ln 1 − a ln (t + 2

1041微微微甲甲甲06-10班班班期期期末末末考考考解解解答答答和和和評評評分分分標標標準準準 1. (14%)

(a) (7%) Find the indefinite integrals

ˆ 1

√x²+ 1dx (5%) and

ˆ 1

x + 2dx (2%). (You may use the integral formula of

ˆ

sec θ dθ.)

(b) (4%) Find the value of the constant a for which the improper integral ˆ ∞

0

 1

√x²+ 1 − a x + 2



dx converges.

(c) (3%) Evaluate the improper integral for this a.

Solution:

(a)(7pts) [part1–(5pts)]

ˆ 1

√x²+ 1 dx

let x = tan θ ⇒ dx=sec²θ dθ (2pts)

ˆ 1

√x²+ 1 dx

=

ˆ sec²θ sec θ dθ

= ˆ

sec θ dθ

=ln | sec θ + tan θ | + c

=ln |p

x²+ 1 + x | + c

[part2–(2pts)]

ˆ 1

x + 2 dx

=ln | x + 2 | + c

(b)(4pts) ˆ ∞

0

( 1

√x²+ 1− a x + 2)dx

Let I=

ˆ ∞ 0

( 1

√x²+ 1− a x + 2)dx I=

ˆ ∞ 0

( 1

√x²+ 1− a x + 2)dx

= lim

t→∞

ˆ t 0

( 1

√x²+ 1 − a x + 2)dx

= lim

t→∞( ln(p

t²+ 1 + t) − ln 1 − a ln (t + 2) + a ln 2 )

= lim

t→∞ln(

√t²+ 1 + t

(t + 2)^a ) + a ln 2 —————————————— (2pts)

Thus we only need to consider lim

t→∞ln

√t²+ 1 + t (t + 2)^a

let L= lim

t→∞ln

√t²+ 1 + t (t + 2)^a I=

ˆ ∞ 0

( 1

√x²+ 1− a

x + 2)dx=ln L+ ln 2 (case1) a > 1

L=0

I= ln L +a ln 2 → −∞ dverges

(case2) a < 1 then L→ ∞ , and I= ln L +a ln 2 → ∞ dverges (case1)and(case2)——————————————————— (1pts)

(case3) a = 1

L= lim

t→∞ln

√t²+ 1 + t (t + 2) = 2 I=ln L+ ln 2 = ln 2 + ln 2 = 2 ln 2

Thus I converges when a=1——————————————— (1pts)

(c)(3pts) When a=1 L= lim

t→∞ln

√t²+ 1 + t

(t + 2) = 2———————————————- (2pts) I=ln L+ ln 2 = ln 2 + ln 2 = 2 ln 2————————————– (1pts)

Page 2 of 10

2. (12%) Find the following indefinite integrals:

(a) (6%)

ˆ 3t²+ t + 4 t³+ t dt.

(b) (6%) ˆ

cos√ x dx

Solution:

(a) 3t²+ t + 4 t³+ t =A

t +Bt + C

t²+ 1 (1 point)

⇒ A = 4, B = −1, C = 1(1 point) ˆ 4

t +−t + 1

t²+ 1dt = 4 ln |t| −1

2ln(t²+ 1) + tan⁻¹t + c(1 pointt for each) (b) Let√

x = u, ⇒ dx = 2udu.

ˆ cos√

xdx = 2 ˆ

u cos udu(2 points)

= 2 ˆ

ud sin

= 2(u sin u − ˆ

sin udu)(2 points)

= 2(u sin u + cos u) + c

= (2√ x sin√

x + 2 cos√

x) + c(1 point for each)

3. (12%)

(a) (10%) Solve the initial-value problem: dx

dt = (a − x)(b − x), where a > b > 0, for x = x(t) with x(0) = 0.

(b) (2%) Find lim

t→∞x(t).

Solution:

(a) dx

(a − x)(b − x) = dt

⇒

ˆ dx

(a − x)(b − x) = ˆ

1dt

⇒ 1

b − a ˆ

( 1

a − x− 1

b − x)dx = t + C

⇒ 1

b − a(− ln |a − x| + ln |b − x|) = t + C

⇒ 1

a − bln    

a − x b − x    

= t + C

∵x(0) = 0

∴C = 1 a − bln

  a b    =

1 a − blna

b

⇒ ln    

a − x b − x   

= (a − b)t + lna b

⇒ a − x b − x = a

be^(a−b)t

⇒ abe^(a−b)t− ae^(a−b)t= ab − bx

⇒ x(t) = ab(e^(a−b)t− 1) ae^(a−b)t− b (b) lim

t→∞x(t) = lim

t→∞

ab(e^(a−b)t− 1) ae^(a−b)t− b = lim

t→∞

ab(a − b)e^(a−b)t a(a − b)e^(a−b)t = b 評分標準

(a) 變數分離得 2 分 積分出來得4 分 把C 求出來得 2 分 帶回C 並解出 x 得 2 分 (b) 沒過程不給分

Page 4 of 10

4. (12%)

(a) (10%) Solve the initial-value problem: xy^′− y = x²sin x, with y(π) = 0.

