108101-04IIIʟʟʟʠʠʠὃὃὃYYYȤȤȤᑖᑖᑖ 1. (10%)  F(x)=S x

1081微微微乙乙乙01-04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%) 令 F (x) =∫

x²

0

cos(t²+t) dt.

(a) (6%) 求 F^′(x). (b) (4%) 求極限 lim_x→0F (x) x² .

Solution:

(a) Let G(x) = ∫

x

0

cos(t² +t)dt and h(x) = x². Then F (x) = G(h(x)). By Fundamental Theorem of Calculus, one has G^′(x) = cos(x²+x). (2%) By Chain Rule, we have

F^′(x) = G^′(x)(h(x)) ⋅ h^′(x) = cos(x⁴+x²) ⋅2x. (2%, 2%)

評分原則：

(i) 直接寫出正確答案者給滿分。

(ii) 知道要用微積分基本定理並列出相關函數者得兩分。

(iii) 沒使用連鎖律或使用錯誤者，依上面配分給分。

(b) By using L’Hospital rule (for the 0

0 form), (1%) we have limx→0

F (x) x² =lim

x→0

F^′(x)

2x (2%)

=lim

x→0cos(x⁴+x²) =1. (1%)

2. (14%) 計算下列不定積分。

(a) (7%) ∫ sin

√

2x + 1 dx (b) (7%) ∫ tan⁻¹(2 x)dx

Solution:

(a)

Let u =√

2x + 1, du = 1

√

2x + 1dx or dx = udu

∫ sin

√

2x + 1dx =∫ (sin u) ⋅ udu

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3 pts for corect substitution

= ∫ u ⋅ d(− cos u) = u ⋅ (− cos u) − ∫ (− cos u)du (2 pts for integration by parts.)

= −u cos u + sin u + C

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

1 pt for∫ cos udu = sin u

= −

√

2x + 1 cos(

√

2x + 1) + sin(

√

2x + 1) + C

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

1 pt

(b) ∫ tan⁻¹( 2

x)dx = x ⋅ tan⁻¹( 2

x) − ∫ x ⋅ d (tan⁻¹( 2

x)) (2 pts)

=x ⋅ tan⁻¹(2

x) − ∫ x ⋅ 1 1 + (²_x)

2(− 2 x²)dx

=x ⋅ tan⁻¹( 2 x) + ∫

2x x²+4dx

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2 pts for d

dx(tan⁻¹( 2 x))

=x ⋅ tan⁻¹(2

x) +ln(x²+4) + C (2 pts for∫ 2x

x²+4dx, 1 pt for final answer.)

3. (18%) 計算下列不定積分。

(a) (9%) ∫

1 x²

√

1 + x² dx (b) (9%) ∫

x²+4x − 1 x⁴−1 dx

Solution:

(a)

∫

1 x²√

1 + x²dx (Let x = tan θ, dx = sec²θdθ)

= ∫

sec²θ

tan²θ sec θdθ (3%)

= ∫

cos θ sin²θdθ

= ∫ 1

sin²θd(sin θ) (3%)

= − 1

sin θ +C (1%)

= −

√ 1 + x²

x² +C (2%) (b) Write

x²+4x − 1 x⁴−1 =

a x + 1 +

b x − 1 +

cx + d

x²+1. (1%) Then one has

x²+4x − 1 = a(x − 1)(x²+1) + b(x + 1)(x²+1) + (cx + d)(x²−1).

Put x = 1, b = 1. Put x = −1, a = 1. Compare the coefficients of x³ and the constants on the both sides, one has c = −1 and d = 1. (4%; 每個變數各一分) So we have

x²+4x − 1 x⁴−1 = 1

x + 1+ 1

x − 1 +−2x + 1 x²+1 . Than

∫

x²+4x − 1

x⁴−1 dx =∫ ( 1

x + 1+ 1 x − 1+

−2x

x²+1+ 1 x²+1)dx

=ln ∣x + 1∣ + ln ∣x − 1∣ − ln(x²+1) + tan⁻¹x. (4%; 每個積分各一分)

4. (18%) 已知a > 0. 令 Ωa: y = x

x + 1, x-軸, x = a 圍成的區域；Σa: y = x

x + 1, y-軸, y = a

a + 1 圍成的區 域.(Ωa和Σa合成一長方形如圖). 令 U(a) 為 Ωa 繞 x-軸旋轉之旋轉體體積；V (a) 為 Σa 繞 y-軸旋轉之

旋轉體體積。

b

)b!

*

͈

͂

b b

y z

,

(a) (8%) 求 U(a).

(b) (8%) 求 V (a).

(c) (2%) 求 U(a) − V (a).

Solution:

([8%]+[8%])會圓盤法或殼形法公式2%；積出不定積分部分4%；算值正確2%。

U (a) = ∫

a

0

π ( x x + 1)

2

dx[2%]

= π∫

a

0

(1 − 1 x + 1)

2

dx

= π∫

a

0

1 − 2 x + 1+

1

(x + 1)²dx[6%]

= π (a − 2 ln(a + 1) + (− 1

a + 1 +1))

= π ((a + 1 − 1

a + 1) −2 ln(a + 1)) [8%]

V (a) = ∫

a

0

2πx( a a + 1 −

x

x + 1)dx[2%]

= 2π (∫

a

0

a

a + 1x dx −∫

a

0

x² x + 1dx)

= 2π (∫

a

0

a

a + 1x dx −∫

a

0

x − 1 + 1

x + 1dx) [6%]

= π ((a + 1 − 1

a + 1) −2 ln(a + 1)) [8%]

[2%]所以對任何a, U(a) − V (a) = 0.

