1091 ®Y01-03í +ãTŒU– 1. Compute the integrals. (a) (6%) S x(sin 2x+ cos x)dx. (b) (9%) S 8x− 4 x

1091 ®Y01-03í +ãTŒU–

1. Compute the integrals.

(a) (6%) ∫ x(sin 2x + cos x)dx. (b) (9%) ∫

8x − 4 x²(x²+4)dx.

Solution:

(a)

∫ x (sin(2 x) + cos x) dx

= ∫ x sin(2 x) dx + ∫ x cos x dx

=

−1

2 ∫ x d cos(2 x) +∫ x d sin x d cos(2 x) = −2 sin(2 x) dx (1pt) d sin x = cos x dx (1pt)

=

−1

2 (x cos(2 x) −∫ cos(2 x) dx) IBP (2pts)

+x sin x −∫ sin x dx ∫ cos(2 x) dx = 1

2 sin(2 x) + C (0.5pt)

=

−1

2 (x cos(2 x) −1

2 sin(2 x) dx) ∫ sin x dx = − cos x + C (0.5pt) +x sin x + cos x + C C (1pt)

(b)

8x − 4 x²(x²+4)

=

A + B x x²+4 +

C x² +

D

x partial fractions

=1 − 2x x²+4− 1

x² +2

x Solve ⇒ A = 1, B = −2, C = −1, D = 2 (4pts)

∫

8x − 4 x²(x²+4)dx

= ∫ ( 1 x²+4−

2x x²+4 −

1 x² +

2

x) dx use (a)

= 1

2tan⁻¹( x

2) −ln (x²+4) + 1

x+2 ln x + C 4 integrals (4pts); C (1pt)

2. (a) (3%) Evaluate and simplify d dxln (

√

x²+1 + x).

(b) (5%) Evaluate ∫ sec x dx.

(c) (7%) Use (b) and trigonometric substitution to find ∫

1 0

1

√

x²+1dx.

(d) (7%) Use (a) and integration by parts to evaluate the integral ∫

1 0

ln (

√

x²+1 + x) dx.

Solution:

(a) d dxln (

√

x²+1 + x) =

1

2(x²+1)⁻¹² (2 x) + 1

√

x²+1 + x (ln x)^′= 1

x (1pt)and chain rule (1pt)

=

√

x²+1 + x (

√

x²+1 + x)(√

x²+1) multiply (x²+1)¹² (1pt)

= 1

√ x²+1 (b)

∫ sec x dx =∫

sec x (sec x + tan x)

sec x + tan x dx multiply (sec x + tan x) (2pts)

= ∫

d(sec x + tan x) sec x + tan x

d

dxsec x = sec x tan x (1pt), d

dxtan x = sec²x (1pt)

=ln ∣ sec x + tan x∣ + C ∫ 1

xdx = ln ∣x∣ + C (1pt) (c)

∫

1 0

1

√

x²+1dx x ∶= tan θ ⇒ dx = sec²θ dθ (2pt)

x = 1 ⇒ θ = π/4 (0.5pt), x = 0 ⇒ θ = 0 (0.5pt)

= ∫

π/4 0

1

sec θ sec²θ dθ simplify (1pt)

= ∫

π/4 0

sec θ dθ use (b) (1pt)

=ln ∣ sec θ + tan θ∣ ∣

π/4

0

simplify (1pt)

=ln ∣ sec(π/4) + tan(π/4)∣ sec(π/4) =

√

2 (0.5pt); tan(π/4) = 1 (0.5pt)

=ln(

√ 2 + 1)

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(d)

∫

1 0

ln (

√

x²+1 + x) dx

=x ln (

√

x²+1 + x) ∣

1 0− ∫

1 0

x d ln (

√

x²+1 + x) IBP: u = ln (

√

x²+1 + x), v = x (2pts)

=ln(

√

2 + 1) −∫

1 0

x dx

√

x²+1 evaluate bdy. term (1pt); use (a) (1pt)

=ln(

√

2 + 1) − (x²+1)¹²∣

1

0 ∫

x dx

√

x²+1 = (x²+1)¹² (2pts)

=ln(

√

2 + 1) −

√

2 + 1 evaluate bdy. term (1pt)

3. (a) (10%) Let R be the region bounded by y = cos x, y = sin 2x and x = 0 in the first quadrant.

