(1 point). Then the following two statements hold true:

1061微微微乙乙乙01-05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. (15%)

(a) (5%) 請精確陳述微積分基本定理。

(b) (10%) 函數 f (x) 滿足

∫

x 0

f (t)dt =∫

1 x

t²f (t)dt +x¹⁶ 8 +

x¹⁸

9 +C 對所有的 x, 其中 C 是常數。求出函數 f (x) 及常數 C.

Solution:

(a) (5 points) Assume that f (x) is continuous on [a, b]

(1 point). Then the following two statements hold true:

(Part I) Let F (x) =∫

x

a f (t)dt. Then F^′(x) = f (x).

(2 points)

b a

f (t)dt = G(b) − G(a).

(2 points)

F (x) =∫ f (x)dx implies F^′(x) = f (x)

is just a definition for anti-derivatives but not a part of Fundamental Theorems of Calculus. You will get no points if you write down this as part I in your answer.

(b) (10 points) Do the derivatives on the both sides of

∫

x 0

f (t)dt =∫

1 x

t²f (t)dt +x¹⁶ 8 +

x¹⁸ 9 +C, we obtain

f (x) = −x²f (x) + 2x¹⁵+2x¹⁷. Hence

f (x) =2x¹⁵+2x¹⁷

1 + x² =2x¹⁵.

(7 points)

x 0

f (t)dt =∫

1 x

t²f (t)dt +x¹⁶ 8 +

x¹⁸

9 +C, we have 0 =∫

1 0

t²f (t)dt + C

= ∫

1

0 2t¹⁷dt + C

= t¹⁸

9 ∣

1

0

+C

=

−1 9 +C.

Hence C =−1

9 .

(3 points)

註: Doing the derivatives incorrectly such as writing down

f (x) − f (0) = f (1) − x²f (x) + 2x¹⁵+2x¹⁷,

or making any similar mistakes in this step, you will get no more then 5 points for this question.

Page 1 of 10

2. (7%) 求 lim_x→∞ ln(1 + x) ln(1 + x²)

.

Solution:

(7 points) By using L’Hospital rule (for the ∞

∞

form), we have

x→∞lim

ln(1 + x) ln(1 + x²)

= lim

x→∞

1 1+x

2x 1+x²

= lim

x→∞

x²+1

2x(x + 1)

(5 points)

= lim

x→∞

x²+1 2x²+2x

= 1

2.

(2 points)

(ln(1 + x²))

′

= 1 1 + x² you will get no points for this question.

3. (7%) 求 lim_x→0( 1 x²−

1 sin²x).

Solution:

Solution 1.

cosx = 1 −x² 2! +

x⁴ 4! −

x⁶ 6! +...

cos2x = 1 −2²∗x² 2! +

2⁴∗x⁴ 4! −

2⁶∗x⁶ 6! +...

sin²x = 1 − cos2x

2 =

2 ∗ x² 2! −

2³∗x⁴ 4! +

2⁵∗x⁶ 6! +...

x→0lim

2∗x² 2! −²

3∗x⁴ 4! +²

5∗x⁶

6! +... − x² x²∗ (^2∗x

2

2! −²

3∗x⁴ 4! +²

5∗x⁶ 6! +...)

xlim→0

−²

3∗x⁴ 4! +²

5∗x⁶ 6! +...

2∗x⁴ 2! −²

3∗x⁶ 4! +²

5∗x⁸ 6! +...⇒

−2 3

2 =

−1 3

Solution 2.

x→0lim

sin²x − x²

x²sin²x because equ. satisfy L’H (0 0)

⇒lim

x→0

sin2x − 2x

2xsin²x + x²sin2x using equ. sin2x = 2sinxcosx and L’H (0 0)

⇒lim

x→0

2cos2x − 2

2sin²x + 4xsin2x + 2x²cos2x using equ. L’H (0 0)

⇒lim

x→0

−4sin2x

6sin2x + 12xcos2x − 4x²sin2x using equ. L’H (0 0)

⇒lim

x→0

−8cos2x

24cos2x − 32sin2x − 8x²cos2x

⇒

−8 24 ⇒

−1 3

score(solution1): Final answer 1 point. Using cosx to find sin2x 3 points. Find cos2x 1 point. Find sin2x 2 point.

score(solution2): Final answer 1 point. Using L’H 3 points. Correct 1st differential equ 1 point. Correct 2nd differential equ 2 point.