5. (12%) 求下列極限。

(a) (6%) lim

x→0

cos x²−1

sin x⁴ (b) (6%) lim

x→1( 2

x²−1− 1 ln x)

Solution:

(a) 使用羅必達時要說明type 0 0, ∞

∞

,沒寫 type 扣一分, 因 type 沒寫單小題最多扣兩分。

Method 1.

limx→0

cos x²−1 sin x⁴

=lim

x→0(

cos x²−1 sin x⁴

sin x⁴

x⁴ ) 2 point

=lim

x→0

cos x²−1

x⁴ 1 point

=lim

x→0

−2x sin x²

4x³ 1 point

= −1 2lim

x→0

sin x²

x² 1 point

= − 1

2 1 point.

Method 2.

limx→0

cos x²−1 sin x⁴

=lim

x→0

−2x sin x²

4x³cos x⁴ 2 point

= −1 2lim

x→0(sin x² x²

1

cos x⁴) 2 point

= − 1 2lim

x→0( sin x²

x² )lim

x→0( 1

cos x⁴) 1 point

= −1

2 1 point.

Method 3.

limx→0

cos x²−1 sin x⁴ =lim

x→0

−^x

4

2! +^x

8

4! − ⋯ x⁴−^x

12

3! + ⋯

4 point

lim−_2!¹ +^x

4

4! − ⋯

1 point

(b) Method 1

limx→1( 2 x²−1−

1

ln x) =lim

x→1

2 ln x − (x²−1)

(x²−1) ln x 2 point.

=lim

x→1

2/x − 2x

2x ln x + (x²−1)/x 2 point.

=lim

x→1

−2/x²−2

2 ln x + 2 + 1 + 1/x² 1 point.

= −1 1 point.

此處必須說明 (各一分)

limx→1(x²−1) ln x = 0 = lim

x→1x ln x = 0.

才能直接在上面計算用羅必達.

Method 2

limx→1( 2 x²−1−

1

ln x) =lim

x→1

ln x²− (x²−1)

(x²−1) ln x 1 point

=lim

x→1

−¹₂(x²−1)²+¹₃(x²−1)³− ⋯

(x²−1) ln x 2 point limx→1

(x²−1) ln x ⋅lim

x→1(−

1 2+

1

3(x²−1)²) − ⋯ 1 point

=lim

x→1

2x 1/x ⋅

−1

2 1 point

= −1 1 point.

6. (a) (4%) 寫下 −1

√

1 − x² 對 x = 0 的泰勒展式。

(b) (6%) 求 cos⁻¹x 對 x = 0 的泰勒展式。

(c) (4%) 求 cos⁻¹(x²)在 x = 0 的 10 次泰勒多項式。

Solution:

(a)

−1

√

1 − x² = −(1 − x²)^−1/2

= −

∞

∑

k=0

(−1)^kC_k^−1/2x^2k 4 point.

正負號不對扣一分.

(b) Take integral on both sides above (2 point) , we have cos⁻¹x = c −

∞

∑

k=0

1

2k + 1(−1)^kC_k^−1/2x^2k+1 3 point.

Let x = 0, we have c = π/2 1 point.

(c) From (b), we have

cos⁻¹x² = π 2 −

∞

∑

k=0

(−1)^kC_k^−1/2 x^4k+2 2k + 1.

P₁₀(x) = π 2 −

2

∑

k=0

(−1)^kC_k^−1/2x^4k+2 2k + 1

= π

2 −x²− 1 6x⁶−

3

40x¹⁰ 1 point for each term.

7. (14%) 若 f(x) = 1 2x²+

1

3 ⋅ 2x³+ ⋯ + 1

n(n − 1)xⁿ+ ⋯, 在 ∣x∣ < 1.

(a) (2%) 求 f⁽¹⁰⁾(0)

(b) (6%) 求 f^′(x) 和 f^′′(x) 在 x = 0 的泰勒展式，並認出它們是什麼基本函數。

(c) (6%) 把 f (x) 表示成基本函數。

Solution:

(a) The coefficient of the term x¹⁰ in the Taylor series of f (x) at x = 0 is f⁽¹⁰⁾(0) 10! =

1 10 ⋅ 9 ⇒ f⁽¹⁰⁾(0) = 10!

10 ⋅ 9 =8! (2 pts)

(b) The Taylor series of f^′(x) at x = 0 is x + 1

2x²+ ⋯ + 1

nxⁿ+ ⋯ (2 pts) x + 1

2x²+ ⋯ + 1

nxⁿ+ ⋯ = −ln(1 − x) (1 pt) The Taylor series of f^′′(x) at x = 0 is 1 + x + x²+ ⋯ +xⁿ+ ⋯ (1 pt)

1 + x + x²+ ⋯ +xⁿ+ ⋯ = 1

1 − x (2 pts) (c) f (x) =∫

x

0

f^′(t)dt + f (0) (1 pt) Since f (0) = 0, wehave

f (x) =∫

x

0

f^′(t)dt

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(1 pt for f (0) = 0)

= ∫

x

0

−ln(1 − t)dt = − [t ln(1 − t)∣

t=x t=0− ∫

x

0

t −1

1 − tdt] (2 pts for integration by parts)

= −x ln(1 − x) +∫

x

0

t t − 1dt

= −x ln(1 − x) +∫

x

0

1 + 1 t − 1dt

= −x ln(1 − x) + x + ln ∣x − 1∣

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(2 pts)