(b) (2%) Find lim

x→0⁺

y(x) x² .

Solution:

(a) The linear differential equation is y^′− 1

xy = x sin x. We multiply the integrating factor e^{´ −x1dx}= e^{− ln |x|}= x⁻¹ (積分因子有算出來得 5 分，有錯整題最多給 2 分。) on both sides of the differential equation and get

1 xy^′− 1

x²y = sin x ⇒ d dx

 1 xy



= sin x ⇒ 1 xy =

ˆ

sin x dx = − cos x + C.

So y(x) = x(− cos x + C). (寫到這裡可得 8 分。) The initial condition y(π) = 0 implies 0 = π(1 + C) and we get C = −1, (解對積分常數再得 1 分。) so the solution of the differential equation is

y(x) = x(− cos x − 1). (最後完整的函數寫出再得 1 分。) (b) The limit is

x→0lim⁺ y(x)

x² = lim

x→0⁺

x(− cos x − 1)

x² = lim

x→0⁺

− cos x − 1

x = −∞.

• 注意到，若用以上觀察，寫「−∞」或「不存在」或「發散」都可以算對，得 2 分。 若是用羅必達法則計算該極

限，最後的答案只能寫「−∞」，而不能寫「不存在」或「發散」。 因為使用羅必達法則後，若分子、分母各自微

分後的極限「不存在」或「發散」，並不能對原極限下結論； 於是判定你對羅必達法則使用錯誤或是沒有正確理

解，無法給分。 課本第 302 頁指出，使用羅必達後，唯有「極限存在」或是「±∞」這種狀況可以反推原極限的

結果。

5. (10%) Evaluate lim

x→0

1 x

ˆ x

0 (1 − tan t)^1/tdt.

Solution:

First, note that lim

x→0

´x

0(1 − tan t)^1t dt

x is of the form 0 0 . We can use l’Hospital’s rule to evaluate the limit.

x→0lim

´x

0(1 − tan t)^1t dt

x = lim

x→0(1 − tan x)^1x By fundamental theorem of calculus. (5 pts) Then we take log to evaluate the last limit.

x→0lim

ln(1 − tan x)

x is of the form 0 0

= lim

x→0

− sec²x

1 − tan x = −1 by l’Hospital’s rule Thus we have lim

x→0

´x

0(1 − tan t)^1t dt

x = e⁻¹= 1

e (5 pts)

Note lim

x→0

´x

0(1 − tan t)^1tdt

x 6= f^′(0) where f (x) =

ˆ x

0 (1 − tan t)^1t dt FTC:If f (x) =

ˆ x

0 g(t) dt for x ∈ [0, a]

then f^′(x) = g(x) for x ∈ (0, a) but not for x = 0, a

Page 6 of 10

6. (12%)

(a) (4%) Show that the area of an ellipse with the semi-major axis of length a and the semi-minor axis of length b is abπ. See Figure 1(a).

(b) (8%) A toothpaste tube is modeled in Figure 1(b).

x y x

y

a z b

z = 0 z = 20

(a) (b)

Figure 1: (a) The area of an ellipse is abπ. (b) Find the volume of the modeled toothpaste tube.

• One side is flat and is located at −π ≤ x ≤ π, y = 0, z = 0.

• The other side is a circle with radius 2, so the equation of the circle is x²+ y²= 4, z = 20.

• Each cross-section for 0 < z < 20 is an ellipse with the semi-major axis of length a and the semi-minor axis of length b, where

a = π + (2 − π) z

20 and b = 1 10z.

Find the volume of the modeled toothpaste tube with 0 ≤ z ≤ 20.

Solution:

(a)

Ellipse equation: x² a²+y²

b² = 1 =¿ y = ±b s

1 − x² a² (1%) Aera

= 4 ˆ a

0

b s

1 − x²

a²dx (1%)= 4b a

ˆ a 0

pa²− x²dx

= 4ab ˆ ^π₂

0

cos²θdθ (let x = a sin θ, dx = a cos θdθ) (1%)

= 2ab ˆ ^π₂

0

(1 + cos 2θ)dθ (1%)

= 2ab(θ +1 2sin 2θ)|

π 2

0

= abπ (1%) (b)

Volume

= ˆ 20

0 π(π + (2 − π) z 20) 1

10zdz (4%)

7. (12%)

(a) (3%) Find all intersection points of the two curves r =√

2 sin θ and r²= cos 2θ in their polar equations.

Figure 2: (a) Find all intersection points of the two curves. (b)Find the area of the shaded region.

(b) (9%) Find the area of the shaded region in Figure 2.

