93ç,ç‚Bù ‚5

93ç,ç‚Bù ‚5

1. (14}) ^I R [ýâ y W¸s( y = cos x, y = sin x, 0 ≤ x ≤ π

4, ^FˇA5–, ^t°

â R ÷ y WìžF) ñ5ñ
Solution:

÷ y ^W

V = Z ^π₄

0 2πx(cos x − sin x)dx

= 2π Z ^π₄

0 x cos xdx − Z ^π₄

0

x sin xdx

!

Z ^π₄

0 x cos xdx = x sin x|

π

04 − Z ^π₄

0

sin xdx

= π 4

√2

2 − (− cos x)|

π

04

=

√2π 8 +

√2 2 − 1

− Z ^π₄

0

x sin xdx = Z ^π₄

0

xd(cos x)

= x cos x|

π

04 − Z ^π₄

0

cos xdx

=

√2π 8 −

√2 2

∴V =

√2π 4 − 1

!

· 2π

2. (15})ÊÇ2â,ƒ-ªõƒú‘(, ( C : y = f (x), Ó( y = 2x², £Ó(

y = x²
( C íÔ44uÊÓ( y = 2x² ,LSøõ P åò(£®(, F² ì|Ví– A £ B íÞ ·uó
tŽâ¥_Ô4V²ì( C íj˙
(Tý:

Þ ílªUày = f (x) C x = f⁻¹(y) VªW)

3. (10})t°.ì }

Z dx x³+ 1. Solution:

1

1

x³+ 1 = 1

3(x + 1)+ −x + 2 3(x²− x + 1)

= 1

3(x + 1)− (2x − 1)

6(x² − x + 1)+ 1 2(x² − x + 1) So

Z dx x³ + 1 = 1

3

Z dx x + 1 − 1

6

Z d(x²− x + 1) x²− x + 1 +1

2

Z dx

x² − x + 1

= 1

3ln |x + 1| − 1

6ln |x²− x + 1| + 1

√3tan⁻¹ x − 1/2√ 3/2 + C

4. (10})t°.ì } Z

x ln x dx.

solution:

Z

x ln xdx = Z

ln xd x² 2



= x²

2 ln x − Z x²

2 · 1

xdx = x²

2 ln x −1

4x²+ C

5. (15})I R [ýâ y W, ( y = sin x, £(¨ 0 ≤ x ≤ π

2 FˇA5–, I S H[

â R ÷ x WìžF)5 ñ

(a) t°.ì } Z

sec³x dx

(b) t° S í[Þ
(7l¶íÞ )

6. (10})q a, b Ñb/ b > a > 0
I F (x) = b^x+1 − a^x+1

x + 1
t°”Ì lim

x→−1F (x) 5 M

7. (12})

(a) tŸ| tan⁻¹x 5 Maclaurin b

solution:

tan⁻¹x = x − x³ 3 + x⁵

5 − · · · + (−1)²ⁿ⁻¹ x²ⁿ⁻¹

2n − 1 + · · · . (b) q f (x) = x tan⁻¹x, t° f⁽¹⁶⁾(0)

8. (15})ç 1 ≤ p ≤ 2 v, nb

∞

X

n=1

ln n

n^p uY¹Cuêà

2