981ʟ ᑖᵬʟʠὃᑖ

98學年度第1學期 微積分甲一組期末考評分標準

1. (16%) (a) Evaluate I₁ =

Z e^x+ 1 e^2x− 9dx.

(b) Evaluate I2 =

Z sec³x tan x dx. 評分標準:

(a) (8%)

Z e^x+ 1 e^2x− 9dx=

Z 2 3

1

e^x− 3 +1 3

1

e^x+ 3dx (1%)

u=e^x

=

Z 2 3

1 u− 3

du u + 1

3 1 u+ 3

du

u (3%)

= 2 9

Z 1

u− 3− 1

udu + 1 9

Z 1

u − 1

u+ 3du (2%)

= 2

9ln |u− 3 u | + 1

9ln | u

u+ 3| + C = 2

9ln |e^x− 3 e^x | + 1

9ln | e^x

e^x+ 3| + C (2%)

(b) (8%)

Z sec³x tan x dx =

Z sec³xtan x tan²x dx=

Z sec²x

sec²x− 1dsec x (4%)

= Z

1 + 1 2

1

sec x − 1 − 1 2

1

sec x + 1 dsec x (2%)

= sec x + 1

2ln |sec x − 1

sec x + 1| + C (2%)

2. (8%) Find the integers A, B, and C such that Z 4

2

(ln x)²dx= A(ln 2)²+ B ln 2 + C.

評分標準:

Z 4 2

(ln(x))²dx= x(ln x)²  

4 2− 2

Z 4 2

ln xdx

= x(ln x)²  

4

2− 2(x ln x − x)  

4 2

= (x ln(x)²− 2x ln(x) + 2x)  

4 2

= 4 ln(4)²− 8 ln(4) + 8 − (2 ln(2)²− 4 ln(2) + 4)

= 4 ln(4)²− 8 ln(4) − 2 ln(2)²+ 4 ln(2) + 4

= 14 ln(2)²− 12 ln(2) + 4

(a) If you reached the second line, you will get 2 points.

(b) If you reached the third line, you will get another 2 points.

(c) If you rearranged the result well, you will get another 2 points even if you lost the correct answer.

(d) The final answer is worth 2 points.

3. (10%) Find the length of the curve y =√

x− 1 from x = 1 to x = 5 4. 評分標準:

From (1, 0) to (5 4,1

2), x = y²+ 1.

Its length = Z ¹₂

0

s

1 + (dx dy)²dy

= Z ¹₂

0

p1 + 4y²dy (let y = 1 2tan θ)

= 1 2

Z ^π₄

0

sec³θdθ

By

Z

sec³θdθ = Z

sec θ(tan θ)⁰dθ = sec θ tan θ − Z

(sec θ)⁰tan θdθ

= sec θ tan θ − Z

sec θ tan²θdθ = sec θ tan θ − Z

sec³θdθ+ Z

sec θdθ

=⇒

Z

sec³θdθ = 1

2sec θ tan θ + 1 2

Z

sec θdθ

= 1

2sec θ tan θ + 1

2ln | sec θ + tan θ| + c, so

the length = 1 2

Z ^π₄

0

sec³θdθ

= 1 2(1

2sec θ tan θ + 1

2ln | sec θ + tan θ|)  

π 4

0

= 1 4

√2 + 1 4ln(√

2 + 1).

4. (16%) Evaluate each integral, or show that the integral diverges.

(a) Z 1

1 2

arcsin√x px(1 − x)dx.

(b) Z _∞

0

1 + x²

1 + x⁴ dx. ( Hint. A particular substitution can be applied by observing that 1 + x²

1 + x⁴ = 1 + _x¹2

x² +_x¹2

. )

評分標準:

(a) (2%): Let√x= u, x = u², dx = 2udu Write lim

k→1⁻

Z k

1

√2

2 arcsin udu p(1 − u²) (2%): = lim

k→1⁻(arcsin(u))²  

k

1

√2

(2%): = lim

k→1⁻(arcsin(k))²− (π

4)² = (π

2)²− (π 4)² (2%): = 3

16π² (b) (2%): Let x − 1

x = u, du = (1 + _x¹2)dx Write lim

m→∞

Z 0

−m

du

u²+ 2 + lim

n→∞

Z n 0

du u²+ 2 (2%): = lim

m→∞( 1

√2arctan( u

√2))  

0

−m+ lim

n→∞( 1

√2arctan( u

√2))  

n 0

(2%): = 1

√2 π 2 + 1

√2 π 2 (2%): = π

2 =

√2 2 π

5. (16%) A ring with height 2h is made by drilling a hole through a ball with radius R > h.

(a) Find the total area of the inner and the outer surface of the ring.

(b) Find the value of h such that the volume of the ring is half of the volume of the whole ball.

