MAT 296
Exam 2
18 March 2014
Name:
Instructions: Answer all questions and justify your claims. The test is out of 70 points. Show all of your
work or you will not receive credit. Electronic devices and all other resources are not allowed.
1. (10 points) Find the volume of the solid obtained by rotating, about the y axis, the region bounded by
the curves y = 0 and y = −x2 + x.
Z
V =
Z
2πrh = 2π
0
1
x(−x2 + x) dx = 2π
1
π
= .
12
6
3
2. (10 points) Find the length of the curve given by the equation y = 1 + 6x 2 , 0 ≤ x ≤ 1.
Z
L=
Z
ds =
0
1
Z 1
q
√
√ 2
1 + (9 x) dx =
1 + 81x dx
0 3
2 1
(1 + 81x) 2 ]10
=
3 81
3
2
=
(89 2 − 1)
81(3)
√
2(89 89 − 1)
=
.
243
√
3. (10 points) Find the area of the surface obtained by rotating the curve given by y = 1 + ex , 0 ≤ x ≤ 1
about the x axis.
s
2
Z
Z
Z 1
√
dy
x
dx
SA = 2πr ds = 2π y ds =2π
1+e
1+
dx
0
s
2
Z 1
√
ex
√
dx
=2π
1 + ex 1 +
2 1 + ex
0
s
Z 1
√
e2x
=2π
1 + ex 1 +
dx
4(1 + ex )
0
r
Z 1
√
4 + 4ex + (ex )2
x
=π
1+e
dx
1 + ex
0
Z 1p
=π
(ex + 2)2 dx
0
Z
=π
1
ex + 2 dx
0
=π(e + 1).
4. (10 points) Determine whether the sequence converges or diverges. If it converges, find the limit.
tan
Note that f (x) = tan(x) is continuous at
π
4
lim
2nπ
1 + 8n
since tan(x) =
n→∞
sin(x)
cos(x)
and cos( π4 ) =
√1
2
2nπ
2π
π
=
=
1 + 8n
8
4
so that
lim tan
n→∞
2nπ
1 + 8n
In particular, the sequence converges.
= tan
2nπ
n→∞ 1 + 8n
lim
= tan
π
4
= 1.
6= 0. Also
5. (10 points) Determine whether the series is convergent or divergent.
∞
X
1
ln(n + 1)
n=1
By L’Hospital’s Rule
lim
x→∞
ln(x)
1
= lim
=0
x→∞ x
x
so that for large enough x we have
1
ln(x)
≤ .
x
2
In particular ln(x) ≤
and hence
x
2
≤ x for large enough x. Then for n sufficiently large we have ln(n + 1) ≤ n + 1
1
1
≥
.
ln(n + 1)
n+1
Since the first few terms do not affect the convergence of a series, we see that
∞
X
∞
X
1
1
=
n
+
1
n
n=1
n=2
diverges since p = 1. By the comparison test
∞
X
1
ln(n
+ 1)
n=1
also diverges.
6. (10 points) Determine whether the series is convergent or divergent. If it is convergent, find its sum.
1 2
1
2
1
2
+ +
+
+
+
+ ...
3 9 27 81 243 729
We make use of the following fact: If an ≥ 0 for all n and if we can ”rearrange” the order in which we
sum the numbers an to get a convergent series, then the original series also converges to the same sum.
That is,
1
2
1
2
1
1
1
2
2
2
1 2
+ +
+
+
+
+ ··· =
+
+
+ ... +
+
+
+ ...
3 9 27 81 243 729
3 27 243
9 81 721
∞
∞
X
X
2
1
+
=
2n+1
n
3
9
n=1
n=0
n
∞ n
1X 1
2X 1
=
+
3 n=0 9
9 n=0 9
1
5
=
9 1 − 91
5
= .
8
In particular, the series converges.
7. (10 points) Determine whether the series is convergent or divergent.
∞
X
n=2
Note that for x ≥ 2 we have f (x) =
1 √
x ln(x) 2
1
n(ln(n))
√
2
> 0, f is continuous, and
√
√
x < y ⇒ x ln(x) 2 < y ln(y) 2
1
1
√ >
√
⇒
2
x ln(x)
y ln(y) 2
⇒ f (x) > f (y),
so that f is decreasing and we may apply the integral test.
Z
2
∞
1
x ln(x)
which converges since p =
Z
√
2
√
dx = lim
t→∞
2
t
1
Z
√
x ln(x)
2
ln(t)
dx = lim
t→∞
du
√
ln(2)
u
2
Z
∞
=
du
√
ln(2)
u
2 > 1. Therefore the series converges by the integral test.
2
BONUS. (10 points) If f 0 is continuous, use L’Hospital’s Rule to show that
f (x + h) − f (x − h)
= f 0 (x)
h→0
2h
lim
By L’Hospital’s Rule and the chain rule, differentiating the numerator and denominator inside the limit
with respect to h gives
f 0 (x + h) − f 0 (x − h)(−1)
f (x + h) − f (x − h)
= lim
h→0
h→0
2h
2
1
0
= lim f (x + h) + f 0 (x − h).
2 h→0
lim
Since f 0 is continuous at x we have limh→0 f 0 (x + h) = f (x) = limh→0 f 0 (x − h). In particular
1
1
lim f 0 (x + h) + f 0 (x − h) = 2f 0 (x) = f 0 (x).
2 h→0
2