AP Calculus BC Practice Exam 1 393
- The correct answer is choice B.
Examine the absolute values of
∑∞
n= 1
∣∣
∣∣(−1)
n
n+ 3
∣∣
∣∣=
∑∞
n= 1
1
n+ 3
=
1
4
+
1
5
+
1
6
+.... Since
∑∞
n= 1
1
n+ 3
is a harmonic series that diverges, the
series
∑∞
n= 1
∣∣
∣∣(−1)
n
n+ 3
∣∣
∣∣diverges, which means the
series
∑∞
n= 1
(−1)n
n+ 3
does not converge absolutely.
Now examine the alternating series
∑∞
n= 1
(−1)n
n+ 3
=−
1
4
+
1
5
−
1
6
+.... Since
∑∞
n= 1
(−1)n
n+ 3
is
an alternating harmonic series,
∑∞
n= 1
(−1)n
n+ 3
converges (but not absolutely as shown earlier).
Thus the series
∑∞
n= 1
(−1)n
n+ 3
converges
conditionally.
- The correct answer is (A).
∫ 3
− 1
f(x)dx=
∫ 1
− 1
f(x)dx+
∫ 3
1
f(x)dx
=
1
2
π(1)^2 −
1
2
π(1)^2 = 0
- The correct answer is (A).
We knowf(x)=cosx=
∑∞
n= 0
(−1)nx^2 n
(2n)!
= 1 −
x^2
2!
+
x^4
4!
−
x^6
6!
+···.
Substitute
√
tforx, and cos
√
t
= 1 −
t
2!
+
t^2
4!
−
t^3
6!
+···+
(−1)ntn
(2n)!
+···.The
integral,
∫x
0
cos
√
tdt, will be equal to
∫x
0
(
1 −
t
2!
+
t^2
4!
−
t^3
6!
+···+
(−1)ntn
(2n)!
+···
)
dt,
which, when integrated term by term, is
t−
t^2
2 ·2!
+
t^3
3 ·4!
−
t^4
4 ·6!
+···
+
(−1)ntn+^1
(n+1)(2n)!
+···
∣∣
∣∣
x
0
=x−
x^2
2 ·2!
+
x^3
3 ·4!
−
x^4
4 ·6!
+···
+
(−1)nxn+^1
(n+1)(2n)!
+···.
The series expansion for
∫ x
0
cos
√
tdtis
x−
x^2
2 ·2!
+
x^3
3 ·4!
−
x^4
4 ·6!
+···
+
(−1)nxn+^1
(n+1)(2n)!
+···.
- The correct answer is (B).
∫k
−k
f(x)dx= 2
∫ 0
−k
f(x)dx⇒f(x) is an even
function, i.e.,f(x)=f(−x).
The graph in (B) is the only even function.
