(a) (10%) Solve the initial value problem

1031微微微甲甲甲07-11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. (10%) Solve the initial value problem: dy

dx = e^y

1 + x², y(0) =−1.

Solution:

dy

dx = e^y

1 + x² (separable)

⇒ e^−ydy = 1

1 + x²dx (2%)

⇒ −e^−y = tan⁻¹x + C (3% for each side) With y(0) =−1, we have −e = tan⁻¹0 + C = C.

⇒ C = −e (2%)

∴ The solution is e^−y= e− tan⁻¹x

2. (a) (10%) Solve the initial value problem:





2x(x + 3)y⁰+ (4x + 3) y = 2x¹²(x + 3)¹², y(1) = 1

2, x > 0.

(b) (3%) Find lim

x→∞y(x) and lim

x→0⁺y(x).

Solution:

(a)

Divide the equation by 2x (x + 3) We get y⁰+ 4x + 3

2x(x + 3)y = 1 px(x + 3)

Multiply integration factor I(x) on both side, then Iy⁰+ I 4x + 3

2x(x + 3)y = I 1 px(x + 3) Solve that I⁰= I 4x + 3

2x(x + 3), we let I = e^{´ 2x(x+3)4x+3 dx} Where

ˆ 4x + 3

2x(x + 3)dx = 1 2

ˆ (1

x+ 3

x + 3)dx = 1

2ln x +3

2ln (x + 3) + C, when x > 0 Hence I(x) = e^Cp

x(x + 3)³, Let C = 0⇒ I(x) =p

x(x + 3)³ Iy⁰+ I 4x + 3

2x(x + 3)y = Iy⁰+ I⁰y = (Iy)⁰= I 1

px(x + 3) = x + 3 Iy =

ˆ

(x + 3)dx = 1

2x³+ 3x + D⇒p

x(x + 3)³y = 1

2x³+ 3x + D Bring y(1) = 1

2 into the equation, we find D =1 2 Hence y = x²+ 6x + 1

2p

x(x + 3)³ (b)

xlim→∞y = lim

x→∞

s(1 + ⁶_x+_x¹2)² 4(1 +³_x)³ =

r1 4 = 1

2 Because lim

x→0⁺x²+ 6x + 1 = 1, and lim

x→0⁺2p

x(x + 3)³= 0 So lim

x→0⁺y→ +∞

評分標準如下，

(a)

解出 I(x) +5分 解出y 的不定積分 +3分

代入初始條件求出完整解答 +2分

計算錯誤該部分加分折半 (b)

(a)解出的 y 錯誤不管答案一律0分

(有些同學計算錯誤會影響答案，有些不會，覺得不應該同樣因為計算錯誤而有差別) 此外對一題2分，兩題3分

3. (a) (7%) Find the length of the curve in polar coordinates: r =√

1 + sin 2θ, 0≤ θ ≤ 2π.

(b) (7%) Find the area enclosed by the curve given in (a).

Solution:

(a)

dr dθ = d

dθ

√1 + sin 2θ = cos 2θ

√1 + sin 2θ (2 points)

The length of the curve L is L =

ˆ 2π 0

s r²+

dr dθ

2

dθ (2 points)

= ˆ _2π

0

s

1 + sin 2θ + cos²2θ 1 + sin 2θdθ

= ˆ 2π

0

s

1 + 2 sin 2θ + sin²2θ + cos²2θ 1 + sin 2θ dθ

= ˆ 2π

0

r2 + 2 sin 2θ 1 + sin 2θ dθ

= ˆ 2π

0

√2 dθ

= 2√

2π (3 points)

(b) Because r≥ 0 for 0 ≤ θ ≤ 2π (or due to observation from the figure below), the area A of the enclosed region is

A = ˆ 2π

0

1

2r²dθ (4 points)

= 1 2

ˆ 2π 0

(1 + sin 2θ) dθ

= 1 2

 θ−1

2cos 2θ

2π 0

= 1 2

 2π−1

2− 0 + 1 2



= π (3 points)

4. (a) (7%) Find the length of the parametric curve C: x = ln(sec t + tan t)− sin t, y = cos t, 0 ≤ t ≤ π 3. (b) (7%) Rotate the curve C about x-axis. Find the surface area.

