12.
∞
∞
X
X
sin(n + 12 )π
(−1)n
=
converges by the alternating series test but only condiln ln n
ln ln n
n=10
n=10
∞
X
1
1
diverges to infinity by comparison with
.
tionally since
ln ln n
n
n=10
n=10
(ln ln n < n for n ≥ 10.)
∞
X
25. Let u = x − 1. Then x = 1 + u, and
x ln x = (1 + u) ln(1 + u)
∞
X
un
= (1 + u)
(−1)n−1
n
(−1 < u ≤ 1)
n=1
∞
X
=
n−1 u
(−1)
n
n
n=1
+
∞
X
(−1)n−1
n=1
u n+1
.
n
Replace n by n − 1 in the last sum.
x ln x =
∞
X
n−1 u
(−1)
n
n=1
=u+
∞
X
n
+
n−1
(−1)
n=2
∞
X
(−1)
n=2
un
n−1
un
n−2
1
1
−
n
n−1
∞
X
(−1)n
= (x − 1) +
(x − 1)n
n(n − 1)
n=2
(0 ≤ x ≤ 2).
49. The Fourier sine series of f (t) = π − t on [0, π ] has coefficients
2
bn =
π
∞
X
2
sin(nt).
The series is
n
n=1
Z
0
π
(π − t) sin(nt) dt =
2
.
n
17.
π
x = sin4 t, y = cos4 t, 0 ≤ t ≤
.
2
Z π/2
(cos4 t)(4 sin3 t cos t) dt
Area =
0
Z π/2
cos5 t (1 − cos2 t) sin t dt
=4
Let u = cos t
du = − sin t dt
Z 1
1 1
1
=4
(u 5 − u 7 ) du = 6
−
= sq. units.
6 8
6
0
0
y
x=sin4 t
y=cos4 t
0≤t≤π/2
A
x
Fig. 4-17
19.
a) The density function for the uniform distribution on [a, b] is given by f (x) = 1/(b−a),
for a ≤ x ≤ b. By Example 5, the mean and standard deviation are given by
µ=
b+a
,
2
σ =
b−a
√ .
2 3
b−a
b+a
> b, and similarly, µ − 2σ < a, therefore
+ √
2
3
Pr(|X − µ| ≥ 2σ ) = 0.
1
b) For f (x) = ke−kx on [0, ∞), we know that µ = σ = (Example 6). Thus µ−2σ < 0
k
3
and µ + 2σ = . We have
k
3
Pr(|X − µ| ≥ 2σ ) = Pr X ≥
k
Z ∞
e−kx d x
=k
3/k
∞
−kx = −e = e−3 ≈ 0.050.
Since µ + 2σ =
3/k
c) For f µ,σ (x) =
have
1
2
2
√ e(x−µ) /2σ , which has mean µ and standard deviation σ , we
σ 2π
Pr(|X − µ| ≥ 2σ ) = 2Pr(X ≤ µ − 2σ )
Z µ−2σ
1
2
2
=2
√ e−(x−µ) /2σ d x
σ 2π
−∞
x −µ
Let z =
σ
1
dz = d x
σ
Z −2
2
2
e−z dz
=√
2π −∞
= 2Pr(Z ≤ −2) ≈ 2 × 0.023 = 0.046
from the table in this section.
15.
y
x 2 +y 2 =a 2
ds
θ dθ
−a
s
xa
Fig. 4-15
Consider the area element which is the thin half-ring shown in the figure. We have
dm = ks π s ds = kπ s 2 ds.
kπ 3
Thus, m =
a .
3
Regard this area element as itself composed of smaller elements at positions given by the
angle θ as shown. Then
d M y=0 =
Z
π
(s sin θ )s dθ ks ds
0
3
M y=0
= 2ks ds,
Z a
ka 4
.
s 3 ds =
= 2k
2
0
ka 4
3
3a
Therefore, y =
·
=
. By symmetry, x = 0. Thus, the centre of mass of the
3
2π
2 kπ a
3a
plate is 0,
.
