Solution: We have f′(x

1081模模模組組組03-07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. (6 pts) Find f^′(2) if f (x) = e^g(x)and g(x) =∫

x² 4

t 1 + t⁴dt.

Solution:

We have f^′(x) = e^g(x)g^′(x)(1 point). From Fundamental Theorem of Calculus(2 point),

g^′(x) = d dx∫

x² 4

t

1 + t⁴dt = x²

1 + x⁸⋅2x = 2x³

1 + x⁸(1 point).

We get e^g(2)=e⁰=1 (1 point)and g^′(2) = 2⁴ 1 + 2⁸ =

16

257(1 point). Therefore f^′(2) = 16 257.

2. Compute the following integrals.

(a) (6 pts) ∫

1 0

sin⁻¹(x) dx (b) (10 pts) _∫ ^√1 + x²dx (c) (12 pts) _∫ ^x

3+4x²+4x + 2 x⁴+2x³+2x² dx .

Solution:

(a) ∫

1

0 sin⁻¹xdx = x sin⁻¹x∣

1 0− ∫

1

0 x(sin⁻¹x)^′dx ⋯⋯⋯⋯(2pts for integration by parts)

= π 2 − ∫

1 0

x

√

1 − x²dx ⋯⋯⋯⋯(1pt for sin⁻¹1 = π

2 and 1pt for the other)

= π 2 +

1 2∫

0 1

du

√u (Let u = 1 − x²⇒du = −2xdx)⋯⋯⋯⋯(1pt for correct substitution)

= π 2 −

√u∣

1 0

= π

2 −1 ⋯⋯⋯⋯(1pt) (b) ∫

√

1 + x²dx =∫

√

sec²θ sec²θdθ (Let x = tan θ for −π

2 <θ <π

2 ⇒dx = sec²θdθ)

⋯⋯⋯⋯(3pts for trigonometric substitution)

= ∫ sec³θdθ (∵ sec θ > 0 for −π

2 <θ < π 2)

∫ sec³θdθ =∫ sec θ sec²θdθ = sec θ tan θ −∫ tan θ(sec θ)^′dθ ⋯⋯⋯⋯(2pts for integration by parts)

=sec θ tan θ −∫ sec θ ⋅ tan²θdθ = sec θ tan θ −∫ sec²θ(sec θ − 1)dθ

=sec θ tan θ −∫ sec³θdθ +∫ sec θdθ = sec θ tan θ + ln ∣ sec θ + tan θ∣ −∫ sec³θdθ.

Thus, ∫ sec³θdθ = 1

2sec θ tan θ +1

2ln ∣ sec θ + tan θ∣ + C

⎫⎪

⎪⎪

⎪⎪

⎪⎪

⎬

⎪⎪

⎪⎪

⎪⎪

⎪⎭ (3pts)

Hence ∫

√

1 + x²dx =1

2sec θ tan θ +1

2ln ∣ sec θ + tan θ∣ + C =1 2

√

1 + x²⋅x +1 2ln ∣

√

1 + x²+x∣ + C (2 pts) (∵ tan θ = x, −π

2 <θ <π

2 ∴sec θ =

√ 1 + x²) (c) First, we factorize the denominator:

x⁴+2x³+2x²=x²(x²+2x + 2).(1 point) We write the integrand as follows

x³+4x²+4x + 2 x⁴+2x³+2x² =

A x +

B x² +

Cx + D

x²+2x + 2(2 point).

We have

x³+4x²+4x + 2 = Ax(x²+2x + 2) + B(x²+2x + 2) + (Cx + D)x² (1) Put x = 0 into Equation (1), we get B = 1.

Compare the coefficient of x in Equation (1), we have 2A + 2B = 4 ⇒ A = 1.

Compare the coefficient of x² in Equation (1), we have that 2A + B + D = 4 ⇒ D = 1.

Compare the coefficient of x³ in Equation (1), we have that A + C = 1 ⇒ C = 0.

Thus x³+4x²+4x + 2 x⁴+2x³+2x² =

1 x+

1 x²+

1

x²+2x + 2(3 points). Since

∫ 1

xdx = ln ∣x∣ + C(1 point)

∫ 1

x²dx = −x⁻¹+C(1 point)

∫ 1

x²+2x + 2dx = ∫

1

1 + (x + 1)²dx = tan⁻¹(x + 1) + C(2 points), we have

∫

x³+4x²+4x + 2

x⁴+2x³+2x² dx = ln ∣x∣ −1

x+tan⁻¹(x + 1) + C.

