(24%) Evaluate the integral

1. (24%) Evaluate the integral.

(a) ˆ

sin²√

xdx (b)

ˆ 2√ x + 2

(x + 1)²dx (c)

ˆ xdx

√x²+ x + 1 sol. (a) Let √

x = u , dx = 2udu , ˆ

sin²√ xdx =

ˆ

sin²u · 2udu

=

ˆ 1 − cos 2u

2 · 2udu

= ˆ

udu − ˆ

u cos 2udu

= u² 2 − (u

2 · sin 2u − ˆ 1

2sin 2udu)

= u² 2 − u

2 · sin 2u − 1

4cos 2u + C

= x 2 −

√x

2 · sin 2√ x − 1

4cos 2√ x + C (b)Let √

x + 2 = u , dx = 2udu , ˆ 2√

x + 2 (x + 1)²dx =

ˆ 2u

(u²− 1)² · 2udu =

ˆ 4u² (u²− 1)²du

= ˆ

( 1

u − 1 + −1

u + 1 + 1

(u − 1)² + 1

(u + 1)²)du

= ln |u − 1| − ln |u + 1| − 1

u − 1− 1

u + 1 + C

= ln |√

x + 2 − 1| − ln |√

x + 2 + 1| − 1

√x + 2 − 1 − 1

√x + 2 + 1+ C

= −2(tanh⁻¹(√

x + 2) +

√x + 2 x + 1 ) + C (c)

ˆ xdx

√x²+ x + 1 = 1 2

ˆ 2x + 1

√x²+ x + 1dx −1 2

ˆ dx

√x²+ x + 1

= 1 2

ˆ d(x²+ x + 1)

√x²+ x + 1 −1 2

ˆ dx

q (

√ 3

2 ) + (x + ¹₂)²

=√

x²+ x + 1 − 1

2sinh⁻¹(2x + 1

√3 ) + C

1

your answers.

(a) ˆ ∞

0

√ dx

x + x³ (b)

ˆ ₁

−1

x + 1

√5

x⁶ dx sol.

(a) 0 ≤ ˆ ∞

0

√ dx x + x³

=´∞ 1

√dx

x+x³ +´1 0

√dx x+x³

≤´∞ 1

dx x³ +´1

0

√dx x

= −_2x¹2|^∞₁ + 2√ x|¹₀

= ¹₂ + 2

= ⁵₂ < ∞ Then

ˆ ∞ 0

√ dx

x + x³ converges (b)

ˆ 1

−1

x + 1

√5

x⁶ dx

= ˆ 0

−1

x + 1

√5

x⁶ dx + ˆ 1

0

x + 1

√5

x⁶ dx

= ˆ 0

−1

(x⁻¹⁵ + x⁻⁶⁵ )dx + ˆ 1

0

(x⁻¹⁵ + x⁻⁶⁵ )dx

= (⁵₄x^4/5− 5x^−1/5)|⁰₋₁+ (⁵₄x^4/5− 5x^−1/5|¹₀)

= (0 − lim

x→0⁻5x^−1/5) − (5

4 + 5) + (5

4 − 5) − (0 − lim

x→0⁺5x^−1/5)

= ∞ Then

ˆ ₁

−1

x + 1

√5

x⁶ dx diverges

3. (28%)

(a) Find the interval of convergence of

∞

X

n=1

(x + 4)ⁿ n · 2ⁿ sol.

Ratio Test _n+1·2^(x+4)n+1n+1

(x+4)ⁿ

n·2ⁿ = |^x+4₂ | · |_n+1ⁿ | −→ |^x+4₂ | < 1 Then |x + 4| < 2 , that is −6 < x < −2

But when x = −6,

∞

X

n=1

(−1)ⁿ

n converges by Alternating Series Test when x = −2,

∞

X

n=1

2ⁿ n · 2ⁿ =

∞

X

n=1

1

n diverges So convergent interval is −6 ≤ x < −2

(b) Determine whether the series

∞

X

n=1

√1

nsin( 1

√n) tan( 1

√n) is convergent or di- vergent. Give reasons for your answers.

sol.

