1. (10%)

1. (10%) 求極限: lim

n→∞

√n

n!

n (提示: 取對數) 解:

ln

√n

n!

n = 1 nln (n!

nⁿ) = 1 n{ln1

n + ln 2

n + ... + lnn n} Hence

n−>∞lim ln

√n

n!

n = ˆ 1

0

(ln x)dx = lim

δ−>0⁺

ˆ 1 δ

(ln x)dx = lim

δ−>0⁺x ln x|¹_δ− ˆ 1

δ

x ·dx x

= lim

δ−>0⁺x ln x|¹_δ−x|¹_δ = lim

δ−>0⁺−δ ln δ − (1 − δ) = −1+ lim

δ−>0⁺δ − δ ln δ = −1− lim

t−>∞

− ln t t = −1 where t = ¹_δ. And since e^x is continuous

n−>∞lim

√n

n!

n = lim

n−>∞e^ln

n√ n!

n = e^{limn−>∞ln}

n√ n!

n = e⁻¹

1

2. (10%) 求極限: lim

x→∞

´x²

0 e^t−x2(2t² + 1)dt x⁴

解:

x→∞lim

´x²

0 e^t−x2(2t²+ 1)dt x⁴

= lim

x→∞

(2t²+ 1)e^t−x2|^x₀² −´_x²

0 4te^t−x2dt x⁴

= lim

x→∞

(2t²+ 1)e^t−x2|^x₀² − (4te^t−x2|^x₀² − 4´x²

0 e^t−x2dt) x⁴

= lim

x→∞

(2t²− 4t + 5)e^t−x2|^x₀² x⁴

= lim

x→∞

2x⁴− 4x²+ 5 − 5e^−x2 x⁴

=2.

2

3. (10%)

求心臟線 r = 1 + cos θ 在 x 軸上方之最高高度。

解:

令y = (1 + cos θ) sin θ . 0 = dy

dθ

= cos²θ + cos θ − sin²θ

= cos²θ + cos θ − (1 − cos²θ)

= 2 cos²θ + cos θ − 1

所以 cos θ = ¹₂, −1(不合), 得到 θ = ^π₃。 因此最高點之高度為 (1 + cos θ) sin θ = ³

√3 4 .

3

(a)

2 2

2 2

2 2 2

0 0 0

0

1 cos 2 sin

sin ( ) ( )

2 2 4

x x x

y dx x dx dx

π π π π

4

π = π = π ⁻ =π − π

∫ ∫ ∫

(b)

Method1: (剝殼)

2 2

0 0

2 xsinxdx 2 ( xcosx sin )x 2

π π

π

∫

Method2: (反扣)

2 1 2

( ) 0 2

2 x dy

π π

π ∫

^π

Method 1 : 利用轉動慣量

2

0 sin 1

Area xdx

π

=

∫

2

0 sin 1

Mx x xdx

π

=

∫

2 2

0 2 8

y

M y dx

π π

=

∫

Method2: 利用 Pappus theorem

2πM A_x = × 繞 軸體積 x Y 2πM A_y = × 繞 軸體積y X (d)

Pappus theorem

2 2

8 1 6 25

8 17 3 6 8 10

π π

× + × − = −

+

17 3 17 3 2

2 1

10 5 20

π× ⁻ π × = π− π

5. (10%) 求積分:

ˆ ¹

2

1 4

sin⁻¹√ x px(1 − x)dx 解:

因為

d

dxsin⁻¹√

x = 1

√1 − x· 1 2√

x 所以

ˆ ¹

2

1 4

sin⁻¹√ x

px(1 − x)dx = ˆ ¹

2

1 4

2 sin⁻¹√

x d sin⁻¹√ x

= (sin⁻¹√ x)²]

1 2 1 4

= π 4

2

−π 6

2

= 5

144π².

5

6. (10%) 求積分:

ˆ 1

2 + sin xdx 解:

Let u = tan(^x₂) , then sin x = _1+u^2u2 , dx = _1+u² 2du

´ ₁

2+sin x = ´ ₁

2+ ^2u

1+u2

2

1+u²du = ´ _du

u²+u+1 = ´ _du

(u+¹₂)²+³₄ = ⁴₃´ _du

(^2u+1√

3 )²+1

Let ^2u+1√

3 = tan θ , then du =

√ 3

2 sec²θdθ

´ ₁

2+sin x = ⁴₃´ _du

(^2u+1√

3 )²+1 = ⁴₃´ ^√3

2 sec²θdθ

tan²θ+1 = ^√2

3

´ 1dθ

= ^√2

3(θ + C₁) = ^√2

3(tan⁻¹(^2u+1√

3 ) + C₁) = ^√2

3tan⁻¹(^{2 tan(}

x 2)+1

√

3 ) + C

6

7. (10%) 求積分:

ˆ dx

x³√ x²+ 3 解:´ _dx

x³√

x²+3 =´ _x

x⁴√

x²+3dx =´ ₁

(u²−3)²du Since _(u2−3)^{1 2} = ¹

12√ 3(u+√

3) − ¹

12√

3(u−3) + ¹

12(u+√

3)² + ¹

12(u−√ 3)²,

´ _dx

x³√ x²+3

=´ ₁

12√ 3(u+√

3)du −´ ₁

12√

3(u−3)du +´ ₁

12(u+√

3)²du +´ ₁

12(u−√ 3)²du

= ¹

12√

3(ln(u +√

3) − ln(u −√

3)) − ₁₂¹( ¹

u+√

3 + ¹

u−√ 3)

= ¹

12√ 3ln(

√x²+3+√

√ 3 x²+3−√

3) − ¹₆

√x²+3 x²

7

8. (10%) 設瑕積分

ˆ ∞

√2

( a

√x²− 1 − x

x²+ 1)dx 為收斂。 試決定a值, 並求其積分值。

解:´√∞ 2(^{√ a}

x²−1 − _x2^x+1)dx

= lim_b→∞´√b 2(^√a

x²−1 −_x2^x+1)dx

= lim_b→∞(a ln |x +√

x²− 1| −¹₂ ln |x²+ 1|)|^b√

2

= limb→∞ln^(|b+

√ b²−1|)^a

√b²+1 − ln |1 +√

2| + ¹₂ln 3 and

lim_b→∞ln^(|b+

√b²−1|)^a

√ b²+1

need to be convergent.

If a > 1,it is divergent as b → ∞ ;

If a < 1,it is divergent since ln 0 → −∞as b → ∞;

The value of a must be 1.

and

lim_b→∞ln^(|b+

√b²−1|)

√

b²+1 = ln 2

The value of the improper integral is ln 2 − ln |1 +√

2| + ¹₂ln 3 = ln( ²

√ 3 1+√

2)

8

9. (10%)

解微分方程 xy⁰− 2y = x³sec x tan x, x > 0, y(π

4) = 0 。 解:

xy⁰ − 2y = x³sec x tan x;

同除x³,(y/x²)⁰ = y⁰/x²− 2y/x³ = sec x tan x;

故y/x² = sec x + C;

因y(^π₄) = 0, 故C = −√ 2。

得y = x²(sec x −√ 2) 評分標準:

y⁰ − 2y/x = x²sec x tan x: 至此一分;

用積分因子法, 令(vy)⁰ = vy⁰− 2vy/x = vx²sec x tan x, 並解出v = 1/x² : 至此五分;

求得 y/x² = sec x + C: 至此八分;

得到 y = x²(sec x + C); 而 C 解錯, 至此九分。

若用參數變異法, 則求出齊次解等同解出積分因子, 給五分。

積分因子法推導中間有誤, 酌量扣分, 但直接用錯誤的公式不給分。

9