AP® Exam Practice Questions for Chapter 4 1

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AP® Exam Practice Questions for Chapter 4

1. Use the Trapezoidal Rule to find an approximation of ( )4

0.g x dx

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

( )

4

0

4 00 2 3 2 7 2 12 2 18 2 25 2 33 2 42 52

2 8

1332 83

4

g x dx −≈ + + + + + + + +

= =

So, the answer is B.

2. ( ) ( ) ( )

( ) ( )

5 3 5

0 0 3

13 5 3 2 2

28

f x dx f x dx f x dx= +

= + + −

=

So, the answer is C.

3. ( )2

4

5 9dx

x − +

Let 5, , and 3.u x du dx a= − = =

2 2

1

1 14 4 arctan

1 54 arctan

3 3

4 5tan

3 3

udx Cu a a a

x C

x C−

= + + − = +

−= +

So, the answer is A.

4. ( ) 34 4 , 0 2v t t t t= − ≤ ≤

( ) ( )

( )

2 2 3

0 0

24 2

0

1 14 4

2 0 21

221

16 824 units sec

v t dt t t dt

t t

= −−

= −

= −

=

So, the answer is A.

5. ( ) ( )

( )

( ) ( ) ( )

( ) ( )( ) ( )

1

4

4

1

0 3 4

1 0 3

1 1 12 2 2

1

2 4 2 2

4

g f t dt

f t dt

f t dt f t dt f t dt

−

−

−

− =

= −

= − − −

= − − − − −

=

So, the answer is C.

6.

[ ]

sin 0.4

cos 0.4

cos cos 0.4

1 cos 0.4

cos 0.6

2.214

b

b

x dx

x

bbbb

π

π

π

=

− =

− + =+ =

= −≈

So, the answer is D.

7. ( ) 3 6

1, 5

f x x

a b

′ = +

= =

( ) ( ) ( )b

af b f a f x dx′= +

( ) ( ) 5 3

15 1 6

2 24.672

26.672

f f x dx= + +

≈ +=

So, the answer is D.

8. ( )

( )

3 23 2 2

00

2

1 2 1sin cos

3 3 202

2 90 0 1

3 8

2.568

x x dx x xπ

π

π π

ππ

+ = − −

= − − −

≈

So, the answer is C.

2 AP® Exam Practice Questions for Chapter 4

© 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.

9. (a) ( ) ( ) ( )12

012 0

40 65

25 C

C t dt C C′ = −

= −= − °

The total temperature lost from 0t = to 12 is 25 C.t = °

(b) ( )12

0

1

12C t represents the average temperature for

0 12.t≤ ≤

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

( )

( )

( )

12 3 5 7

0 0 3 5

8 12

7 8

1 1

12 12

1 3 0 5 365 57 57 50

12 2 2

7 550 46

28 7

46 442

12 844 40

249.917 C

C t dt C t dt C t dt C t dt

C t dt C t d

= + +

+ +

− −= + + +−+ +

−+ +

−+ +

≈ °

So, the average temperature of the coffee is about 49.917 C.°

(c) ( ) ( ) ( )5 3 50 574

5 3 23.5

C CC

− −′ ≈ =−

= −

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

changing about 3.5 C− ° per minute.

(d) ( ) ( )

( )( )

( )

2 cos 0.5

12 cos 0.5 0.5

0.5

4 sin 0.5

4 sin 0.5

C t C t dt t dt

t dt

t K

t K

′= = −

= −

= − +

= − +

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

to find K.

( )4 sin 0.5 12 40

38.8823

K

K

− + = ≈

So, ( ) ( )15 4 sin 0.5 15 38.8823

35.130 C.

C = − ⋅ +

≈ °

Note: Could simply explain as “change in temperature from time

1 pt: difference quotient with units

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.

Reminder: Round each answer to at least three decimal places to receive credit on the exam.

