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8/7/2019 Tut6_Soln
http://slidepdf.com/reader/full/tut6soln 1/2
Hints/Solutions for Tutorial Problem Set - 6
1. (a) Since the Riemann integral1 0
sin(x2) dx exists,∞ 0
sin(x2) dx is convergent if ∞ 1
sin(x2) dx
is convergent. Let f (x) = 12x and g(x) = 2x sin(x2) for all x ∈ [1, ∞). Then f is de-
creasing on [1, ∞) and limx→∞
f (x) = 0. Also
x 1
g(t) dt
= | cos1 − cos(x2)| ≤ 2 for all
x ∈ [1, ∞). Hence by Dirichlet’s test,∞ 1
f (x)g(x) dx, i.e.
∞ 1
sin(x2) dx is convergent.
Consequently∞ 0
sin(x2) dx is convergent.
(b) Let f (x) = log x√x
and let g(x) = 1x3/4
for all x ∈ (0, 1]. Then limx→0+
f (x)g(x)
= limx→0+
logxx−1/4
=
limx→0+
1/x
−(1/4)x−5
/4
= limx→0+−
4x1/4 = 0. Since1
0
g(x) dx is convergent, by the limit com-
parison test,1 0
f (x) dx is convergent.
2. (a) Answer : 0 < p < 1(b) Answer : p > −1
3. Since | sinxxp
| ≤ 1xp
for all x ≥ 1 and since∞ 1
1xp
dx converges if p > 1, by the comparison
test,∞ 1
| sinxxp
| dx converges if p > 1. Hence∞ 1
sinxxp
dx converges absolutely if p > 1.
Now we assume that 0 < p ≤ 1. Let f (x) = 1xp and g(x) = sin x for all x ≥ 1. Then
f is decreasing on [1, ∞) and limx→∞
f (x) = 0. Also
x 1
g(t) dt
= | cos1 − cos x| ≤ 2 for
all x ≥ 1. Hence by Dirichlet’s test,∞ 1
f (x)g(x) dx, i.e.
∞ 1
sinxxp
dx converges. If pos-
sible, let∞ 1
| sinxxp
| dx converge. Since sin2 x ≤ | sin x| for all x ≥ 1, by the comparison
test,∞ 1
sin2 xxp
dx converges. Also, by Dirichlet’s test,∞ 1
cos2xxp
dx converges. (This can be
shown in a similar way as was shown earlier.) Hence
∞ 1
2sin2 x+cos2x
xp dx converges,i.e.
∞ 1
1xp
dx converges, which is not true (since 0 < p ≤ 1). Therefore∞ 1
| sinxxp
| dx cannot
converge and consequently∞ 1
sinxxp
dx converges conditionally.
4. At a point of intersection of the cardioid r = a(1 + cos θ) and the circle r = 32
a, wehave a(1 + cos θ) = 3
2a. So θ = π
3corresponds to a point of intersection. Hence the
area of the region that is inside the cardioid r = a(1 + cos θ) and inside the circle
r = 32
a is 2[12
π/3
0
(32
a)2 dθ + 12
π
π/3
a2(1 + cos θ)2 dθ] = (7π4
−9√3
8)a2. Also, the area of
the region that is inside the cardioid r = a(1 + cos θ) and outside the circle r = 32
a is
2[12
π/3 0
a2(1 + cos θ)2 dθ − 12
π/3 0
(32
a)2 dθ] = (9√3
8− π
4)a2.
8/7/2019 Tut6_Soln
http://slidepdf.com/reader/full/tut6soln 2/2
5. Let y = f (x) =x 0
√cos2t dt for all x ∈ [0, π
4]. Then f (x) =
√cos2x for all x ∈ [0, π
4]
(by the first fundamental theorem of calculus). Hence the length of the given curve
is
π4 0
1 + (f (x))2 dx =
π4 0
√1 + cos 2x dx = 1.
6. We have V a = π
a
1
1
x2 dx = π(1 −1
a) → π as a → ∞. Also, S a = 2π
a
1
1
x
1 +
1
x4 dx ≥2π
a 1
1x
dx = 2π log a → ∞ as a → ∞. So S a → ∞ as a → ∞.