微甲統一教學二組95上期末考參考答案 1. (14%) 求極限 (limit) :
(a) lim
x→0
2sin x− 1 ex− 1 Solution:
由 L’Hˆopital Rule
x→0lim
2sin x−1
ex−1 = lim
x→0
ln 2·2sin x
ex = ln 2.
(b) lim
x→0
Rx2 0
√t 1+t3dt x4 Solution:
x→0lim
Rx2 0
√t 1+t3dt t4 =lim
x→0
√x2 1+x6·2x
4x3 =lim
x→0 2
4(x4+1) =12.
2. (6%)求 y = x sin−1x +√1 − x2 之導函數 (derivative)。 Solution:
dy
dx = √x
1−x2 + sin−1x +√−x
1−x2 = sin−1x.
3. (40%) 求積分 (integral) : (a)
Z e 1
t(ln t)2dt Solution:
Re
1 t(ln t)2dt
= 12Re
1(ln t)2dt2
= 12{t2(ln t)2|e1− 2Re
1 t(ln t)dt}
= e22 − 12Re
1 ln td(t2)
= e22 − 12{t2ln t|e1−Re 1 tdt}
= e22 − e22 +12(t22|e1)
= e42 − 14.
1
(b)
Z 2x + 1 (x2+ 1)2dx Solution:
R 2x+1
(x2+1)2dx =R 2x
(x2+1)2dx +R 1
(x2+1)2dx = −(x21+1)+R 1
(x2+1)2dx.
( R 1
(x2+1)2dxx=tan θ= R dθ
sec2θ =R cos2θdθ = R 1+cos 2θ
2 dθ
= θ2 + 14sin 2θ = 12tan−1x +12x2x+1 + C
)
= 12tan−1x +12x21+1 − x21+1 + C.
(c)
Z xdx
√8 − 2x2− x4 Solution:
R xdx
√
8−2x2−x4 u=x2
= 12R du
√8−2u−u2
= 12R √ du
9−(u+1)2
= 12R √d(u+13 ) 1−(u+13 )2
= 12sin−1(u+13 ) + C
= 13sin−1(x23+1) + C.
(d) Z ∞
0
x2e−xdx Solution:
利用 R xne−xdx = −xne−x+ nR xn−1e−xdx 則 R∞
0 x2e−xdx = lim
a→∞
Ra
0 x2e−xdx
= lim
a→∞(−x2e−x|a0 + 2Ra
0 xe−xdx)
= 2 lim
a→∞
Ra
0 xe−xdx
= 2 lim
a→∞(−e−x|a0 +Ra
0 e−xdx)
= 2 lim
a→∞
Ra 0 e−xdx
= 2 lim
a→∞(−e−x|a0)
= 2 lim
a→∞(−e−a+ 1)
= 2.
2
4. (10%)
(a) 求兩曲線 r = cos θ 及 r = 1 − cos θ 之所有交點。
Solution:
cos θ = 1 − cos θ ⇒ cos θ = 12 ⇒ θ = ±π3. 故交點 (1
2, ±π3)。 再由圖可知原點為交點。
(b) 求在圓 r = cos θ 之內部, 且在心臟線 (cardioid) r = 1 − cos θ 之外部的區域 面積。
Solution:
A =Rπ3
0 cos2θ − (1 − cos θ)2dθ.
=R π3
0 (2 cos θ − 1)dθ = 2 sin θ − θ|
π 3
0 =√ 3 − π3. 5. (10%) 求曲線 x23 + y23 = 1 之全長。
Solution:
x = cos3θ
y = sin3θ , θ ∈ [0, 2π]
dx
dθ = −3 cos2θ sin θ.
dy
dθ = 3 sin2θ cos θ.
ds = 3| cos θ sin θ|dθ S = R ds = 4 R0π2 q
(dxdθ)2+ (dydθ)2dθ
= 12R π2
0 cos θ sin θdθ = 6R π2
0 sin 2θdθ = −3 cos 2θ|
π 2
0 = 3 + 3 = 6
6. (10%) 令 R 為曲線 y = 1 + sin x, 0 ≤ x ≤ 2π, 之下方及 x-軸 上方的區域。 將 R 繞 y-軸 旋轉所得之旋轉體體積為何?
Solution:
V =R2π
0 2πx(1 + sin x)dx == 2π(π22 − x cos x sin x|2π0 ) = 4π2(π − 1)
7. (10%) 解微分方程 (differential equation) :
sec x dy
dx = ey+sin x , x ∈ (−π2,π6).
y(0) = 0 Solution:
R dy
ey =R cos xesin xdx ⇒ −e−y = esin x+ C.
因為 y(0) = 0 , 故 −1 = C + 1 ⇒ C = −2
⇒ −e−y = esin x− 2 ⇒ e−y = −esin x+ 2 ⇒ −y = ln(2 − esin x)
⇒ y = −ln(2 − esin x).
3