微甲

1001微甲^01-05班期末考

1. (10%) Evaluate the integral Z ^π

2

0

| cos²x− 3 sin²x| dx.

Sol:

We have known

cos²x = 1 + cos 2x

2 (1 point) , sin²x = 1− cos 2x 2 Compute the integration

Z

cos²x− 3 sin²x dx =−x + sin 2x + C , C ∈ R

In [0, π

2], compute

cos²x− 3 sin²x≥ 0 ⇒ 0 ≤ x ≤ π 6 Then

Z ^π

2

0

¯¯cos²x− 3 sin²x¯¯ dx

= Z ^π

6

0

cos²x− 3 sin²x dx + Z ^π

2

π 6

3 sin²x− cos²x dx

= (−x + sin 2x)¯¯

¯¯

π 6

0

+ (x− sin 2x)¯¯

¯¯

π 2

π 6

= π 6 +√

3

2. (15%) Compute

Z 1

e^2x+ e^x+ 1dx.

Sol:

Let t = e^x ⇒ x = ln t ⇒ dx = dt t Hence

Z 1

e^2x + e^x+ 1dx =

Z 1

t²+ t + 1 dt

t

Suppose that 1

t(t²+ t + 1) = A

t + P (t) t²+ t + 1 A(t²+ t + 1) + P (t)t = 1

Let t = 0 ⇒ A = 1

And then P (t)t = −t²− t ⇒ P (t) = −t − 1 Consider the integral

Z 1 t²+ t + 1

dt t =

Z 1 tdt +

Z −t − 1 t²+ t + 1dt

= Z 1

tdt + (−1 2 )

Z 2t + 1 t²+ t + 1dt +

Z ₋₁

2

t²+ t + 1dt

= ln|t| − 1

2ln|t² + t + 1| +

Z ₋₁

2

t²+ t + 1dt Now we have to compute the third term

Z ₋₁

2

t² + t + 1dt = (−1 2 )

Z dt

(t + ¹₂)²+ ³₄ = −1√

3arctan(2t + 1

√3 ) So

ln|t| − 1

2ln|t²+ t + 1| +

Z ₋₁

2

t² + t + 1dt

= ln|t| − 1

2ln|t²+ t + 1| − 1

√3arctan(2t + 1

√3 ) + C

= x− 1

2ln|e^2x+ e^x+ 1| − 1

√3arctan(2e^x+ 1

√3 ) + C

3. (15%) Let I(p) = Z _∞

0

x^pe^−x2dx, where p is a real number.

(a) Find p so that it converges. (8%) (b) Find I(3). (7%)

Sol:

(a) I(p) = Z _∞

0

x^pe^−x2dx = Z ₁

0

x^pe^−x2dx + Z _∞

1

x^pe^−x2dx

Claim:

Z _∞

1

x^pe^−x2dx converges for all p∈ R.

If p≤ 0, then Z _∞

1

x^pe^−x2dx ≤ Z _∞

1

e^−x2dx≤ Z _∞

1

e^−xdx = e < ∞.

Suppose p > 0. Then there exists integers M1 and M2 such that x^p

e^x ≤ 1 ∀x ≥ M1 and e^−x2 ≤ e^−2x ∀x ≥ M2. Let M = max {M1, M₂}. Then

Z _∞

1

x^pe^−x2dx = Z M

1

x^pe^−x2dx + Z _∞

M

x^pe^−x2dx

≤ Z M

1

x^pe^−x2dx + Z _∞

M

x^pe^−2xdx

= Z M

1

x^pe^−x2dx + Z _∞

M

x^p e^xe^−xdx

≤ Z M

1

x^pe^−x2dx + Z _∞

M

e^−xdx <∞

Claim:

Z 1 0

x^pe^−x2dx <∞ if and only if p > −1.

Since e⁻¹ ≤ e^−x2 ≤ e, ∀x ∈ [0, 1], we have

e⁻¹ Z ₁

0

x^pdx≤ Z ₁

0

x^pe^−x2dx≤ e Z ₁

0

x^pdx

By the fact,

Z 1 0

x^pdxconverges if and only if p >−1, we have

Z 1 0

x^pe^−x2dx <∞ if and only if p > −1.

(b) Let u = x². Then du = 2xdx. So

I(3) = Z _∞

0

x³e^−x2dx = 1 2

Z _∞

0

ue^−udu

= −1

2 [u(e^−u)|^∞0 − Z _∞

0

e^−udu]

= 1 2

Z _∞

0

e^−udu = 1 2

4. (15%) Let the continuous function f (x) be defined in x > 0, and satisfy f (x)−x⁴ = Z x²

1

f (√ t)t dt.

Find f (x).

