Apply the partial fraction decomposition of the integrand, we have 1 x3− 1 = a x − 1+ bx + c x2+ x + 1

98學年度第1學期 微積分甲二組期末考解答

1. (10%) Evaluate the integral

Z dx x³− 1. Sol:

Step 1. Apply the partial fraction decomposition of the integrand, we have 1

x³− 1 = a

x − 1+ bx + c x²+ x + 1. Determine the coefficients a, b, c by the following identity

1 = a(x²+ x + 1) + (bx + c)(x − 1).

Note that a + b = 0. Let x = 1, we have 1 = 3a. Hence,

a = 1

3 and b = −1 3. Set x = 0, we have 1 = a − c or c = −2

3. Step 2. Evaluate the integral a

Z 1

x − 1dx.

Note that

1 3

Z 1

x − 1dx = 1

3ln |x − 1| + C1. Step 3. Evaluate the integral

Z bx + c

x²+ x + 1dx by writing 1

3

Z x + 2

x²+ x + 1dx = 1 6

Z 2x + 1

x²+ x + 1dx +1 2

Z 1

x²+ x + 1dx.

Note that

1 6

Z 2x + 1

x²+ x + 1dx = 1

6ln(x²+ x + 1) + C₂. Write

x²+ x + 1 =

 x + 1

2

2

+ 3 4 and denote x + 1

2 by

√3

2 tan θ. Hence, x²+ x + 1 = 3

4(tan²θ + 1) = 3 4sec²θ

and dx =

√3

2 sec²θ dθ.

1 2

Z 1

x²+ x + 1dx = 1 2

√2 3

Z

dθ = 1

√3θ + C₃ = 1

√3tan⁻¹θ + C₃ = 1

√3tan⁻¹ x +¹₂

√3 2

+ C₃.

Z dx x³− 1 = 1

3ln |x − 1| − 1

6ln (x²+ x + 1) − 1

√3tan⁻¹ x +¹₂

√ 3 2

+ C.

2. (10%) Evaluate the integral

Z x

√x²+ 2x + 2dx.

Sol:

Z x

√x²+ 2x + 2dx =

Z x + 1

p(x + 1)²+ 1 − 1

p(x + 1)²+ 1dx

=

Z u

u²+ 1du −

Z du

√u²+ 1

=√

u²+ 1 − ln   u +√

1 + u²   + c

=√

x²+ 2x + 2 − ln  

(x + 1) +√

x²+ 2x + 2   + c Where the 3rd equality follows by:

u = tan θ; du = sec θ; tan²θ + 1 = sec²θ

Z sec²θ sec θdθ =

Z

sec θdθ

= ln |sec θ + tan θ| + c

= ln   u +√

1 + u²   + c

3. (10%) Let I_n = Z ∞

0

xⁿe^−xdx.

(a) Find the recursive relation between I_n and I_n−1.

(a)

I_n = Z ∞

0

xⁿe^−xdx = lim

a→∞

Z a 0

xⁿe^−xdx

= lim

a→∞−xⁿe^−x   

a 0+ n

Z a 0

xⁿ⁻¹e^−xdx

= lim

a→∞−xⁿe^−x   

a

0+ nI_n−1 Since lim

a→∞−xⁿe^−x  

a 0

= lim

a→∞

aⁿ e^a = 0 Hence I_n= nI_n−1

(b) Follow (a) we have I3 = 3I2 = 3 · 2I1 = 3 · 2 · 1I0

Moreover, we have I₀ = Z ∞

0

e^−xdx = lim

a→∞

Z a 0

e^−xdx = lim

a→∞−e^−x  

a 0

= 1 Hence I₃ = 3! = 6.

(c) Same discuss as (b) we can get the general formula of In

I_n= nI_n−1 = n · (n − 1) · I_n−1 = · · · = n!I₀ = n!

4. (15%) Given 0 < a < b, find the arc length of y = lne^x+ 1

e^x− 1 for a ≤ x ≤ b.

Sol:

dy

dx = (^e_e^xx⁺¹−1)⁰

e^x+1 e^x−1

=

e^x(e^x−1)−e^x(e^x+1) (e^x−1)²

e^x+1 e^x−1

= −2e^x

(e^x− 1)² ×e^x− 1

e^x+ 1 = −2e^x e^2x− 1

L = Z b

a

r

1 + (dy

dx)²dx = Z b

a

r

1 + ( −2e^x e^2x− 1)²dx

= Z b

a

s

e^4x+ 2e^2x+ 1 (e^2x− 1)² dx =

Z b a

e^2x+ 1 e^2x− 1dx

Let t = e^x, then dt

dx = e^x → dx = dt t .

x = a → t = e^a; x = b → t = e^b L =

Z e^b e^a

t²+ 1 t(t + 1)(t − 1)dt

= Z e^b

e^a

(−1 t + 1

t + 1+ 1 t − 1)dt

= (− ln t + ln (t + 1) + ln (t − 1))   

e^b e^a

= ln e^2b− 1

e^2a− 1 + a − b

5. (10%) The region in the first quadrant enclosed by x = y², x-axis and x = 1 is rotated about the line y = 3. Find the volume of the solid.

