Tut6_Soln

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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.

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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 → ∞.