ç,ç‚Bù ‚5

94

ç,ç‚Bù ‚5

(12%) 1. Evaluate the integral

Z 2x²+ 5x + 3

(x²+ 2x + 2)(x − 1)dx.

Ans: tan⁻¹(x + 1) + 2 ln |x − 1| + C.

(10%) 2. Evaluate the integral Z

e^sin−1xdx.

Solution:

Z

e^sin−1xdx^y=sin=^−1x Z

e^ydsin y = e^ysin y − Z

sin y · e^ydy

= e^ysin y + Z

e^ydcos y = e^ysin y + e^ycos y − Z

cos y · e^ydy

|Z {z }

e^ydsin y

So Z

e^ydsin y = e^y

2(sin y + cos y) + C = e^sin−1x 2

 x+√

1 − x² + C.

(12%) 3. Evaluate the integral Z 2

1

dx x²√

1 + x². Solution:

cos θ = 1

√1 + x² , tan θ = x ⇒ sec²θ dθ = dx

Z dx

x²√

1 + x² = Z cos θ · sec²θ dθ tan²θ

=

Z 1

cos²θ sin²θ cos²θ

cos θ dθ

=

Z cos θ sin²θdθ

= Z

csc²θcos θ dθ

Let u = cos θ ⇒ du = − sin θ dθ dv = csc²θ dθ⇒ v = − cot θ

= − cos θ · cot θ − Z

(− cot θ)(− sin θ) dθ

= − cos θ cot θ − sin θ + C

= −

√1 + x² x

1

(12%) 4. Find the length of the curve y = Z _x2¹

1

√t³+ 1 dt, 1

2 ≤ x ≤ 1.

Solution:

y⁰ = −2x⁻³√

1 + x⁻⁶, 1 + y⁰² = 4x⁻¹²+ 4x⁻⁶+ 1 s=

Z 1

1 2

p1 + y⁰²dx= Z 1

1 2

(2x⁻⁶+ 1) dx =

 x−2

5x⁻⁵

1

1 2

= 129 10. (12%) 5. For what values of a and b is the following equation true?

x→0lim

 sin 2x

x³ + a + b x²



= 0 Solution:

L= lim

x→0

 sin 2x

x³ + a + b x²



= lim

x→0

sin 2x + ax³+ bx x³

= limH

x→0

2 cos 2x + 3ax²+ b

3x² .

As x → 0, 3x² → 0, and (2 cos 2x + 3ax²+ b) → b + 2, so the last limit exists only if b + 2 = 0, that is b = −2. Thus,

x→0lim

2 cos 2x + 3ax² − 2 3x²

= limH

x→0

−4 sin 2x + 6ax 6x

= limH

x→0

−8 cos 2x + 6a

6 = 6a − 8

6 , which is equal to 0 if and only if a = 4

3. Hence, L = 0 if and only if b = −2 and a= 4

3.

(12%) 6. Let f (x) = sin²x

(a) Find the Maclaurin series for f (x).

Solution:

sin²x= 1

2(1 − cos 2x) cos x = 1 − x²

2! + x⁴

4! − · · · = X∞ n=0

(−1)ⁿx²ⁿ (2n)!

cos(2x) = X∞ n=0

(−1)ⁿ2²ⁿx²ⁿ (2n)!

sin²x = 1 2

"

1 − X∞ n=0

(−1)ⁿ2²ⁿx²ⁿ (2n)!

#

= − X∞ n=1

(−1)ⁿ2²ⁿ⁻¹x²ⁿ (2n)!

= X∞ n=1

(−1)ⁿ⁻¹2²ⁿ⁻¹x²ⁿ (2n)!

2

(b) Find f⁽⁹⁴⁾(0).

(10%) 7. Discuss the convergence of the the series X∞

n=1

n!

nⁿ⁻¹. Solution:

ratio test

a_n= n!

nⁿ⁻¹

n→∞lim  

a_n+1 a_n

 = lim

n→∞

(n+1)!

(n+1)ⁿ n!

nⁿ⁻¹

= lim

n→∞

(n + 1)!

n!

nⁿ⁻¹ (n + 1)ⁿ

= lim

n→∞(n + 1) nⁿ⁻¹ (n + 1)ⁿ

= lim

n→∞

 n n+ 1

n−1

= e⁻¹

(10%) 8. Discuss the convergence of the the series X∞

n=1

(−1)ⁿ⁻¹

√n+ 1 −√ n− 1

n .

(absolutely convergent, conditionally convergent, or divergent) (10%) 9. Discuss the convergence of the the series

X∞ n=2

(−1)ⁿ n(ln n).

(absolutely convergent, conditionally convergent, or divergent)

3