(c) Evaluate the limit L0 = lim x→0+x1/6 Z 1 √x cos t t4/3 dt

(95 上微甲一組 部分期末考參考答案) 1. (15%)

(a) Find a nonzero number a 6= 0 that satisfies lim

x→∞

 3x + a 3x − a

2x+1

= 4.

(b) Determine whether the improper integral I₀ = Z 1

0

cos t

t^4/3 dt is convergent or divergent.

(c) Evaluate the limit L₀ = lim

x→0⁺x^1/6 Z 1

√x

cos t t^4/3 dt.

Ans. (a) a = ,

(b) I0 is (convergent/divergent), (c) L0 = .

Solution:

(a) Method 1:

x→∞lim(^3x+a_3x−a)^2x+1

= lim

x→∞(1+_3x−a^2a )^2x+1= lim

x→∞(1+_3x−a^2a )^{3x−a2a ·2a(2x+1)(3x−a)} =[ lim

x→∞(1+_3x−a^2a )^3x−a2a ]^x→∞lim

2a(2x+1) 3x−a

=e^43a =4.

⇒ ⁴₃a = 2 ln 2 ⇒ a = ³₂ln 2 Method 2:

x→∞lim[(1 + _3x−a^2a )^3x−a]²³ · (1 + _3x−a^2a )^1+23a

= [ lim

x→∞(1 + _3x−a^2a )^3x−a]²³ · lim

x→∞(1 + _3x−a^2a )^1+23a

= (e^2a)²³ · 1 = e^43a = 4.

⇒ ⁴₃a = 2 ln 2 ⇒ a = ³₂ln 2.

Method 3:

x→∞lim[(1 + _3x−a^2a )^3x−a]²³ · (1 + _3x−a^2a )^1+23ae^{x→∞lim(2x+1) ln(}

3x+a 3x−a)

= 4 By L’ Hˆopital Rule lim

x→∞3a^4x_9x^2+4x+12−a² = ^4a₃ = ln 4

⇒ a = ³₂ ln 2.

(b) lim

t→0⁺ cos t t^4/3 = ∞.

Method 1:

Apply limit comparison test lim

x→0⁺

cos x x4/3 1 x4/3

= lim

x→0⁺cos x = 1 Since R1

0 dx

x^p div. ⇔ p ≥ 1. so R1 0

cos t

t^4/3 dt div.

Method 2:

Apply comparison test ^{cos x}_x4/3 ≤ ^{cos 1}_x4/3, x ∈ (0, 1]

Since R1 0

dx

x^p div. ⇔ p ≥ 1. so R1 0

cos t

t^4/3 dt div.

(c) lim

x→0⁺ R√1

x cos t t4/3dt

x^{− 16} (由(b) 知為^∞_∞ type)

= lim

x→0⁺

− ^cos

√x

(√ x)4/3·¹

2x^−1/2

−¹₆x^−7/6 (L’Hˆopital’s rule)

= lim

x→0⁺3 · cos√ x

= 3.

2. (24%) Evaluate the following three integrals.

I₁ = Z e³

e

(ln x)²dx, I₂ = Z 1/4

0

r x

1 − xdx, I₃ =

Z sin x(1 − cos x)

(sin²x + 2 cos²x)(1 + cos x)² dx .

Ans. I₁ = ,

I₂ = ,

I3 = .

Solution:

(1) (Let u = (ln x)², dv = dx, ⇒ du = 2 ln x · ¹_xdx, v = x.) I₁ =Re³

e (ln x)²dx = x(ln x)²|^e_e³ −Re³

e 2 ln xdx = 9e³− e − 2Re³

e ln xdx (Let u = ln x, dv = dx, ⇒ du = _x¹, v = x.)

=9e³− e − 2(x ln x|^e_e³ −Re³

e 1dx) =9e³− e − 2(3e³− e − e³+ e) = 5e³− 3e.

(2) I2 =R ¹₄

0

p _x

1−x (Let x = sin²θ,√

x = sin θ,√

1 − x = cos θ)

=R ^π₆

0 sin θ

cos θ · 2 sin θ cos θdθ =2R ^π₆

0 sin²θdθ =2R ^π₆

0

1−cos 2θ 2 dθ

=[θ − ^{sin sin 2θ}₂ ]|

π 6

0 = ^π₆ −

√3 4 . (3) I₃ =R sin x(1−cos x)

(sin²x+2 cos²x)(1+cos x)² dx

= −R _{1−cos x}

(1+cos²x)(1+cos x)² d cos x (Let u = cos x)

= −R ¹₂

1+u +_(1+u)¹ 2 + ⁻

1 2(u+1) 1+u² du

= −¹₂ln |1 + u| − _1+u¹ + ¹₂R _u

1+u²du + ¹₂R ₁

1+u²du

= −¹₂ln |1 + u| − _1+u¹ + ¹₄ln |1 + u²| +¹₂ tan⁻¹u + C

= −¹₂ln |1 + cos x| + _{1+cos x}¹ +¹₂ln |1 + cos²x| +¹₂tan⁻¹(cos x) + C

3. (10%) Find the area of the region in polar coordinates that is inside the curve r = 2 + 2 cos θ and outside the curve r = 3.

Ans. The area = .