2π
6.
q
2(x + 1)3 = 3(y − 1)2 , y = 1 + 23 (x + 1)3/2
q
′
y = 32 (x + 1)1/2 ,
r
r
3x + 3
3x + 5
ds = 1 +
dx =
dx
2
2
√
0
Z 0√
2
1
3/2 3x + 5 d x =
(3x + 5) L= √
9
2 −1
−1
√
2 3/2
=
5 − 23/2 units.
9
9. a) About the x-axis:
1
1
4−
V =π
dx
Let x = tan θ
(1 + x 2 )2
0
d x = sec2 θ dθ
Z π/4
2
sec θ
dθ
= 4π − π
sec4 θ
0
Z π/4
cos2 θ dθ
= 4π − π
0
π/4
π
= 4π − (θ + sin θ cos θ )
0
2
π2
π
15π
π2
= 4π −
− =
−
cu. units.
8
4
4
8
Z
b) About the y-axis:
Z
1
x 2−
1
1 + x2
dx
1
1
2 2
= 2π x − ln(1 + x ) 2
0
1
= 2π 1 − ln 2 = 2π − π ln 2 cu. units.
2
V = 2π
0
y
y=2
y=
x
1
Fig. 1-9
1
1+x 2
x
21.
Z
=
ln(ln x)
dx
x
Z
Let u = ln x
dx
du =
x
ln u du
U = ln u
du
dU =
Zu
= u ln u −
d V = du
V =u
du = u ln u − u + C
= (ln x)(ln(ln x)) − ln x + C.
27.
P
P
(−1)n
is a couna) “ an converges implies (−1)n an converges” is FALSE. an =
n
terexample.
P
P
P
b) “ an converges and (−1)n an converges implies
an converges absolutely” is
FALSE. The series of Exercise 25 is a counterexample.
P
P
c) “ an converges absolutely implies (−1)n an converges absolutely” is TRUE, because
|(−1)n an | = |an |.
P
P
28. “If an and bn both diverge, then so does
P
P
P
1
1
(an + bn )” is FALSE. Let an = and bn = − , then an = ∞ and bn = −∞ but
n
n
P
P
(an + bn ) = (0) = 0.
36.
a) “If lim an = ∞ and lim bn = L > 0, then lim an bn = ∞” is TRUE. Let R be an
arbitrary, large positive number. Since lim an = ∞, and L > 0, it must be true that
L
2R
for n sufficiently large. Since lim bn = L, it must also be that bn ≥
for
an ≥
L
2
2R L
= R for n sufficiently large. Since R is
n sufficiently large. Therefore an bn ≥
L 2
arbitrary, lim an bn = ∞.
b) “If lim an = ∞ and lim bn = −∞, then lim(an + bn ) = 0” is FALSE. Let an = 1 + n
and bn = −n; then lim an = ∞ and lim bn = −∞ but lim(an + bn ) = 1.
c) “If lim an = ∞ and lim bn = −∞, then lim an bn = −∞” is TRUE. Let R be an
arbitrary, large
Since lim an = ∞ and lim bn = −∞, we must
√ positive number.
√
have an ≥ R and bn ≤ − R, for all sufficiently large n. Thus an bn ≤ −R, and
lim an bn = −∞.
d) “If neither {an } nor {bn } converges, then {an bn } does not converge” is FALSE. Let
an = bn = (−1)n ; then lim an and lim bn both diverge. But an bn = (−1)2n = 1 and
{an bn } does converge (to 1).
e) “If {|an |} converges, then {an } converges” is FALSE. Let an = (−1)n . Then limn→∞ |an | = limn→∞ 1 = 1,
but limn→∞ an does not exist.
n 2
∞ X
n
39.
converges by the root test of Exercise 31 since
n+1
n=1
σ = lim
n→∞
n
n+1
n 2 1/n
= lim n→∞
1
1+
=
1 n
n
1
< 1.
e