微甲

991微甲^08-12班期末考解答和評分標準 1. (7%) Evaluate

∫ ln 7 0

e^2x

(e^x+ 1)¹³ dx.

Sol:

Let u = e^x+ 1, du = e^xdx, x = 0, u = 2, x = ln 7, u = 8.

∫ _ln7

0

e^2x

(e^x+ 1)¹³ dx =

∫ ₈

2

u− 1

u¹³ du (3pt up)

=

∫ 8 2

u²³ − u¹³ du = 3

5u⁵³ − 2 3u²³

= 66 5 + 3

102²³ 2. (10%) Evaluate

∫ _∞

0⁺

√ 4dx

x(4 + x). Caution! lim

x→0⁺

√ 4

x(4 + x) =∞.

Sol:

Let u =√

x, du = 1 2√

x, x = 0⁺, u =√ 0⁺

lim

b→∞

∫ _b

0⁺

√ 4dx

x(4 + x) = lim

b→∞

∫ ^√_b

√0⁺

8

4 + u² du (5pt up) Let u = 2 tan θ, du = 2 sec²θdθ,

blim→∞

∫ ^√b

√0⁺

8

4 + u² du =

∫ ^π

2

0

4 dθ = 2π θ 變換好^{7pt up}

一點點計算錯誤扣一分^, 上下限扣兩分。

3. (8%) Evaluate

∫ ^π

2

0

dx

3− 5 sin x. Hint: Set t = tan(x 2

) and express sin x and dx in terms of t.

Sol:

Wrong answer (you will get points even if the answer here is incorrect)Let t = tan(x 2)⇒

sin x = 2t

1 + t² and dx = 2 1 + t² dt,

∫ ^π

2

0

dx

3− 5 sin x =

∫ ₁

0

1 3− _1+t^5·2t2

· 2 1 + t² dt

= 2

∫ 1 0

1

3t²− 10t + 3dt (3%)

= 2

∫ ₁

0

1

(3t− 1)(t − 3)dt

= 2

∫ ₁

0

( −³₈ 3t− 1 +

1 8

t− 3 )

dt

= 1

4[− ln |3t − 1| + ln |t − 3|]¯¯¯¹

0

=−1

4ln 3 (5%) Correct answer

2

∫ 1 0

1

3t²− 10t + 3dt = lim

a→¹₃⁻2

∫ a 0

1

3t²− 10t + 3dt + lim

b→¹₃⁺2

∫ 1 b

1

3t²− 10t + 3dt Since

lim

a→¹3

−2

∫ a 0

1

3t² − 10t + 3dt = lim

a→¹3

−2

∫ a 0

( −³₈ 3t− 1+

1 8

t− 3 )

dt

= lim

a→¹₃⁻

1

4[− ln |3t − 1| + ln |t − 3|]¯¯¯^a

0 =∞

and

lim

b→¹₃⁺

2

∫ 1 b

1

3t² − 10t + 3dt = lim

b→¹₃⁺

2

∫ 1 b

( −³₈ 3t− 1+

1 8

t− 3 )

dt

= lim

b→¹3 +

1

4[− ln |3t − 1| + ln |t − 3|]¯¯¯¹

b =−∞

∴ 2

∫ 1 0

1

3t²− 10t + 3dt does not exist.

4. (10%) Let Ω be the region enclosed by y = 2e^√x

2x− 1, y = 0, x = 1 and x = 4. Find the volume generated by revolving Ω about the line x = 1

2. Sol:

∫ ₄

1

2π(x− 1 2)

( 2e^√x 2x− 1

)

dx(3%) = 2π

∫ ₄

1

e^√xdx

= 4π

∫ 2 1

ue^udu (u =√

x⇒ dx = 2u du)

= 4π(ue^u− e^u)¯¯¯²

1 = 4πe² (7%).

5. (10%) Evaluate

∫ ₂

1

x²+ 4

x⁴+ 3x³+ 2x² dx.

Sol:

x² + 4

x⁴+ 3x³+ 2x² = a x + b

x² + c

x + 1 + d

x + 2 (4 pts)

⇒ x²+ 4 = ax(x + 1)(x + 2) + b(x + 1)(x + 2) + cx²(x + 2) + dx²(x + 1) Put x = 0: 4 = 2b ⇒ b = 2.

Put x =−1: 5 = c.

Put x =−2: 8 = −4d ⇒ d = −2.