(c) (7%) Find the volume of the solid bounded by the surface given in (b) and the planes x = 0, x = ln(2+√ 3)−

√3 2 . Solution:

(a).

d

dt(x(t)) = sec(t)− cos(t), d

dt(y(t)) =−sin(t). . . . (1 points) ˆ π3

0

ds = ˆ π3

0

tan(t)dt . . . (2 points)

=−ln|cos(t)|^π/30 . . . (2 points)

= ln(2) ] . . . (2 points)

, where ds = sd

dt(x(t))

² +

d dt(y(t))

²

dt = tan(t)dt.

(b).

ˆ π3

0

2πy(t)ds = ˆ π3

0

2πcon(t)tan(t)dt . . . (2 points)

=−2πcos(t)|^π/30 . . . (3 points)

= π ] . . . (2 points)

(c).

ˆ

πy²(t)d(x(t)) = ˆ π3

0

πcon²(t)[sec(t)− cos(t)]dt . . . (2 points)

= ˆ π3

0

πsin²(t)con(t)dt

= π

3sin³(t)|^π/30 . . . (3 points)

=

√3

8 π ] . . . (2 points) , where d(x(t)) = [sec(t)− cos(t)]dt.

5. (10%) Find

ˆ 1

x²(x²+ x + 1)dx.

Solution:

Let 1

x²(x²+ x + 1) = a x+ b

x² + cx + d

x²+ x + 1, where a, b, c, d are determined. (1 point) Then solve a = −1,b = 1,c = 1,d = 0. (2 points, if solve partially 1 point)

so we have

ˆ 1

x²(x²+ x + 1)dx = ˆ −1

x dx + ˆ 1

x²dx +

ˆ x

x²+ x + 1dx

=− ln |x| − 1 x+

ˆ x

x²+ x + 1dx (2 points)

ˆ x

x²+ x + 1dx =

ˆ x +¹₂

x²+ x + 1dx−1 2

ˆ 1

x²+ x + 1dx = 1

2ln(x²+ x + 1) (2 points)

ˆ 1

(x +¹₂)²+ (^√₂³)² = 2

√3arctan2x + 1

√3 (3 points)

so Answer is− ln |x| − 1 x+1

2ln(x²+ x + 1)− 1

√3arctan2x + 1

√3 + C If you don’t wirte +C, you will lose 1 point.

6. (10%) Find

ˆ 1

sec x− 1dx.

Solution:

Solution 1.

ˆ 1

sec x− 1dx =

ˆ 1

sec x− 1 ·sec x + 1 sec x + 1dx =

ˆ sec x + 1

sec²x− 1dx (2分)

=

ˆ sec x + 1

tan²x dx (3分) =

ˆ cos x

sin²x+cos²x sin²x

 dx

=

ˆ 1

sin²xd sin x +

ˆ 1− sin²x sin²x dx

=

ˆ 1

sin²xd sin x + ˆ

csc²x dx− ˆ

1 dx

=− 1

sin x− cot x − x + C.

後面的三項積分個別計分，第一項積出 − 1

sin x 再得 4 分，其它兩項及常數 C 一共佔 3 分。

Solution 2. Let t = tanx

2 (佔 1 分), then we have cos x =1− t²

1 + t² (佔 1 分), sin x = 2t

1 + t² (沒有用到), and dx = 2

1 + t²dt. (佔 1 分) So

ˆ 1

sec x− 1dx =

ˆ cos x

1− cos xdx (2分) = ˆ 1−t²

1+t² ·1+t²²

1−^1−t1+t²²

dt =

ˆ 1− t²

t²(1 + t²)dt (5分)

= ˆ 1

t²− 2 t²+ 1



dt (7分) = −1

t − 2 tan⁻¹t + C (9分)

=− 1

tan^x₂ − 2 tan⁻¹ tanx

2

+ C =− cotx

2 − x + C. (10分)

前面兩種解法是大部分人採用的積分策略，所以在此列出詳細配分標準，以下還有數種解法，處理方式都很漂亮，在此也 完整呈現供大家參考及學習，配分以個案處理， 但原則上和前兩種配分方式雷同。

Solution 3.

