡ä( x) = f ( x) does not change from negative

2 pts: answer with reason

does not change from negative to positive on this

to positive at any point.

(c) Because F ¡ä( x) = f ( x) changes from increasing to

2 pts: answer with reason

decreasing at x = 0, an inflection point of the graph

of F ( x) is x = 0.

(d) Because F ¡ä( 2) = f ( 2) = 1, the slope of the tangent

line is 1. Use F ( 2) =

2

?3 f (t ) dt

= ? 12 and m = 1

to write the equation of the tangent line.

y +

1

2

= 1( x ? 2)

y = x ?

5

2

So, the equation of the tangent line at F ( 2)

is y = x ? 52 .

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