Page 3 of 10

4. (14%)

(a) (7%) 導出 tan⁻¹x在 x = 0 處之泰勒展示。(需求出 第 n 項) (b) (7%) 求 lim_x→03 tan⁻¹x − 3x + x³

3x⁵ .

Solution:

(a) Solution 1.

1

1 − x =1 + x + x²+... + xⁿ 1

1 + x² =

dtan⁻¹x dx =

1 1 − (−x²)

=1 + (−x²) + (−x²)²+... + (−x²)ⁿ=1 − x²+x⁴−x⁶+... + (−1)ⁿx²n tan⁻¹x =∫

1

1 + x²dx. = x −x³ 3 +

x⁵

5 +... + (−1)ⁿ 2n + 1x²ⁿ⁺¹=

∞ n=0∑

(−1)ⁿ 2n + 1x²ⁿ⁺¹

Solution 2.

dtan⁻¹x

dx = (1 + x²)⁻¹ (1 + x²)⁻¹=

∞

∑

n=0

C_n⁻¹x²ⁿ

tan⁻¹x =

∞

n=0∑C_n⁻¹x²ⁿ⁺¹ 2n + 1

score: Final answer 1 point. Know tan⁻¹x = 1

1 − x² 2 points.

Know 1

1 − x =1 + x + x²+... 2 points. Correct concept 4 points.

(b) Solution 1.

limx→0

3tan⁻¹x − 3x + x³ 3x⁵ =lim

x→0

3(x −^x₃³+^x

5

5 +...) − 3x + x³

3x⁵ =lim

x→0

3(^x₅⁵−^x

7

7 +...) 3x⁵

= 3 5∗

1 3=

1 5. Solution 2.

limx→0

3tan⁻¹x − 3x + x³

3x⁵ because equ. satisfy L’H (0 0) limx→0

3(_1+x¹2) −3 + 3x² 15x⁴ =lim

x→0

3 − 3 − 3x²+3x²+3x⁴ (1 + x²)15x⁴ =lim

x→0

3x⁴ 15x⁴+15x⁶ ⇒

1 5 score: answer 1 point. Using equ. L’H or upward result of (a.) 6points.

5. (14%) 求出以下兩個不定積分。

(a) (7%) ∫ x sin⁻¹xdx.

(b) (7%) ∫

ln x x ln x + xdx.

Solution:

(a). Let u = sin⁻¹x, dv = xdx then du = dx

√

1 − x², v = x² 2 .

∫ x sin⁻¹x = x²

2 sin⁻¹x − 1 2∫

x²dx

√

1 − x² (1)

For integral in (1), let x = sin θ then dx = cos θdθ

∫

x²dx

√

1 − x² = ∫

sin²θ

√

1 − sin²θ

cos θdθ =∫ sin²θ

cos θ cos θdθ

= ∫ sin²θdθ =∫

1 − cos 2θ 2 dθ

= θ 2 −

sin 2θ 4 Since sin θ = x then cos θ =

√

1 − x², sin 2θ = 2x

√

1 − x² and θ = sin⁻¹x θ

2− sin 2θ

4 = 1

2sin⁻¹x − x 2√

1 − x² Hence (1) become

∫ x sin⁻¹x =x²

2 sin⁻¹x −1

4sin⁻¹x + x 4√

1 − x² +C Grading Critiria:

ˆ 使用分佈積分並正確令u = sin⁻¹x, dv = xdx得2分

ˆ 正確寫下使用分佈積分的結果(式子(1))得

1

ˆ 根據式子(1)的結果，令x = sin θ並正確寫下變數替換後的式子得

2

ˆ 剩下計算過程加解答

2

ˆ 也可以用其他方法做，給分原則不變。

(b). Let u = ln x then du =dx x

∫

ln x

x ln x + xdx =∫ u

1 + udu =∫ (1 − 1 1 + u)du

=u − ln (1 + u) + C = ln x − ln (1 + ln x) + C Grading Critiria:

ˆ 使用變數變換，令u = ln x並正確寫下變換後的式子得

3

ˆ 剩下計算過程和解答共佔

4

Page 5 of 10

6. (6%) 求出 sin(x²)在 x = 0 處之泰勒展示的第 n 項。(可可可直接使用 sin x 及 cos x 之泰勒展示，不必推導。)

Solution:

Write down the tylor expansion of sin x

sin x = x −x³ 3! +

x⁵ 5! −

x⁷ 7! + ⋯⋯

Substitute x with x² then it’s the tylor expansion of sin(x²)

sin x²=x²− x⁶

3! + x¹⁰

5! − x¹⁴

7! + ⋯⋯

Grading Critiria:

ˆ 寫下sin x的泰勒展式得

3

ˆ 把x換成x²帶入sin x的泰勒展式得

3

ˆ 化簡有錯扣

1

ˆ sin x的泰勒展示寫錯直接0分

7. (7%) 求出 sin²x在 x = 0 處之泰勒展示的第 n 項。(可直接使用 sin x 及 cos x 之泰勒展示，不必推導。)

Solution:

Method1

sin²x = 1 − cos 2x

2 (1 point)

cos x =

∞

∑

n=0

(−1)ⁿ x²ⁿ

(2n)! (2 points)

cos(2x) =

∞

n=0∑(−1)ⁿ(2x)²ⁿ

(2n)! (2 points)

sin²x = 1 2−

1 2

∞

n=0∑(−1)ⁿ(2x)²ⁿ (2n)!

= 1 2

∞

∑

n=1

(−1)ⁿ⁻¹2²ⁿ (2n)! x²ⁿ

=

∞

∑

n=1

(−1)ⁿ⁻¹2²ⁿ⁻¹

(2n)! x²ⁿ (2 points)

Method2

d

dxsin²x = 2 sin x cos x = sin(2x) (1 point) sin x =

∞

n=0∑(−1)ⁿ x²ⁿ⁺¹

(2n + 1)! (2 points)

sin(2x) =

∞

n=0∑(−1)ⁿ(2x)²ⁿ⁺¹

(2n + 1)! (2 points)

sin²x =∫

x 0

∞

∑

n=0

(−1)ⁿ(2x)²ⁿ⁺¹ (2n + 1)! dx

=

∞

∑

n=0

(−1)ⁿ2²ⁿ⁺¹ (2n + 2)! x²ⁿ⁺²

=

∞

∑

n=1

(−1)ⁿ⁻¹2²ⁿ⁻¹

(2n)! x²ⁿ (2 points)

Therefore, the n-th term of the Taylor expansion of sin²x centered at x = 0 is

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

0 , n = 0

(−1)ⁿ⁻¹2²ⁿ⁻¹

(2n)! x²ⁿ , n ≥ 1

Note that there will be 1˜2 point(s) deducted if you (to some degree) don’t specify the n-th term or just leave the answer in form of 1

2 − 1 2

∞

n=0∑(−1)ⁿ(2x)²ⁿ (2n)!