ˆ 1

sec x− 1dx =

ˆ cos x 1− cos xdx =

ˆ 1− 2 sin^{2 x}2

2 sin^{2 x}₂ dx

= ˆ

csc²x 2d x

2

− ˆ

1 dx =− cotx

2 − x + C.

Solution 4.

ˆ 1

sec x− 1dx =

ˆ 1

sec x− 1·sec x + 1 sec x + 1dx =

ˆ sec x + 1 sec²x− 1dx =

ˆ sec x + 1 tan²x dx.

Since

ˆ sec x tan²xdx =

ˆ sec x tan x tan³x dx =

ˆ 1

tan³xd sec x = sec x tan³x−

ˆ

sec x d cot³x

= sec x tan³x+

ˆ

3 sec x cot²x csc²x dx = sec x tan³x+ 3

ˆ cos x sin⁴xdx

= sec x tan³x+ 3

ˆ 1

sin⁴xd sin x = sec x tan³x− 1

sin³x+ C,

and ˆ

1

tan²xdx = ˆ

cot²x dx = ˆ

(csc²x− 1) dx = − cot x − x + C, we get

ˆ 1

sec x− 1dx = sec x tan³x− 1

sin³x− cot x − x + C.

Solution 5.

ˆ 1

sec x− 1dx =

ˆ 1

sec x− 1·sec x + 1 sec x + 1dx =

ˆ sec x + 1 sec²x− 1dx =

ˆ sec x + 1 tan²x dx.

Since

ˆ sec x tan²xdx =

ˆ

cot x csc x dx =− csc x + C,

and ˆ

1

tan²xdx =

ˆ cos²x sin²xdx =

ˆ

cos²x csc²x dx =− ˆ

cos²x d cot x

=− cos²x cot x + ˆ

cot x d cos²x

=− cos²x cot x− 2 ˆ

cot x cos x sin x dx

=− cos²x cot x− 2 ˆ

cos²x dx =− cos²x cot x− ˆ

(1 + cos 2x) dx

=− cos²x cot x− x − sin 2x 2 + C

− cos²x cot x− x − sin x cos x + C we get

ˆ 1

sec x− 1dx =− csc x − cos²x cot x− x − sin x cos x + C.

Solution 6. Let t = cotx

2, then we have cos x =t²− 1

1 + t², sin x = 2t

1 + t², and dx =− 2 1 + t²dt.

So

ˆ 1

sec x− 1dx =

ˆ cos x 1− cos xdx =

ˆ t²−1 1+t² ·1+t⁻²²

1−^t1+t²⁻¹²

dt =

ˆ 1− t² 1 + t²dt

= ˆ 

−1 + 2 1 + t²



dt =−t + 2 tan⁻¹t + C

=− cotx

2 + 2 tan⁻¹ cotx

2

+ C.

Solution 7. Let t = sec x+tan x. Since sec²x−tan²x = (sec x+tan x)(sec x−tan x) = 1, we have 1

t = sec x−tan x and

sec x = t²+ 1

2t , tan x = t²− 1

2t , dt = t²+ 1 2 dx.

So ˆ 1

sec x− 1dx =

ˆ 1

t²+1

2t − 1 · 2 t²+ 1dt =

ˆ 4t

(t− 1)²(t²+ 1)dt

=

ˆ  2

(t− 1)² − 2 t²+ 1



dt =− 2

t− 1 − 2 tan⁻¹t + C

=− 2

sec x + tan x− 1 − 2 tan⁻¹(sec x + tan x) + C.

7. Consider the intersection of three circular cylinders with radius R and all axes of cylinders lie in a plane with polar equations θ = 0, θ = π

3, and θ = 2π

3 . The cross-section is a hexagon, and the shape of the solid looks like the union of two umbrellas.

Problem 1. (%) Solve the initial value problem: dy dx= y− 1

1 + x², y(0) =−1.

Problem 2. (%+%)

(a) Solve the initial value problem:







x(x + 3)y^′+

 2x +3

2



y = x¹²(x + 3)¹², y(1) =1

2, x > 0.

(b) Discuss the asymptotic behavior lim_x

→∞y(x) and lim

x→0⁺y(x).