Page 2 of 8

3. (10 pts) Determine whether the following improper integral is convergent or divergent. Evaluate it if it is convergent.

∫

∞ 0

e^−2xcos x dx .

Solution:

Let u = e^−2x and v = sin x, then

du = −2e^−2xdx and dv = cos x dx . Perform integration by parts

∫ e^−2xcos x dx = e^−2xsin x + 2∫ e^−2xsin x dx . 1 pt : 第一次提到/使用 by parts, 1 pt : 操作正確

Similarly, for the last integral, let ˜u = e^−2x and ˜v = − cos x, then

d˜u = −2e^−2xdx and d˜v = sin x dx . Again, use integration by parts

∫ e^−2xsin x dx = −e^−2xcos x − 2∫ e^−2xcos x dx . 1 pt : 第二次提到/使用 by parts, 3 pt : 函數選取正確, 1 pt : 計算正確

Putting these together gives

∫ e^−2xcos x dx = e^−2xsin x − 2e^−2xcos x − 4∫ e^−2xcos x dx ,

⇒ ∫ e^−2xcos x dx =1

5e^−2x(sin x − 2 cos x) + C . Therefore,

∫

t 0

e^−2xcos x dx = 1

5e^−2t(sin t − 2 cos t) +2 5 . 1 pt : 得到正確定積分

Note that ∣ sin t − 2 cos t∣ ≤ 3 for any t and e^−2t →0 as t → ∞. As t → ∞, the limit of the right hand side exists, and is equal to 2/5. That is to say, the improper integral converges, and is 2/5.

1 pt : 正確論證極限過程, 1 pt : 答案正確

4. (10 pts) Consider the crescent-shaped region (called a lune) bounded by arcs of circles with radii r and R, where 0 < r < R. Rotate the region about the y-axis. Find the resulting volume.

x y

Solution:

2π∫

r 0 (

√

r²−x²+

√

R²−r²−

√

R²−x²)x dx = 2

3π (r³−R³+ (R²+ r²

2 )

√

R²−r²)

Page 4 of 8

5. (12 pts) Given an increasing supply function S(q) and a decreasing demand function D(q) where S(q) and D(q) are continuous, we define the total surplus at quantity q as T S(q) =∫

q 0

D(t) − S(t) dt for q ≥ 0.

(a) (4 pts) Show that if D(q^∗) =S(q^∗)for some q^∗>0 then T S(q) obtains the absolute maximum value at q = q^∗. (b) (8 pts) Suppose that D(q) = 6 − (1 +q

2)

2/3

and S(q) = (1 +q 2)

1/3

. Compute T S(q) and find the absolute maximum value of T S(q).

Solution:

(a) T S^′(q) = d dq∫

q 0

D(t) − S(t)dt = D(q) − S(q) ⋯⋯⋯⋯(2pts)

Because D(q) is decreasing and S(q) is increasing, we know that T S^′(q) = D(q) − S(q) is increasing.

Hence D(q^∗) =S(q^∗)implies that T S^′(q^∗) =0. ⋯⋯⋯⋯(1pt) Moreover, T S^′(q) > 0 for 0 < q < q^∗ and T S^′(q) < 0 for q > q^∗.

Hence T S(q) obtains absolute maximum value at q = q^∗. } ⋯⋯⋯⋯(1pt) (b)

T S(q) = ∫

q 0

6 − (1 +t 2)

2 3

− (1 + t 2)

1 3

dt (Let u = 1 + t

2 ⇒du = 1

2dt, dt = 2du) = 2∫

1+q 1

6 − u²³−u¹³du⋯⋯⋯(2pts for correct substitution.)

= 2 [6u −5 3u³⁵ −

3 4u⁴³]

u=1+^q2

u=1

= 6q −6 5(1 + q

2)

5 3

− 3 2(1 +q

2)

4 3

+ 27

10⋯⋯⋯⋯(2pts)

From (a), we know that the absolute maximum value of T S(q) occurs at q = q^∗ if D(q^∗) =S(q^∗)(1pt) D(q^∗) =S(q^∗) ⇒6 − (1 +q^∗

2 )

2 3

= (1 +q^∗ 2 )

1 3

. Let y = (1 +q^∗

2 )

1 3

. 6 − y²=y ⇒ y = 2 or − 3

∵q^∗>0 ∴ (1 +q^∗ 2 )

1 3

=2 ⇒ q^∗=14

⎫⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎬

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎭

⋯⋯⋯⋯(2pts)

T S(q^∗) =6 × 14 −6 5⋅2⁵−

3 2 ⋅2⁴+

27

10=24.3 ⋯⋯⋯⋯(1pt)

6. (a) (6 pts) Solve the initial-value problem: du dt =

2 t + sec²t

2 u , u(0) = −5.