Rotate R about the x-axis. Find the volume of the resulting solid.

Solution:

First, find intersection points of y = cos x and y = sin 2x for 0 ≤ x ≤ π 2. y = cos x = sin 2x = 2 sin x cos x ⇒ sin x = 1

2 or cos x = 0 i.e. x =π 6 or π

2. Hence R is bounded by y = cos x and y = sin 2x, with 0 ≤ x ≤ π

6. (2 pts for the range of x.) By the disc method, the volume is

V = π∫

π 6

0

(cos x)²− (sin 2x)²dx (2 pts for the formula.

1 pt for recognizing cos x ≥ sin 2x for 0 ≤ x ≤ π 6.)

=π∫

π 6

0

1 + cos 2x

2 −

1 − cos 4x

2 dx (2 pts for half angle formulas)

= π

2 [sin 2x

2 +sin 4x 4 ] ∣

x=^π₆

x=0 (2 pts for integrating cos 2x and cos 4x.)

= 3√

3

16 π (1 pt for the final answer.)

If Students write V = ∫

π 6

0

2πx(cos x − sin 2x)dx = ⋯, they get 2 pts for the range of x (0 ≤ x ≤ π

6) and 1 pt for recognizing cos x ≥ sin 2x for 0 ≤ x ≤ π 6. (b) (6%) (A) ∫

1 0

πx dx. (B) ∫

1 0

πx⁴dx. (C) ∫

1 0

πx²dx. (D)∫

1 0

x²dx.

Match each solid with the integral that represents its volume.

Integrals A B C D

Solid 2 1 4 3

1.5 pts for each answers.

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4. (8%) Find the length of the curve

y = f (x) =∫

x 1

√

t³−1 dt, 1 ≤ x ≤ 4.

Solution:

By the fundamental theorem of calculus f^′(x) =

√

x³−1. (2%)

length =∫

4 1

√

1 + f^′(x)²dx (2%)

= ∫

4 1

x^3/2dx (2%)

= 62

5 (2%).

5. Let f (x) = xe^x. (When finding the following Taylor series, you don’t need to specify the range of x for which the series equals the function.)

(a) (4%) Find the Taylor series for f (x) at x = 0.

(b) (7%) Calculate ∫

x

0 te^tdt and find its Taylor series at x = 0.

(c) (4%) Find the sum

∞

∑

n=0

1 n!(n + 2).

Solution:

(a)

f (x) = xe^x=x(1 + x 1!+

x² 2! + ⋯ +

xⁿ

n! + ⋯) (2%)

=x +x² 1! +

x³ 2! + ⋯ +

xⁿ⁺¹

n! + ⋯ (2%).

(b)

∫

x 0

te^t dt =∫

x 0

t de^t (2%)

=xe^x−e^x+1. (2%)

Use term-be-term integration for the Taylor expansion of te^t in (a), we have

∫

x 0

te^t dt =∫

x 0

t dt +∫

x 0

t² 1!dt +∫

x 0

t³

2!dx + ⋯ +∫

x 0

tⁿ⁺¹

n! dt + ⋯ (1%)

= x² 2 + x³

3 ⋅ 1!+ ⋯ + xⁿ⁺²

(n + 2) ⋅ n!+ ⋯. (2%) (c) By (b), we have

xe^x−e^x+1 = x² 2 + x³

3 ⋅ 1! + ⋯ + xⁿ⁺²

(n + 2) ⋅ n!+ ⋯. (1%) Put x = 1 (1%), we have

1 =1 2+

1 3 ⋅ 1! +

1

4 ⋅ 2!+ ⋯ + 1

(n + 2) ⋅ n!+ ⋯. (2%)

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6. (a) (6%) Find the Taylor series for f (x) = ln(1 − x²), g(x) = cos x − 1, and h(x) = sin(2x⁴)at x = 0.