AP® Exam Practice Questions for Chapter 4 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) ( )

( )

6 6

0 0

6

0

6

0

sin6 12

62 sin

12 12

2 cos12

2 0 1

2

tM t dt dt

dt

t

π π

π π

π

=

− =

= −

= − − =

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

(b) ( ) ( ) ( )2

2

40

0.006 0.12 0.87 sin 406 12

0.006 0.12 40.87 sin6 12

I t S t M t

tt t

tt t

π π

π π

= − +

= − + − +

= − + −

(c) ( )

( ) ( )2

0.012 0.12 cos6 12 12

3 0.012 3 0.12 cos72 4

0.181 in. h

tI t t

I

π π π

π π

′ = − −

′ = − −

≈ −

(d) Because I(t) is decreasing over [ ]0, 6 , the maximum

is at 0.t =

( )0 40.87,I = so the maximum amount of snow is

40.87 inches.

Reminder: Round each answer to at least three decimal places to receive credit on the exam.

4 AP® Exam Practice Questions for Chapter 4

© 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) ( ) sin4

tv t π=

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

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

(b) 9

0sin

4

t dtπ

(c) ( )

( ) ( )( )

cos4 4

3 3

3cos

4 4

tv t

a v

π π

ππ

′ =

′=

=

Because 3 4π is in Quadrant II, ( )cos 3 4π is

negative. So, ( )3 0.a < Because ( )3 0,v > the

acceleration and velocity are in opposite directions. This means that the particle is slowing down.

(d) ( ) ( )

sin4

4sin

4 4

4cos

4

s t v t dt

t dt

t dt

t C

π

π ππ

ππ

=

=

=

= − +

Use ( )0 4s = − to find C.

44 cos 0

44

C

C

π

π

− = − +

− + =

So, ( ) 4 4 4cos .

4

ts t π ππ π

−= − +

Therefore, ( ) 4 3 4 43 cos .

4 4s π π

π−= − +

2 pts: answers with reason

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: definitel integral

Reminder: does not need to be evaluated or

simplified. Leaving the answer as or

is sufficient.

Reminder: The answer does not need to be evaluated or simplified.

AP® Exam Practice Questions for Chapter 4 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. ( ) ( )3

xF x f t dt=

(a) ( ) ( )

( )

( ) ( )( )

0

3

3

0

21 14 2

0

2 2 3 1

5.64

F f t dt

f t dt

π

=

= −

= − + + ≈ −

( ) ( )

( ) ( )

( )( )

4

3

12

0 0 3

4

1 1 0.5

F f

F f t dt

′ = =

=

= − = −

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

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

to positive at any point.

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

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

of ( )F x is 0.x =

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

line is 1. Use ( ) ( )2

3

12

2F f t dt= = − and 1m =

to write the equation of the tangent line.

( )12

52

1 2y x

y x

+ = −

= −

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

is 52.y x= −

2 pts: answer with reason

does not change from negative to positive on this

2 pts: answer with reason

6 AP® Exam Practice Questions for Chapter 4

© 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) ( )

( ) ( )( )( )

1

0 5.5

1 0 5.5

9 0 5.5

0 14.5

f x dx

F F

F

F

= −

− = −

− = −

=

( )

( ) ( )( )

( )

3

1 6

3 1 6

3 9 6

3 3

f x dx

F F

F

F

= −

− = −

− = −

=

( )

( ) ( )( )

( )

4

3 15.5

4 3 15.5

4 3 15.5

4 18.5

f x dx

F F

F

F

=

− =

− =

=

So, ( ) ( )0 14.5 and 4 18.5.F F= =

(b) Because ( ) ( ) ( )0 5 3 and F F F x> > is continuous

on [ ]0, 3 , there is at least one x-value in this

interval where ( ) 5F x = by the Intermediate

Value Theorem. Because ( ) ( )3 5 4F F< <

and ( )F x is continuous on [ ]3, 4 , there is at least

one x-value in this interval where ( ) 5.F x = So,

( )F x must equal 5 at least 2 times on [ ]0, 4 .

(c) Because ( ) ( ) 0F x f x′ = > on the interval ( )3, 4 ,

F is increasing on the interval ( )3, 4 .

3 pts: answer with reason

2 pts: answer with reason