By Limit Comparison Test lim_n→∞

√1

nsin(^√1_n) tan(^√1_n) n^{− 32} = lim

n→∞n sin(^√1_n) tan(^√1_n)=limn→∞

sin( 1√ n)

√1 n

·

sin( 1√ n)

√1 n

·sec(^√1

n)

= 1

Since

∞

X

n=1

n⁻³² converges,

∞

X

n=1

√1

nsin( 1

√n) tan( 1

√n) converges

(c) Determine whether the series

∞

X

n=1

(−1)ⁿ

√n + 1 is conditionally convergent, ab- solutely convergent, or divergent. Give reasons for your answers.

sol.

Since each term ^√_n+1¹ is positive and decreasing to zero By Alternating Series Test we have derive that

∞

X

n=1

(−1)ⁿ

√n + 1 is convergent Also, |^(−1)√_n+1ⁿ| = ^√_n+1¹ > ^√_n+^1√_n = ₂^√1_n

and

∞

X

n=1

1 2√

n is divergent p-series for p = ¹₂ Hence

∞

X

n=1

| (−1)ⁿ

√n + 1| diverges

Then

∞

X

n=1

(−1)ⁿ

√n + 1 is not absolutely convergent It’s conditionally convergent

(d) Determine whether the series

∞

X

n=1

(−1)ⁿcos n is conditionally convergent, ab-

3

sol.

Since limn rightarrow∞cos n doesn’t exist and limn rightarrow∞cos n 6= 0 We can not apply by the Alternating Series Test to

∞

X

n=1

(−1)ⁿcos n

Hence

∞

X

n=1

(−1)ⁿcos n is divergent

4. (10%)

(a) Find f (x) such that ˆ

(ln x)ⁿdx = f (x) − n ˆ

(ln x)ⁿ⁻¹dx . (b) Prove that

ˆ ₁

0

(ln x)ⁿdx = (−1)ⁿn! for any positive integer n.

sol.

(a) Use integration by parts, ˆ

(ln x)ⁿdx = x(ln x)ⁿ− n ˆ

(ln x)ⁿ⁻¹dx.

∴ f (x) = x(ln x)ⁿ. (b) First we check that lim

b→0⁺x(ln x)^k|¹_b = 0 for any finite positive integer k.

lim

b→0⁺x(ln x)^k|¹_b = 1 · (ln 1)^k− lim

b→0⁺b(ln b)^k. Use L’Hˆopital’s Rule, lim

b→0⁺b(ln b)^k= lim

b→0⁺

(ln b)^k

1 b

= lim

b→0⁺

k(ln b)^{k−1 1}_b

−_b¹2

= −k lim

b→0⁺

(ln b)^k−1

1 b

= · · ·

= (−1)^k−1k! lim

b→0⁺

(ln b)

1 b

= (−1)^k−1k! lim

b→0⁺ 1 b

−_b¹2

= (−1)^kk! lim

b→0⁺b = 0.

Next We use mathematical induction to prove the result.

For n = 1, ˆ 1

0

(ln x)dx = lim

b→0⁺x(ln x)|¹_b − ˆ 1

0

dx = 0 − 1 = −1 (o.k.) Suppose this is true for n = k, that is,

ˆ 1 0

(ln x)^kdx = (−1)^kk!.

Then for n = k + 1, ˆ ₁

0

(ln x)^k+1dx = lim

b→0⁺

x(ln x)^k+1|¹_b − (k + 1) ˆ ₁

0

(ln x)^kdx

= 0 − (k + 1)(−1)^kk! = (−1)^k+1(k + 1)!

By mathematical induction, this is true for any positive integer n.

