Fall 04 Midterm Answers

Fall 04 Midterm Answers
1
sin(tan−1 x) tan−1 x
1
π
dx = √ (1 − )
2+1
x
4
2
0
−1
u = tan x
dx
du = 2
x +1
Z
1.
1
Z
0
sin(tan−1 x) tan−1 x
dx =
x2 + 1
π/4
Z
u sin u du
0
π/4
= −u cos u + sin u
0
1
π
= √ (1 − )
4
2
Z
2.
x2 − x + 2
dx = 2 ln |x| − 1/2 ln(x2 + 1) − tan−1 x + C
x3 + x
x2 − x + 2
A Bx + C
= + 2
3
x +x
x
x +1
x2 − x + 2 = A(x2 + 1) + (Bx + C)x
A = 2 C = −1 B = −1
Z
√
Z
3.
e
1−x2
x
Z
Z
Z
2
x
dx
x2 − x + 2
dx
=
dx
−
dx
−
x3 + x
x
x2 + 1
x2 + 1
= 2 ln |x| − 1/2 ln(x2 + 1) − tan−1 x + C
dx
· √
= −e
2
x 1 − x2
√
1−x2
x
+C
x = sin θ
dx = cos θ dθ.
√
Z
e
1−x2
x
·
dx
√
=
x2 1 − x2
Z
Z
=
cos θ
e sin θ ·
cos θ dθ
sin2 θ cos θ
ecot θ · csc2 θ dθ
= −ecot θ + C
√
= −e
1−x2
x
+C
Z
π
4.
0
√
sin x
dx
x(1 + x)
Converges.
√
√
As x → 0, sin x ∼ x, so sin x ∼ x.
√
√
sin x
x
1
∼
= 1/2 .
x(1 + x)
x·2
2x
Z π
dx
converges, so by limit comparison test, the original integral
1/2
2x
0
converges.
Z ∞
ln(5 + cos x)
5.
dx
Diverges.
x2
0
Z ∞
Z 1
Z ∞
ln(5 + cos x)
ln(5 + cos x)
ln(5 + cos x)
=
+
2
2
x
x
x2
0
0
1
Z 1
ln 6
ln(5 + cos x)
ln 6
∼ 2 .
dx diverges, so first integral diAs x → 0,
2
x2
x
0 x
verges by limit comparison test. Since one part of the original integral
diverges, the whole integral must diverge.
∞
X
7n ln n
6.
n!
n=3
Converges.
n!
an+1
7n+1 ln(n + 1)
· n
= lim
n→∞ an
n→∞
(n + 1)!
7 ln n
7
ln(n + 1)
= lim
·
n→∞ n + 1
ln n
=0<1
lim
so series converges by Ratio Test.
7.
∞
X
n=1
sin(
1
) ln n
n2
Converges.
As n → ∞, 1/n2 → 0, so sin(1/n2 ) ∼ 1/n2 . ln n < n1 /2, so
∞
∞
∞
X
X
1
ln n X n1/2
<
=
.
2
2
3/2
n
n
n
n=1
n=1
n=1
The right hand series converges, so the left hand series does as well. (Direct
Comparison.) Then the original series converges, by Limit Comparison.
2
8.
∞
X
89
2n + 3n+1
=
n
5
30
n=2
∞
∞
∞
X
X
2n X 3n+1
2n + 3n+1
=
+
5n
5n n=2 5n
n=2
n=2
4
8
27
81
+
+ . . .) + ( +
+ . . .)
25 125
25 125
4
2
4
27
3
9
=
(1 + +
+ . . .) + (1 + +
+ . . .)
25
5 25
25
5 25
4
1
27
1
=
·
+
·
25 1 − 2/5 25 1 − 3/5
89
=
30
=(
9.
∞
X
n=1
(1 +
x n2
) converges when x < 0.
n
Use the root test.
lim a1/n
= lim (1 + x/n)n
n
n→∞
2
/n
n→∞
= lim (1 + x/n)n
n→∞
= ex .
When this limit is < 1, the series converges. Thus the series converges
when ex < 1, i.e. when x < 0. When this limit is > 1, the series diverges.
Thus the series diverges when ex > 1, i.e. when x > 0. When the limit is
equal to 1 i.e. when x = 0, the test doesn’t tell us anything, but in this
case, the series is
∞
∞
X
X
2
(1)n =
1
n=1
n=1
which obviously diverges.
3