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