2π(Rπ 0 x sin xdx− R2π π x sin xdx)

 } (B) ç‚5t

2003  1 ~ 11 n.

1. IR[ýâ y = 1 + sin x D y − 1 s(Ê x = 0 D x = 2π 5ÈFˇ–, t°

âR÷yWìžF) ñ5ñ . (12%) j: V = 2π{Rπ

0 x[(1+sin x)−1]dx+R2π

π x[1−(1+sin x)]dx} = 2π(Rπ

0 x sin xdx−

R2π

π x sin xdx).

Because R x sin xdx = R xd(− cos x) = −x cos x + R cosxdx = −x cos x + sin x + C.

Hence,Rπ

0 x sin xdx = −x cos x + sin x|^π₀ = π, R2π

π x sin xdx = −x cos x + sin x|^2π_π = −2π − π = −3π.

Therefore, V = 2π[π − (−3π)] = 8π²

2. t°Ö( x = t − sin t, y = 1 − cos t ø‘ (0 ≤ t ≤ 2π) 5Å. (12%) j: Because x = t − sin t, y = 1 − cos t. Hence ^dx_dt = 1 − cos t, ^dy_dt = sin t

s = Z 2π

0

q

(1 − cost)²+ sin²tdt

= Z 2π

0

√2 − 2 cos tdt

= Z 2π

0

r 4 sin² t

2dt

= 2 Z 2π

0

sin t 2dt

= 4(− cos t 2)|^2π₀

= 4[1 − (−1)] = 8 3. t°.ì }R e^3x

e^3x+2e^2x+2e⁺1dx. (12%) j: I u = e^x, † x = ln u, dx = ^du_u.

Z e^3x

e^3x+ 2e^2x+ 2e^x+ 1dx

=

Z u³

u³+ 2u²+ 2u + 1·du u

=

Z u²

(u + 1)(u²+ u + 1)du

q _(u+1)(u^u2_2+u+1) = _u+1^A +_u^Bu+C2+u+1, † u²= A(u²+ u + 1) + (Bu + C)(u + 1).

I u = −1, ) A = 1, ] −u − 1 = (Bu + C)(u + 1), ¹ Bu + C = −1. Ĥ,

Ÿ =

Z du u + 1−

Z du

u²+ u + 1 1

= ln |u + 1| −

Z du

(u + ¹₂)²+³₄

= ln |u + 1| − 1

√3 2

tan⁻¹(u +¹₂

√3 2

) + C

= ln(e^x+ 1) − 2

√3tan⁻¹(2e^x+ 1

√3 ) + C

4. t°”Ì limx→0cos 3x cos x

1

x2 5M. (12%) j:

x→0lim(cos 3x cos x)^x21

= lim

x→0exp(ln | cos 3x| − ln | cos x|

x² )

= exp( lim

x→0

−3 sin 3x cos 3x +^{sin x}_{cos x}

2x )

= exp( lim

x→0

−3 tan 3x + tan x

2x )

= exp( lim

x→0

−9 sec²3x + sec²x

2 )C = exp[ lim

x→0(sin 3x 3x · −9

2 cos 3x +sin x

x · 1

2 cos x)]

= exp(−9 2+1

2) = exp(−4) = e⁻⁴. 5. qp > 1, t°¡ }R∞

1 (ln x)²

x^p dx 5M. (12%) j:

Z (ln x)² x^p dx =

Z

x^−p(ln x)²dx

= 1

1 − p Z

(ln x)²dx^1−p

= 1

1 − p[x^1−p(ln x)²− Z

x^1−pd(ln x)²]

= 1

1 − p[x^1−p(ln x)²− 2 Z

x^−pln xdx]

= (ln x)²

(1 − p)x^p−1 − 2 (1 − p)²

Z

ln xdx^1−p

= (ln x)²

(1 − p)x^p−1 − 2

(1 − p)²[x^1−pln x − Z

x^1−pd(ln x)]

= (ln x)²

(1 − p)x^p−1 − 2 ln x

(1 − p)²x^p−1 + 2 (1 − p)²

Z x^−pdx

= (ln x)²

(1 − p)x^p−1 − 2 ln x

(1 − p)²x^p−1 + 2

(1 − p)³x^p−1 + C 2

ÄÑ p > 1, lim_x→∞^{(ln x)}_xp−1² = lim_{x→∞2(p−1)x}^{ln x}_p−1 = lim_{x→∞2(p−1)}¹_2xp−1 = 0.

