and then, convinced of the error, he abruptly dismissed the meeting – to the astonishment of most of the audience.
It was an evidence of intellectual courage as well as honesty and doubtless won for him the supreme admiration
of every person in the group – an admiration which was in no ways diminished, but rather increased, when at a
later meeting he announced that after all he had been able to prove the step to be correct.
—H. E. Slaught
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124
The AP CALCULUS PROBLEM BOOK
4.19
Multiple Choice Problems on Integrals
1102
(AP)
.
For any real number
b
,
b
0

2
x

dx
is
A)
−
b

b

B)
b
2
C)
−
b
2
D)
b

b

E) None of these
1103
(AP)
.
Let
f
and
g
have continuous first and second derivatives everywhere. If
f
(
x
)
≤
g
(
x
)
for all real
x
, which of the following must be true?
I)
f
′
(
x
)
≤
g
′
(
x
) for all real
x
II)
f
′′
(
x
)
≤
g
′′
(
x
) for all real
x
III)
1
0
f
(
x
)
dx
≤
1
0
g
(
x
)
dx
A) None
B) I only
C) III only
D) I and II
E) I, II, and III
1104
(AP)
.
Let
f
be a continuous function on the closed interval [0
,
2]. If 2
≤
f
(
x
)
≤
4, then
the greatest possible value of
2
0
f
(
x
)
dx
is
A) 0
B) 2
C) 4
D) 8
E) 16
1105
(AP)
.
If
f
is the continuous, strictly increasing function on the interval [
a, b
] as shown
below, which of the following must be true?
I)
b
a
f
(
x
)
dx < f
(
b
)(
b
−
a
)
II)
b
a
f
(
x
)
dx > f
(
a
)(
b
−
a
)
III)
b
a
f
(
x
)
dx
=
f
(
c
)(
b
−
a
) for some
c
in [
a, b
].
a
b
f
(
x
)
A) I only
B) II only
C) III only
D) I and II
E) I, II, and III
1106
(AP)
.
Which of the following definite integrals is
not
equal to zero?
A)
π
−
π
sin
3
x dx
B)
π
−
π
x
2
sin
x dx
C)
π
0
cos
x dx
D)
π
−
π
cos
3
x dx
E)
π
−
π
cos
2
x dx
Archimedes will be remembered when Aeschylus is forgotten, because languages die and mathematical ideas
do not.
—G. H. Hardy
CHAPTER 4.
INTEGRALS
125
1107.
π
/
2
π
/
6
cot
x dx
=
A) ln
1
2
B) ln 2
C)
1
2
D) ln(
√
3
−
1)
E) None of these
1108.
3
−
2

x
+ 1

dx
=
A)
5
2
B)
17
2
C)
9
2
D)
11
2
E)
13
2
1109.
2
1
(3
x
−
2)
3
dx
=
A)
16
3
B)
63
4
C)
13
3
D)
85
4
E) None of these
1110.
π
/
2
π
/
4
sin
3
θ
cos
θ
d
θ
=
A)
3
16
B)
1
8
C)
−
1
8
D)
−
3
16
E)
3
4
1111.
1
0
e
x
(3
−
e
x
)
2
dx
=
A) 3 ln(
e
−
3)
B) 1
C)
1
3
−
e
D)
e
−
1
2(3
−
e
)
E)
e
−
2
3
−
e
1112.
0
−
1
e
−
x
dx
=
A) 1
−
e
B)
1
−
e
e
C)
e
−
1
D) 1
−
1
e
E)
e
+ 1
1113.
1
0
x
x
2
+ 1
dx
=
A)
π
4
B) ln
√
2
C)
1
2
(ln 2
−
1)
D)
3
2
E) ln 2
Anyone who cannot cope with mathematics is not fully human. At best he is a tolerable subhuman who has
learned to wear shoes, bathe, and not make messes in the house.
—Robert A. Heinlein
126
The AP CALCULUS PROBLEM BOOK
1114.
The acceleration of a particle moving along a straight line is given by
a
= 6
t
. If, when
t
= 0 its velocity
v
= 1 and its distance
s
= 3, then at any time
t
the position function is given
by
A)
s
=
t
3
+ 3
t
+ 1
B)
s
=
t
3
+ 3
C)
s
=
t
3
+
t
+ 3
D)
s
=
1
3
t
3
+
t
+ 3
E)
s
=
1
3
t
3
+
1
2
t
2
+ 3