Solution:

The curves intersect when r = 2 + 2 cos θ = 3

⇔ 2 cos θ = 1 ⇔ cos θ = ¹₂ ⇔ θ = ±^π₃. Let f (θ) = 3, g(θ) = 2 + 2 cos θ

Then A = ¹₂R ^π₃

−^π₃[g²(θ) − f²(θ)]dθ

=R ^π₃

0 [g²(θ) − f²(θ)]dθ

=R ^π₃

0 [4 + 8 cos θ + 4 cos²θ − 9]dθ

=R ^π₃

0 [8 cos θ + 4 cos³θ − 5]dθ

=R ^π₃

0 [8 cos θ + 4(^{1+cos 2θ}₂ ) − 5]dθ

=R ^π₃

0 (8 cos θ + 2 cos 2θ − 3)dθ

= [8 sin θ + sin 2θ − 3θ]

π 3

0

= ⁹

√3 2 − π.

4. (10%) Find the arc length parameter s = s(θ) along the curve (in polar coordi- nates) r = 3 cos² θ

2



, −π/2 ≤ θ ≤ π/2, from the point where θ = 0. Then solve for θ as a function of s : θ = θ(s).

Ans. s(θ) = ,

θ(s) = .

Solution:

r(u) = 3 cos^{2 u}₂, −^π₂ ≤ u ≤ ^π₂ s(θ) =Rθ

0

q

r²+ (_du^dr)²du

=Rθ

0 p(3 cos^{2 u}₂)²+ (−3 cos^u₂sin^u₂)²du

=Rθ 0

q

9 cos^{4 u}₂ + 9 cos^{2 u}₂ sin^{2 u}₂du

=Rθ 0

q

9 cos^{2 u}₂(cos^{2 u}₂ + sin^{2 u}₂)du

=Rθ

0 p9 cos^{2 u}₂du

=Rθ

0 3 cos^u₂du

= 6 sin^θ₂, −^π₂ ≤ θ ≤ ^π₂ θ(s) = 2 sin⁻¹(₆^s).

5. (15%) Find the unit tangent vector T, the principal unit normal vector N, and the curvature κ of the curve r(t) = (e^tsin 2t)i + (e^tcos 2t)j + (2e^t)k, t ∈ R.

Ans. T = ,

N = ,

κ = .

Solution:

V = ^dr_dt = e^t(sin 2t + 2 cos 2t)i + e^t(cos 2t − 2 sin 2t)j + 2e^tk,

|V | =pe^2t(sin 2t + 2 cos 2t)²+ e^2t(cos 2t − 2 sin 2t)²+ 4e^2t= 3e^t T = _{|V |}^V = ¹₃(sin 2t + 2 cos 2t)i + ¹₃(cos 2t − 2 sin 2t)j +²₃k,

dT

dt = ²₃(cos 2t − 2 sin 2t)i + ²₃(− sin 2t − 2 cos 2t)j,

|^dT_dt| =q

4

9(cos 2t − 2 sin 2t)²+⁴₉(− sin 2t − 2 cos 2t)² = ²₃√ 5, N = ^dTdt

|^dT_dt| = ^√1

5(cos 2t − 2 sin 2t)i + ^√1

5(− sin 2t − 2 cos 2t)j, κ = _|v|¹ |^dT_dt| = _3e¹t · ²₃√

5 = ²

√5 9e^t .

Hence T = ¹₃(sin 2t + 2 cos 2t)i +¹₃(cos 2t − 2 sin 2t)j +²₃k, N = ^√1

5(cos 2t − 2 sin 2t)i + ^√1

5(− sin 2t − 2 cos 2t)j, κ = ²

√5 9e^t .

6. (16%) Let y = f (x) = 1

2(x − 1)² and r(x) = (x, f (x)), 1 ≤ x ≤ 2, be the curve representing the graph of the function y = f (x) for x ∈ [1, 2].

(a) Find the arc length of this curve.

(b) Find the area of the surface generated by revolving this curve about the y- axis.

Ans. (a) The arc length = ,

(b) The area of the surface of revolution = .

Solution:

(a) Method 1:

Arc length=R2

1 p1 + (x − 1)²dx

=R1 0

√1 + u²du (Let u = x − 1).

=R sec θ sec²θdθ =R sec³θdθ. (Let u = tanθ, du = sec²θdθ).

=sec θ tan θ −R (sec²θ − 1) sec θdθ

=¹₂ sec θ tan θ +¹₂ln| sec θ + tan θ|

=¹₂√

u² + 1 · u +¹₂ln |√

u²+ 1 + u|¹₀ = ¹₂√

2 + ¹₂ln(√ 2 + 1).

(b) Surface area=R 2πsds = R₁²2πxp1 + f⁰(x)²dx =R2

1 2πxp1 + (x − 1)²dx

x−1=t

= R1

0 2π(t + 1)√

1 + t²dt = 2πR1 0 t√

1 + t²dt + 2πR1 0

√1 + t²dt

= 2π · ¹₃(1 + t²)³²|¹₀+ 2π · ¹₂{√

2 + ln(1 +√ 2)}

= ²₃π{2³² − 1} + π{√

2 + ln(1 +√ 2)}

7. (10%) Find the area of the infinite region bounded by the positive x- and y-axis and the graph of f (x) = 1

√x(2 + x), 0 ≤ x < ∞.

Ans. The area = .

Solution:

Area=R∞ 0

√ dx

x(2+x) = lim

M →∞

RM 0

√ dx x(2+x)

x=t²,dx=2tdt

= R∞

0 2tdt

t(2+t²) = 2R∞ 0

dt t²+(√

2)²

= 2{^√1₂tan⁻¹(^√t₂)}^∞₀ =√

2 ·^π₂ = ^√π₂