(You don’t need to specify the range of x for which the function equals its Taylor series.) (b) (4%) Evaluate lim

x→0

(cos x − 1) ln(1 − x²) sin(2x⁴)

.

Solution:

(a) ln(1 − y) = −y −y² 2 −

y³ 3 − ⋯ −

yⁿ⁺¹

n + 1− ⋯, for −1 ≤ y < 1.

f (x) = ln(1 − x²) = −x²− x⁴

2 − x⁶

3 − ⋯ − x^2x+2

n + 1 − ⋯, for −1 < x < 1

(1 pt for knowing the Taylor series for ln(1 − y). 1 pt for the Taylor series for f (x).) cos x = 1 − x²

2! + x⁴

4! − x⁶

6! + ⋯ +

(−1)ⁿx²ⁿ

(2n)! + ⋯, for x ∈ R g(x) = cos x − 1 = −x²

2! + x⁴

4! − x⁶

6! + ⋯ +

(−1)ⁿx²ⁿ

(2n)! + ⋯ for x ∈ R

(1 pt for knowing the Taylor series for cos x. 1 pt for the Taylor series for g(x).) sin y = y −y³

3! + y⁵ 5! + ⋯ +

(−1)ⁿ

(2n + 1)!y²ⁿ⁺¹+ ⋯ h(x) = sin(2x⁴) =2x⁴−

2³x¹² 3! + ⋯ +

(−1)ⁿ2²ⁿ⁺¹

(2n + 1)! x⁸ⁿ⁺⁴+ ⋯

(1 pt for knowing the Taylor series for sin y. 1 pt for the Taylor series for h(x).) (b) Solution 1:

(cos x − 1) ln(1 − x²) sin(2x⁴)

= (−^x

2

2 +^x

4

4! − ⋯)(−x²−^x

4

2 − ⋯) 2x⁴−⁴₃x¹²+ ⋯

(2 pts for plugging in Taylor series)

=

1

2x⁴+ (¹₄ −_4!¹)x⁶⋯ 2x⁴−⁴

3x¹²+ ⋯

→ 1

4 as x → 0 Hence lim

x→0

(cos x − 1) ln(1 − x²) sin(2x⁴)

= 1

4 (2 pts for final answers)

Solution 2:

limx→0

(cos x − 1) ln(1 − x²) sin(2x⁴)

0 0

ÔÔ

L’H lim

x→0

−sin x ⋅ ln(1 − x²) + (cos x − 1)_1−x^−2x2

cos(2x⁴) ⋅8x³

=lim

x→0− ( sin x

x

ln(1 − x²) x²

1 8 cos(2x⁴)

) + (

cos x − 1 x²

−1

4(1 − x²)cos(2x⁴) )

= 1 8+

1 8 =

1 4.

(2 pts for using L’Hospital’s Rule and obtaining correct first derivatives. 2 pts for final answers.)

7. (a) (6%) Compute lim

x→0

∫

4x²

0 cos(√ t)dt

x² by L’Hospital’s Rule.

(b) (8%) Compute lim

x→∞(1 + x)^{1/ ln x}. Solution:

(a) As x → 0, ∫

4x² 0

cos(

√

t)dt → 0 and x²→0.

Hence we can apply L’Hospital’s rule.

limx→0

∫

4x²

0 cos(√ t)dt x²

0 0

ÔÔ

L’H lim

x→0

cos(2∣x∣) ⋅ 8x

2x =4

(1 pt for using L’H rule. 3 pts for d dx∫

4x² 0 cos(

√ t)dt) (2 pts for the final answer.)

(b)

ln(1 + x)^{ln x1} = 1

ln x⋅ln(1 + x) (2 pts)

x→∞lim

ln(1 + x) ln x

∞

∞

ÔÔ

L’H lim

x→∞

1 1+x

1 x

(2 pts)

= lim

x→∞

x

1 + x =1 (2 pts)

Hence lim

x→∞(1 + x)^{ln x1} =e¹ =e (2 pts)

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