FJ R∞ 1

(ln x)²

x^p dx = _(1−p)x^{(ln x)}_p−1² −_(1−p)^{2 ln x}_2xp−1 + _(1−p)²_3xp−1|^∞₁ = −_(1−p)¹ ₃ =

1 (p−1)³

Y L’Hˆopital ¶† (^∞_∞$) C x^α>> (ln x)^β (as x → ∞), α, β > 0 ).

6. qa, bÑb, / limx→0(x⁻³sin 3x + ax⁻²+ b) = 0, t°a, b 5M. (12%) j: x⁻³sin 3x + ax⁻²+ b = sin 3x+ax+bx³

x³ .

I f (x) = sin 3x + ax + bx³, g(x) = x³,

† ^f0(x)

g⁰(x) = 3 cos 3x+a+3bx² 3x² f⁰⁰(x)

g⁰⁰(x) = −9 sin 3x+6bx

6x = −⁹₂·^{sin 3x}_3x + b → −⁹₂+ b, ç x → 0.

Y L’Hˆopital ¶†, limx→0f (x)

g(x) = limx→0f⁰(x)

g⁰(x) = limx→0f⁰⁰(x)

g⁰⁰(x) = −⁹₂+ b, ]) −⁹₂+ b = 0, ¹ b = ⁹₂.

¢Ä lim_x→0^f_g⁰0^(x)(x) = 0, / lim_x→0g⁰(x) = 0, ] lim_x→0f⁰(x) = 0, 7 f⁰(x) = 3 cos 3x + a + 3b²→ 3 + a, ç x → 0, ]) 3 + a = 0, ¹ a = −3.

7.

(1) tŸ| tan⁻¹x 5 MacLaurin b£wY¹¸ˇ. (4%) (2) t°bP∞

n=0 (−1)ⁿ

3ⁿ(2n+1) 5¸. (8%) j:

(1) tan^−x=P∞ n=0

(−1)ⁿ

2n+1x²ⁿ⁺¹, −1 ≤ x ≤ 1.

(2) I x = ^√1

3, †

tan⁻¹ 1

√3 =

∞

X

n=0

(−1)ⁿ 2n + 1· 1

3ⁿ · 1

√3,

])

∞

X

n=0

(−1)ⁿ

3ⁿ(2n + 1) =√

3 tan⁻¹ 1

√3 =√ 3 ·π

3 = π√ 3 6 . 8. q f (x) = xq

1−x

1+x, −1 < x < 1.

(a) t° f (x) 5 MacLaurin b. (10%) (b) t° f⁽ⁿ⁾(0). (6%)

j:

3

(a)

f (x) = xr 1 − x

1 + x = x(1 − x)

√1 − x² = (x − x²)(1 − x²)⁻¹²

= (x − x²)[1 + (−1

2)(−x²) +(−¹₂)(−³₂)

1 · 2 (−x²)²+ · · · + (−¹₂)(−³₂) · · · (²ⁿ⁻¹₂ )

n! (−x²ⁿ) + · · ·]

= (x − x²)(1 − x²

2 + 1 · 3

2²· 2!x⁴+ · · · +1 · 3 · · · (2n − 1)

2ⁿ(n!) x²ⁿ+ · · ·)

= x + x³

2 + 1 · 3

2²· 2!x⁵+ · · · + 1 · 3 · · · (2n − 1)

2ⁿ(n!) x²ⁿ⁺¹+ · · ·

−(x²−x⁴

2 + 1 · 3

2²· 2!x⁶+ · · · + 1 · 3 · · · (2n − 1)

2ⁿ(n!) x²ⁿ⁺²+ · · ·)

= x − x²+x³ 2 −x⁴

2 + · · · +1 · 3 · · · (2n − 1)

2ⁿ(n!) x²ⁿ−1 · 3 · · · (2n − 1)

2ⁿ(n!) x²ⁿ⁺²+ · · ·) (b) I a_n [ý f (x) 5 MacLaurin b2 xⁿ á5[b, †

a_n=

( 1·3···(2m−1)

2^m(m!) , J n = 2m + 1, m = 0, 1, 2, · · ·

−1·3···(2m−1)

2^m(m!) , J n = 2m + 2, m = 0, 1, 2, · · · Ä an= ^f(n)_n!⁽⁰⁾, ] f⁽ⁿ⁾(0) = an(n!), Ĥ

f⁽ⁿ⁾(0) =

( 1·3···(2m−1)

2^m(m!) [(2m + 1)!], J n = 2m + 1, m = 0, 1, 2, · · ·

−1·3···(2m−1)

2^m(m!) [(2m + 2)!], J n = 2m + 2, m = 0, 1, 2, · · ·

4