Sol:

f (x) is a continuous function, therefore we differentiate both sides and get

f⁰(x)− 4x³ = d dx

Z x² 1

f (√ t)tdt

= d dy

Z _y

1

f (√

t)tdtdy dx

= f√ yydx²

dx

= f (x)x²2x

= 2x³f (x)

Accordingly,

f⁰(x)− 2x³f (x) = 4x³ Integrating factor:

e^R−2x3dx = e^{−12 x4}

e^{−12 x4}[f⁰(x)− 2x³f (x)] = e^{−12 x4}4x³ d

dx[e^{−12 x4}f (x)] = e^{−12 x4}4x³

e^{−12 x4}f (x) = Z

e^{−12 x4}4x³dx (Let u = −1

2 x⁴, then du =−2x³dx, − 2du = 4x³dx)

= Z

e^u− 2du

=−2e^u+ C

=−2e^{−12 x4} + C

Consequently,

f (x) =−2 + Ce^12x4

In addition,

f (1)− 1⁴ = Z 1

f (√ t)tdt

1 = f (1) =−2 + Ce¹² C = 3

e¹² = 3e⁻¹² Finally, we find:

f (x) =−2 + 3e^{−12 +12x4}

ps. Other methods (for example, seperation variables) solve the differential equation f⁰(x)− 2x³f (x) = 4x³ are also permitted.

5. (15%) Consider the plane curve 3ay² = x(a− x)² where a > 0 is a constant.

(a) Find the arc length of the loop defined by the curve. (7%)

(b) Find the surface area of the surface obtained by rotating the loop around x-axis. (8%) Sol:

(a) Find the arc length 3ay² = x(a− x)²

=⇒ 3a(2y)dy

dx = 2x(a− x)(−1) + (a − x)² =⇒ dy

dx = (a− x)(a − 3x) 6ay S =

Z _a

0

ds

= Z _a

0

r

1 + (dy

dx)²dx = Z _a

0

s

1 + (a− x)²(a− 3x)² 36a²y² dx

= Z _a

0

s

1 + (a− x)²(a− 3x)² 3ay²(12a) dx =

Z _a

0

s

1 + (a− x)²(a− 3x)² x(a− x)²(12a) dx

= Z _a

0

r(a + 3x)² 12ax dx =

Z _a

0

r 1 12a

(a + 3x)

√x dx

= Z _a

0

r 1 12a

√a xdx +

Z _a

0

r 1

12a3x¹²dx = a

√3 + a

√3 = 2√ 3a 3 Length=2× S = 4√

3 3 a

(b) Find the surface area

A = Z a

0

2πyds

= Z a

0

2π r

1 + (dy

dx)²ydx = 2π Z a

0

√1 12a

(a + 3x)

√x

√x(a√ − x) 3a dx

= 2π Z _a

0

1

6a(a + 3x)(a− x)dx = π 3a

Z _a

0

(a²− ax + 3ax − 3x²)dx

= π

3a(a³− 1 2a³+3

2a³− a³) = π 3a²

6. (15%) The solid in the figure is cut from a vertical cylinder of radius 10 cm by two planes making angles of 60^◦ with the horizontal. Find its volume.

Sol:

The cross section perpendicular to the base of cylinder and parallel to the intersection of two planes is rectangle with width √

10²− x² and height √

3(10− x), where x is the distance to center of the cylinder, so the area of cross section is 2√

3(10− x)√

10²− x². The volume is

2 Z ₁₀

0

2√

3(10− x)√

10²− x²dx

= 4√ 3

Z ^π

2

0

10(1− sin u)10p

1− sin²u10 cos udu

= 4000√ 3

Z ^π

2

0

(1− sin u) cos²udu

= 4000√ 3

Z ^π

2

0

1 + cos 2u

2 du + 4000√ 3

Z ₀

1

v²dv

= 4000√ 3π

2 1

2 + 4000√ 31

2

sin π− sin 0

2 − 4000√ 31

3(1³− 0³)

= 1000√

3π− 4000√ 3 3

(b) Find the area of the plane region bounded by the curve. (5%) (c) Find the arc length of the curve. (5%)

Sol:

(a) x = r cos θ, y = r sin θ, (r²)² = 2r²cos θ sin θ, r² = sin 2θ,

(b)

Area = 1 2

Z ^π

2

0

r²dθ + 1 2

Z ^3π

2

π 2

r²dθ

= 2× 1 2

Z ^π

2

0

sin 2θdθ = 1

(c)

Arc length =Z p

(r⁰θ)²+ r²

=Z √

csc θdθ

= Z ^π

2

0

√csc θdθ + Z ^3π

2

π 2

√csc θdθ