評分標準:

(a) The area of the inner surface = 2π√

R²− h²· 2h = 4πh√

R² − h² (2%) There are two meathod to compute the area of the outer surface:

method 1 : Let x = R cos(θ), y = R sin(θ); dx

dθ = −R sin(θ), dy

dθ = R cos(θ);

hence ds = r

(dx

dθ)²+ (dy dθ)²dθ Z arcsin(R^h)

− arcsin(R^h)

2π(R cos(θ))p

R²sin²θ+ R²cos²θdθ (4%)

=

Z arcsin(R^h)

− arcsin(^hR)

2πR²cos θdθ

= 4πRh (2%)

method 2 : x² + y²= R², therefore dx

dy = −y

pR²− y², and ds = s

1 + (dx dy)²dy Z h

−h

2πxds

= 2πp

R²− y² s

1 + y²

dy (4%)

method 1 :

V = 2 Z h

0

(p

R²− y²)²πdy− (√

R²− h²)²π· (2h) (4%)

= 2π(R²y− y³

3)|^h0 − 2h(R²− h²)π

= 4

3h³π (2%) method 2 :

V = 2 Z R

√R²−h²

2πx√

R² − x²dx (4%)

= −2π Z R

√R²−h²

√R²− x²d(R²− x²)

= 4

3h³π (2%) Volume of the ring = 1

2 Volume of the whole ball (2%)

⇒ 4

3h³π = 1 2 ·4

3πR³

⇒ h = 2⁻¹³R

6. (14%) Let Γ1 be the curve (x²+ y²)² = a²(x²− y²), and Γ2 be the curve x²+ y² = a²

2, a > 0 . (a) Find all points of intersection of Γ₁ and Γ₂.

(b) Find the area of the region that lies inside Γ1 and Γ2. ( Hint. Use polar coordinates. ) 評分標準:

(a) Using polar coordinates to change both eqations to r² = a²cos 2θ and r = a

√2. (4 pts) After solving ( a

√2)² = r² = a²cos 2θ, we have cos 2θ = 1

2 which implies θ = ±π 5, ±5π

6 . They intersect at the points

(r, θ) = ( a

√2,π 6), ( a

√2,−π 6 ), ( a

√2,5π 6 ), ( a

√2,−5π

6 ). (4 pts, Each 1 pt) (b) By symmetry, the area is equal to

4hZ ^π₆

0

1 2( a

√2)²dθ+ Z ^π₄

π 6

1

2a²cos 2θdθi

= 4hπa² 24 +a²

4 sin 2θ  

π 4

π 6

i

= a²(1 −

√3 2 +π

6).

For (b), listing the integral gets 4 pts and calculating the exact value gets 2 pts.

7. (10%) Solve the differential equation y⁰− (sec x)y = (cos²x)y² with initial condition y(0) = 1.

評分標準:

Let u = y¹⁻²= y⁻¹, u⁰ = −y⁻²y⁰ (2%) y⁰

−y² − (sec x) y

−y² = (cos²x) y²

−y² u⁰+ (sec x)u = −(cos²x) (3%) e

R

sec xdx = eln | tan x+sec x| = | tan x + sec x| = tan x + sec x As x near 0⁺, y(0) = 1 > 0, tan x + sec x > 0 (2%)

(u(sec x + tan x))⁰ = − cos²x(tan x + sec x) = − sin x cos x − cos x (2%) u(sec x + tan x) = 1

2cos²x− sin x + C u(0) = 1,C = ¹₂

y(x) = 1

u(x) = tan x + sec x

1

2 + ¹₂cos²x− sin x (1%)

8. (10%) (a) Find integer k such that lim

x→0

x− sin x

x^k exists and is non-zero.

(b) Apply Mean Value Theorem for Integrals and the result in (a) to determine the range of p such that lim

x→0⁺

Z x sin x

t^pf(t) dt exists, where f (t) is a continuous function in t and f(0) 6= 0.

評分標準:

(a)

xlim→0

x− sin x

x^k = lim

x→0

1 − cos x kx^k−1

= lim

x→0

sin x

k(k − 1)x^k−2 = lim

x→0

cos x

k(k − 1)(k − 2)x^k−3

∵cos 0 = 1, ∴k− 3 = 0, k = 3 (4%)

xlim→0

x− sin x x^k = 1

6 6= 0 If k < 3 then lim

x→0

x− sin x

x^k = 0 If k > 3 then lim

x→0

x− sin x x^k = ∞ (b) Mean value theorem of integral:

Z x sin x

t^pf(t)dt = c^pf(c)(x − sin x) for some c ∈ (sin x, x) (4%) Note that x > sin x for x > 0

xlim→0

Z x sin x

t^pf(t)dt = lim

x→0c^pf(c)(x − sin x) = lim_x

→0c^p+3(x

c)³(x− sin x x³ )

xlim→0

x− sin x x³ = 1

6 1 = lim

x→0(x

x)³ ≤ lim_x

→0(x

c)³ ≤ lim_x

→0( x

sin x)³ = 1