Solution:

(a). If we solve the equations r =√

2 sin θ and r²= cos 2θ, we get

 r =√ 2 sin θ r²= cos 2θ

⇒ 2 sin²θ = cos 2θ = 1 − 2 sin²θ

⇒ θ = π

6,5π 6 ,7π

6 ,11π 6

⇒ r = 1

√2, 1

√2,−1

√2,−1

√2

We have found two points of intersection:

(r, θ) = ( 1

√2,π

6) (1pt) (r, θ) = ( 1

√2,5π

6 ) (1pt) However, if we plug θ = 0 into r = √

2 sin θ and θ = π/4 into r² = cos 2θ ,we can find one more point of intersection:

r = 0 (1pt)

(b).

A = 2 1

2 ˆ ^π₆

0

(√

2 sin θ)²dθ +1 2

ˆ ^π₄

π 6

cos 2θdθ

!

(6pts)

= ˆ ^π₆

0 (1 − cos 2θ)dθ + ˆ ^π₄

π 6

cos 2θdθ

=

 θ −1

2sin 2θ

    

π 6

0

+1 2sin 2θ

π 4

π 6

(2pts)

= π

6 +1 2−

√3

2 (1pt)

Page 8 of 10

8. (16%) The front door of a school bus is designed as in Figure 3. It is a folding door with AB = BO = 1

2. The door is opened or closed by rotating OB about z-axis, while A is moving along y-axis.replacemen

1 2

x x

y

y

z

O O

A

A

B B

C γ(t)

γ(t)

t

SchoolBus

Figure 3: (a) Find the enclosed area by the curve γ(t) and two axes. (b) Find the length of the curve γ(t).

The region swept out by the bus door on the xy-plane is enclosed by the two axes and the curve γ(t) parametrized by

γ(t) = (x(t), y(t)) =











(sin³t, cos³t) if 0 ≤ t ≤ π 4

 1 2sin t,1

2cos t

 if π

4 ≤ t ≤ π 2, where 0 ≤ t ≤π

2 denotes the angle between the positive y-axis and OB.

(a) (10%) Find the area of this region.

(b) (6%) Find the length of the curve γ(t).

Solution:

The curve consists of two parts: a half of a branch of an astroid (for 0 ≤ t ≤π

4), and one-eighth of a circle (for π

4 ≤ t ≤ π 2).

(a)

(1) Formulation:

A = ˆ

ydx = ˆ ^π₄

0

y(t)x^′(t)dt + ˆ ^π₂

π 4

y(t)x^′(t)dt = ˆ ^π₄

0

cos³t · 3 sin²t cos tdt + ˆ ^π₂

π 4

1

2cos t ·1 2cos tdt

= ˆ ^π₄

0

3 cos⁴t sin²tdt + ˆ ^π₂

π 4

1

4cos²tdt Or, noting that as y goes from 0 to 1, t varies from π

2 to 0, A =

ˆ xdy =

ˆ 0

π 4

x(t)y^′(t)dt + ˆ ^π₄

π 2

x(t)y^′(t)dt

= ˆ ^π₄

0

3 cos²t sin⁴tdt + ˆ ^π₂

π 4

1

4sin²tdt

Alternatively, if polar coordinates are to be used (letting the polar axis be in the +y direction and θ be in the

3

(2) Integration and evaluation:

ˆ

cos⁴t sin²tdt = ˆ

(1 + cos 2t 2 )(1

4sin²2t)dt

= 1 8

ˆ

(1 − cos 4t

2 + sin²2t cos 2t)dt

= 1 16(t −1

4sin 4t +1 3sin³t) ˆ

cos²tdt = ˆ

(1 + cos 2t 2 )dt =1

2t +1 4sin 2t

A = ˆ

ydx = ˆ ^π₄

0

3 cos⁴t sin²tdt + ˆ ^π₂

π 4

1

4cos²tdt

= 3 · 1 16(π

4 − 0 +1 3) +1

4(π 8 −1

4) = ( 3 64π + 1

16) + ( 1 32π − 1

16) = 5 64π (For your reference, the other methods yields

ˆ

ydx = (3 64π − 1

16) + ( 1 32π + 1

16) = 5 64π and

ˆ 1 2r²dθ = 3

64π + 1 32π = 5

64π respectively.)

• Grading policy: for the astroid part: 3% for formulation, 3% for integration and 2% for evaluation; 2% in total for the one-eighth circle part.

(b)

(1) For the astroid part:

L1= ˆ

ds = ˆ ^π₄

0

r (dx

dt)²+ (dy dt)²dt =

ˆ ^π₄

0

q

(3 sin²t cos t)²+ (−3 cos²t sin t)²dt (3%)

= ˆ ^π₄

0 3| sin t cos t|p

sin²t + cos²tdt = 3 [ 1 2 sin²t]

π 4

0 = 4

3 (2%)

(2) For the circle part:

L2= ˆ ^π₂

π 4

r (1

2cos t)²+ (−1

2sin t)²dt = ˆ ^π₂

π 4

1 2dt = π

8 (1%)

Or directly L2=1

8(2 · π · 1 2) = π

8 since it is one-eighth of a circle.

Thus the total length is L = L1+ L2=3 4 +π

8.

Page 10 of 10