Put x = 1: 5 = 6a + 6b + 3c + 2d ⇒ a = −3. (3 pts)

∫ 2 1

x² + 4

x⁴+ 3x³+ 2x² dx =

∫ 2 1

(−3 x + 2

x² + 5

x + 1 + −2 x + 2) dx

=−3 ln x¯¯¯²

1

+ (−2)1 x

¯¯¯²

1

+ 5 ln(x + 1)¯¯¯²

1− 2 ln(x + 2)¯¯¯²

1

=−3 ln 2 + 1 + 5(ln 3 − ln 2) − 2(ln 4 − ln 3)

= 1 + 7 ln 3− 12 ln 2. (3 pts)

6. (15% total) (a) Find

∫

sec³x dx, given

∫

sec x = ln| sec x + tan x| + a constant. (7%)

(b) The curve y = ln x, 0 < x < 1, is rotated about y-axis. Find the area of the resulting surface. (8%)

Sol:

(a) step1: Using integration by parts yields

∫

sec³x dx = tan x sec x−

∫

tan²x sec x dx = tan x sec x−

∫

(sec²x− 1) sec x dx.

step2: By rearrangement of the above result, it follows that

∫

sec³x dx = 1 2

(

sec x tan x + ln¯¯sec x + tan x¯¯) + C, where C is an arbitrary constant.

(b) step3: The area of the resulting surface is

∫

circumference∗ ds =

∫ ₁

0

(2πx)√

1 + (y⁰)²dx = 2π

∫ ₁

0

√1 + x²dx.

step4: Let x = tan θ, then dx

dθ = sec²θ and the previous integral thus becomes 2π

∫ ^π

4

0

(sec θ) sec²θ dθ.

step5: From (a), one can evaluate 2π

∫ ^π

4

0

sec³θ dθ =√

2π + π ln(√

2 + 1), which completes (b).

Grading Policy:

1. 4 points for step1.

2. 3 points for step2 with total 7 points for (a).

3. 4 points for step3.

4. 2 points for step4.

5. 2 points for step5 with total 8 points for (b).

7. (10%) Solve (x²+ 1)²dy

dx + 3x(x²+ 1)y = 2x with the initial condition y(0) = 3.

Sol:

step1: First we derive the integrating factor and take C = 0 to make it simple, I(x) = e

∫ _3x

x2+1dx

= (x²+ 1)³². (5 pts)

step2: By multiplying the above integrating factor, the original ODE can be rearranged as

((x²+ 1)³²y)⁰ = 2x(x²+ 1)⁻¹²

⇒ ((x²+ 1)³²y) = 2(x²+ 1)¹² + C⁰

⇒ ((x²+ 1)³²y) = 2(x²+ 1)¹² + 1 (5 pts) 8. (10%) Solve dy

dx = y²sin x

1 + y³ , y(0) = 1.

Sol:

1 + y³

y² dy = sin xdx (5 pts)

∫ 1 + y³ y² dy =

∫

sin xdx

−1 y +y²

2 =− cos x + c (3 pts)

Since y(0) = 1, we have

−1 + 1

2 =−1 + c 1

2 = c (2 pts)

Hence, the solution is

y² 2 − 1

y + cos x = 1 2. 9. (10%) Consider the parametric curve with x⁰(t) = √

3t, y⁰(t) = 2t√

t− t², 0 ≤ t ≤ 1.

Find the arc length of this curve.

Sol:

Length =

∫ ₁

0

√(x⁰(t))²+ (y⁰(t))²dt (1 pt)

=

∫ 1 0

t√

−4t²+ 4t + 3 dt

=

∫ ₁

0

2t

√

1− (t −1

2)²dt (2 pts) Let θ = arcsin(t−1

2)⇒ t = 1

2 + sin θ, and dt = cos θdθ (1 pt)

Length =

∫ ^π

6

−^π₆

(2 sin θ + 1) cos θ cos θ dθ (2 pts.)

=

∫ ^π

6

−^π₆

2 sin θ cos²θdθ +

∫ ^π

6

−^π₆

cos²θ dθ

= (−2

3cos³θ)¯¯

¯¯

π 6

−^π6

+ (θ

2+ sin 2θ 4 )¯¯

¯¯

π 6

−^π6

(1 pt for each term)

= 0 + π 6 +

√3 4

= π 6 +

√3

4 (2 pts)

10. (10%) Find the surface area of the solid generated by revolving the cardioid r = 1 + sin θ about y-axis.

Sol:

A =

∫ _π

0

2πx ds

=

∫ π/2

−π/2

2πr cos θ

√ r²+

(dr dθ

)2

dθ (x = r cos θ.)

= 2π

∫ _π/2

−π/2

(1 + sin θ) cos θ√

2 + 2 sin θ dθ

= 2π

∫ 2 0

t√

2t dt (Set t = 1 + sin θ.)

= 32 5 π.

Grading Policy:

1. Write down or derive ds =

√ r²+

(dr dθ

)2

dθ. (3 pt)

2. Insert r(θ) (1 pt) and the corresponding range of θ (2 pts).

3. Evaluate this integral by using any proper techniques and taking correct steps. (2 pts. You will get all of these 2 points if you only make a mistake in inserting a range of θ in previous steps.)

4. Give the correct value of the surface area. (2 pts)