Sol:

(1) Disk Method:

Area = Z 1

0

π · 3²− π(3 −√ x)²dx

= Z 1

0

6π√

x − πxdx

= 4πx^3/2− π 2x²

1 0

= 7 2π (2) Shell Method:

Area = Z 1

0

2π(3 − y)(1 − y²)dy

= Z 1

0

2π(3 − y − 3y²+ y³)dy

= 2π



3y − y²

2 − y³+y⁴ 4

  

1 0

= 7 2π

−x²+ c = 3 2y² 3

2y² = −x²+ 7

7. (10%) (a) Find the particular solution of 3xy⁰− y = ln x + 1, x > 0, satisfying y(1) = −2.

(b) Find the value y(e).

Sol:

(a) y⁰− 1

3xy = ln x + 1 3x

integrating factor V = e^R−3x1dx = x⁻¹³ then

x⁻¹³y = 1 3

Z

(ln x + 1)x⁻⁴³dx

= −x⁻¹³(ln x + 1) + Z

x⁻⁴³dx (integration by part)

= −x⁻¹³(ln x + 1) − 3x⁻¹³ + C

So y = −(ln x + 4) + Cx¹³.

When x = 1, y = 2 then −2 = −4 + C so C = 2 i.e. y = 2x¹³ − ln x − 4

(b) y(e) = 2e¹³ − 5

8. (10%) Given a > 0, one arch of the cycloid x = a(θ − sin θ), y = a(1 − cos θ), 0 ≤ θ ≤ 2π, is rotated about the x-axis. Find the area of the resulting surface.

Sol:

A = Z 2π

0

2πa(1 − cos θ) q

a²(1 − cos θ)²+ a²sin²θ

= Z 2π

0

2πa(1 − cos θ)√ 2a√

1 − cos θ

= 2πa² Z 2π

0

4 sin³(θ 2)dθ

= 16πa² Z π

0

sin³αdα

= 16πa² Z π

0

(1 − cos²α) sin αdα

= 16πa² Z 1

−1

(1 − u²)du

= 16πa²4 3

= 64 3 πa²

9. (15%) (a) Plot the region A which is inside the circle r = 6 cos θ and outside the cardioid r = 2(1 + cos θ).

(b) Find the area of A.

(c) Find the length of the boundary of the region A.

Sol:

(a) The graph r = 6 cos θ is a circle with radius 3 and center (3, 0) The graph r = 2 + 2 cos θ is shaped like a heart symmetric at x-axis The graph outside the cardiod and inside the circle is shaped like moon (b) First we find their intersection points.

2(1 + cos θ) = 6 cos θ ⇒ cos θ = 0.5, θ = π

3 and − π 3 By symmetry

Area = 2 × 0.5hZ ^π₃

0

(6 cos θ)²dθ − Z ^π₃

0

4(1 + cos θ)²dθi

= 36 Z ^π₃

0

(cos θ)²dθ − 4 Z ^π₃

0

(1 + cos θ)²dθ

= Z ^π₃

0

h

18 cos 2θ + 18 − 4 − 8 cos θ − 2 cos 2θ − 2i dθ

= (9 sin 2θ + 18θ − 4θ − 8 sin θ − sin 2θ − 2θ)   

π 3

0

= 4π

(c) The boundary of the region has two parts.

The first part is formed by the circle r = 6 cos θ −π

3 ≤ θ ≤ π 3 The length is 2 × 3 × 2π

3 = 4π

The second part is formed by the cardioid. Using the formula ds = r

r²+ (dr dθ)²dθ

ds =p

(2 + 2 cos θ)² + (−2 sin θ)²dθ

=p

4 + 8 cos θ + 4(cos θ)²+ (4 sin θ)²dθ

=√

8 + 8 cos θdθ

The length is

2 Z ^π₃

0

√

8 + 8 cos θdθ = 2 Z ^π₃

0

4 cosθ 2dθ

= 16 sinθ 2   

π 3

0 = 8 So the total length is 8 + 4π