Problem 3. (%+%)

(a) Find the length of the polar curve: r =√

1 + sin 2θ, 0≤ θ ≤ 2π.

(b) Find the area enclosed by the curve given in (a).

Problem 4. (%+%+%)

(a) Find the length of the parametric curve C: x = ln(sec t + tan t)− sin t, y = cos t, 0 ≤ t ≤^π3. (b) Rotate the curve C about the x-axis. Find the surface area.

(c) Find the volume of the solid bounded by the surface given in (b) and the planes x = 0, x = ln(2 +√

3).

Problem 5. (%) Find

Z 1

x²(x²+ x + 1)dx.

Problem 6. (%) Find

Z 1

sec x− 1dx.

Problem 7. (%+%) Consider the intersection of three circular cylinders with radius R and all axes of cylinders lie in a plane with polar equations θ = 0, θ =π

3, and θ =2π

3. The cross-section is a hexagon, and the shape of the solid looks like the union of two umbrellas.

θ =^π₃ θ =^2π₃

θ = 0 R y

(a) (b)

Figure 1: Intersection of three cylinders. (a) Vertical view. (b) The solid.

(a) Find the area of the cross section of the solid when the height is y, y∈ [−R, R].

(b) Find the volume of the solid.

Problem 8. (%) Evaluate the improper integral Z_∞

0

x³e^−x2dx.

1

(a) (6%) Find the area of the cross section of the solid when the height is y, y∈ [−R, R].

(b) (6%) Find the volume of the solid.

Solution:

(a)Observe the figures below.

(a)Observe the figures below.

θ =^π₆

d d

y R

When the height is y, the area of the cross section(hexagon) is decided by d =p

R²− y². Hence,

Area = 6× Area of equilateral triangles = 6 ×1 2× d × 2

√3d

= 2√

3(R²− y²)

(b) V = ZR

−R

A(y)dy = ZR

−R

2√

3(R²− y²)dy = 2√

3(R²y−1

3y³)|^R_−R=8√ 3 3 R³

When the height is y, the area of the cross section(hexagon) is decided by d =p

R²− y².(3%) Hence,

Area = 6× Area of equilateral triangles = 6 ×1

2 × d × 2

√3d

= 2√

3(R²− y²) (3%)

(b) V = ˆ R

−R

A(y)dy(3%)

= ˆ R

−R

2√

3(R²− y²)dy = 2√

3(R²y−1

3y³)|^R−R(2%) = 8√ 3

3 R³(1%)

8. (10%) Evaluate the improper integral ˆ _∞

0

x³e^−x2dx.

Page 6 of 7

Solution:

ˆ _∞

0

x³e^−x2dx = lim

t→∞

ˆ t 0

x³e^−x2dx = lim

t→∞

1 2

ˆ t 0

x²e^−x2dx²= lim

t→∞

1 2

ˆ t² 0

u e^−udu (3分)

= lim

t→∞−1 2

ˆ t² 0

u de^−u = lim

t→∞−1 2

hu e^−ui 

u=t² u=0 −

ˆ t² 0

e^−udu

! (6分)

= lim

t→∞



−1

2t²e^−t2−1 2

he^−ui^u=t

2

u=0



= lim

t→∞



−1

2t²e^−t2−1

2e^−t2+1 2



(7分) = 1 2, where we use the l’Hospital Rule to get find the limit:

t→∞lim t²e^−t2 = lim

t→∞

t² e^t2

(^∞_∞,L⁰)

= lim

t→∞

2t

2t e^t2 = lim

t→∞

1 e^t2 = 0.

這題重點在於是否具有「瑕積分的觀念」，而判定的依據是以有寫出「極限」為準；縱使前面所有的式子都是寫

ˆ _∞

0

或 是上、下限寫 |^∞0 ， 但最後一定要用極限判斷 lim

t→∞t²e^−t2 = 0，就認定你知道「瑕積分的意義」；這個極限值不是顯然 的，一定要討論，佔 3 分， 也就是說，最後有完整討論這個極限就可以拿到滿分 10 分。 另一個極限式 lim

t→∞e^−t2 = 0

可接受直接寫答案，沒有解釋不會扣分。