5

5. (12%) Find the area of the surface generated by revolving the curve y = e , x ≥ 0, about the x-axis. sol. 1

Area = ˆ ∞

0

2πyds = ˆ ∞

0

2πy r

1 + (dy dx)²dx

= ˆ ∞

0

2πe^−2xp

1 + (−2e^−2x)²dx

= −π ˆ ₀

1

√

1 + 4u²du (Let u = e^−2x)

= −π 2

ˆ 0 tan⁻¹(2)

p1 + tan²θ sec²θdθ (Let 2u = tan θ)

= π 2

ˆ tan⁻¹(2) 0

sec³θdθ

= π 2( 1

2sec θ tan θ + 1

2ln | sec θ + tan θ| )|^tan₀ ⁻¹⁽²⁾

= π 4(2√

5 + ln(√

5 + 2))

Here we use ˆ

sec³θdθ = ˆ

sec θ(sec²θdθ)

= ˆ

sec θd tan θ = sec θ tan θ − ˆ

tan θ(tan θ sec θdθ)

= sec θ tan θ − ˆ

(sec²θ − 1) sec θdθ

= sec θ tan θ − ˆ

sec³θdθ + ˆ

sec θdθ

⇒ ˆ

sec³θdθ = 1

2(sec θ tan θ + ln | sec θ + tan θ|)

and

sec(tan⁻¹(2)) =√ 5

sol. 2

Area = ˆ ∞

0

2πyds = ˆ ∞

0

2πe^−2xp

1 + (−2e^−2x)²dx

= −π ˆ ₀

1

√1 + 4u²du (Let u = e^−2x)

= −π ˆ ₀

sinh⁻¹(2)

p1 + sinh²θcosh θ

2 dθ (Let u = sinh θ 2 )

= π 2

ˆ sinh⁻¹(2) 0

cosh²θdθ

= π 2

ˆ _sinh⁻¹₍₂₎

0

cosh 2θ + 1

2 dθ (”double angle formulas”)

= π

2(sinh 2θ

4 + 1

2θ)|^sinh₀ ⁻¹⁽²⁾

= π

2(sinh θ cosh θ

2 + 1

2θ)|^sinh₀ ⁻¹⁽²⁾ (”double angle formulas” again)

= π

2(cosh(sinh⁻¹(2)) + 1

2sinh⁻¹(2))

= π 4(2√

5 + ln(√

5 + 2)) (cosh sinh⁻¹(2) =

q

1 + sinh²(sinh⁻¹(2)) =√

1 + 4 =√ 5) (and sinh⁻¹(x) = ln(x +√

1 + x²))

7

(a) Find the Taylor’s series of sin(ax) − tan⁻¹x − x at x = 0.

(b) Find the value of a for which the limit lim

x→0

sin(ax) − tan⁻¹x − x

x³ + x⁴ is finite and then evaluate the limit.

sol.

(a) sin(ax) =

∞

P

n=0

(−1)ⁿ(ax)²ⁿ⁺¹

(2n+1)! = ax −^(ax)_3!³ + ^(ax)_5!⁵ − . . . tan⁻¹x =

∞

P

n=0

(−1)ⁿx²ⁿ⁺¹

2n+1 = x −^x₃³ +^x₅⁵ − . . .

⇒ sin(ax) − tan⁻¹x − x = (a − 2)x +

∞

P

n=1

(−1)ⁿ[a²ⁿ⁺¹− (2n)!]_(2n+1)!^x2n+1

= (a − 2)x − (a³− 2)^x_3!³ + (a⁵− 4!)^x_5!⁵ − . . . (b) lim

x→0

sin(ax)−tan⁻¹x−x

x⁴+x³ = lim

x→0 a−2

x² −^a3_3!⁻² + ^a5_5!^−4!x²− · · · · _x+1¹ is finite if a = 2

Then lim

x→0

sin(ax)−tan⁻¹x−x

x⁴+x³ = −²³₆⁻² = −1