Page 7 of 10

8. (10%) 求曲線 y = ln sec x 在 0 ≤ x ≤ π

4 間之長度。(任何積分公式都可使用，無須推導。)

Solution:

The arc length of the curve L =∫

b a

√

1 + [f^′(x)]² dx, (2 points) where f (x) = ln sec x and f^′(x) = 1

sec x⋅sec x tan x = tan x (2 points) L =∫

π/4 0

√

1 + tan²x dx (2 points)

= ∫

π/4 0

sec x dx (1 point)

= ∫

π/4 0

sec x ⋅sec x + tan x sec x + tan x dx

= ∫

π/4 0

sec²x + sec x tan x sec x + tan x dx

=ln ∣ sec x + tan x∣∣

π/4 0

(2 points)

=ln ∣

√

2 + 1∣ − ln ∣1 + 0∣

=ln(√

2 + 1) (1 point)

Note that there will be no point if you put the formula in the wrong way at the beginning.

9. (20%) 設 Ω 是由 y = 1

x(3 − x), x = 1, x = 2 及 x軸所包圍的區域。

(a) (7%) 求 Ω 之面積。

(b) (7%) 求 Ω 繞 x 軸旋轉所產生的體積。

(c) (6%) 求 Ω 繞 y 軸旋轉所產生的體積。

Solution:

Ω bounded by (y = 1

x(3 − x)) + (x = 1) + (x = 2) + (y = 0)

0 0.5 1 1.5 2 2.5 3

0 0.1 0.2 0.3 0.4 0.5

x

y

(a)

∫

2 1

1

x(3 − x)dx =∫

2 1

1/3 x +

1/3 3 − xdx = 1

3(ln(x))²₁+ 1

3(−ln(3 − x))²₁= 2 3ln(2) Note:

1

x(3 − x)= A x + B

3 − x⇒1 = A(3 − x) + Bx ⇒ A = 1/3 , B = 1/3 配分:

寫出公式 (1 pt) 求出係數 (3 pt) 積分 (2 pt) 答案 (1 pt)

(b) 法一

V_x= ∫

2 1

π( 1

x(3 − x))²dx =∫

2 1

π2/27 x +

1/9 x² +

2/27 3 − x+

1/9 (3 − x)²dx

=π[2

27(ln(x))²₁+ 1 9

−1 x +

2

27(−ln(3 − x))²₁+ 1 9

1

3 − x] =π( 4

27ln(2) +1 9)

Note:

( 1 x(3 − x))²=

A x +

B x²+

C 3 − x+

D (3 − x)²

Page 9 of 10

⇒1 = Ax(3 − x)²+B(3 − x)²+Cx(3 − x) + Dx² put x = 0, x = 3, x = 1, x = −1, ⇒ A = 2/27, B = 1/9, C = 2/27, D = 1/9

配分:

寫出公式 (1 pt) 求出係數 (3 pt) 積分 (2 pt) 答案 (1 pt)

法二

By (a) we have

V_x= ∫

2 1

π( 1

x(3 − x))²dx =∫

2 1

π(1/3 x +

1/3

3 − x)²dx = π 9∫

2 1

( 1 x+

1

3 − x)²dx = π 9∫

2 1

( 1 x²+

2 x(3 − x)+

1 (3 − x)²)dx

= π 9 ∫

2 1

1 x²dx +π

9 ∫

2 1

2

x(3 − x)dx +π 9∫

2 1

1

(3 − x)²dx = π 9(

−1 x)²₁+

π 9 ⋅2 ⋅2

3ln(2) +π 9(

1

(3 − x))²₁=π(4

27ln(2) + 1 9)

(c) 法一

V_y= ∫

2 1

2πxf (x)dx =∫

2 1

2πx 1

x(3 − x)dx =∫

2 1

2π 1

3 − xdx = 2π∫

2 1

(−ln(3 − x))²₁=2πln(2)

配分:

寫出公式 (1 pt) 積分 (3 pt) 答案 (2 pt)

法二

由 Pappus Thm

(對y軸旋轉之體積)=(Ω質心繞一圈軌跡長)(Ω 面積) 容易看出Ω 對x方向為對稱圖形，故質心之x座標: ¯x =3

2 由(a)

⇒Vy=2π(3

2) ⋅ ∣Ω∣ = 2π ⋅ (3 2) ⋅ (

2

3ln(2)) = 2πln(2)