(b) (10 pts) Solve the differential equation (x²+2)y^′(x) + (4x)y = 2x with y(0) = 2.

Solution:

(a)

2udu = (2t + sec²t)dt ⇒ ∫ 2u du = ∫ (2t + sec²t) dt (1pt)

⇒u²=t²+tan t + C, C is constant (3pts) by initial condition ⇒ C = 25 (2pts)

⇒u²=t²+tan t + 25 ⇒ u = −

√

t²+tan t + 25 (b)

⇒ (x²+2) y^′+ (4 x)y = 2 x

⇒ y^′(x) + 4 x

x²+2y = 2 x

x²+2 (2pts)

⇒ I(x) = e^{∫ x2 +24 x dx}=e^(ln(x2+2)2)= (x²+2)²(4pts)

⇒ ((x²+2)²y)^′=2 x(x²+2) ⇒ (x²+2)²y = 1

2x⁴+2 x²+C, C is constant (2pts) by initial condition ⇒C = 8 ⇒y = x⁴+4x²+16

2 (x²+2)² (2pts)

Page 6 of 8

7. (18 pts) The eight-like curve has the following parametric equation:

x = 2√

2 sin t , y = sin t cos t for 0 ≤ t ≤ 2π .

(a) (4 pts) Find the tangent line at t = 0.

(b) (8 pts) Find its arc length.

(c) (6 pts) Find the shaded area which is enclosed by the curve 0 ≤ t ≤ π

2 and the x-axis.

Solution:

(a) The slope is given by

y^′(t) x^′(t)∣

t=0

=

cos²t − sin²t 2√

2 cos t ∣

t=0

= 1 2√

2 .

2 pts : 知道斜率為 y^′/x^′(若微分計算錯誤，後續均不給分), 1 pt : 正確求得該點斜率 (求得 ± 1 2√

2 不給分) The position at t = 0 is (x(0), y(0)) = (0, 0). It follows that the tangent line is y = 1

2√

2x. 1 pt : 正確求得 該點切線方程式 (求得 y = ± 1

2√

2x不給分) (b)

(x^′)²+ (y^′)²= (2√

2 cos t)²+ (cos²t − sin²t)²

=8 cos²t + (2 cos²t − 1)²

=8 cos²t + 4 cos⁴t − 4 cos²t + 1

= (2 cos²t + 1)².

1 pt : 知道要計算 (x^′)²+ (y^′)², 1 pt : 一個微分計算正確, 2 pts : 整理成完全平方

The arc length is

∫

2π 0

√

(x^′)²+ (y^′)²dt =∫

2π 0

(2 cos²t + 1) dt

= ∫

2π 0

(cos(2t) + 2) dt

= ( 1

2sin(2t) + 2t)∣

2π

t=0

=4π .

1 pt : 曲線弧長列式正確(含正確積分範圍), 2 pts : 不定積分計算正確, 1 pt : 最後帶值計算正確 (c) 解法一 As t goes from 0 to π/2, x(t) is increasing. The area of the shaded region is

∫

π 2

0

y(t) dx(t) =∫

π 2

0

(sin t cos t)(2√

2 cos tdt) .

算 換為 dt 式子正確(含正確積分範圍)

Let u = cos t, du = − sin tdt. The integral becomes

2√ 2∫

1 0

u²du = 2√ 2u³

3 ∣

1

u=0

= 2√

2 3 .

1 pt : 變數變換選取正確, 1 pt : 變數變換範圍判斷正確(若先對不定積分變數變換, 則「不定積分計算正確」

佔2分), 1 pt : 不定積分計算正確, 1 pt : 求值正確

解法二 As t goes from 0 to π/2, x goes from 0 to 2√

2. The area of the shaded region is

∫

2√2 0

y dx =∫

2√2 0

x 2√ 2

√ 1 −x²

8 dx . 1 pt : 知道要計算∫ ydx, 2 pts : 定積分範圍以及 y(x) 表達正確

Let u = x²,

∫

2√ 2 0

x 2√ 2

√ 1 − x²

8 dx = 1 4√

2∫

8 0

√ 1 −u

8du = 1 4√

2

−2 ⋅ 8 3 (1 −u

8)

3 2

∣

8

u=0

= 2√

2 3 . 2 pts : 不定積分計算正確, 1 pt : 求值正確

Page 8 of 8