微乙小考六 (2015/12/24)
1. (7分) 求下列曲線圍成區域的面積 y = 3 sin x, y = sin 2x, −
π
π
≤x≤ .
2
2
sol:
Z
π/2
Z
0
Z
|sin2x − 3sinx|dx =
A=
−π/2
π/2
(sin2x − 3sinx)dx +
−π/2
(3sinx − sin2x)dx
0
1
1
π/2
= (− cos2x + 3cosx)|0−π/2 + (−3cosx + cos2x)|0 = 4
2
2
2. (7分) 計算由曲線 x = 4y 2 − 2, x = 2y 2 , y = 0 所圍成區域對 x 軸旋轉所得的旋轉體體積。
sol: x = 4y 2 − 2 and x = 2y 2 intersects at (x, y) = (2, ±1), and the rotating volume with respect
to x axis = volume rotated by x = 4y 2 − 2 with x axis (V1) minus the volume rotated by
x = 2y 2 with x axis (V2).
x+2
x = 4y 2 − 2 ⇒ y 2 =
4
Z 2
x+2
V1=
π(
)dx = 2π
4
−2
x
x = 2y 2 ⇒ y 2 =
2
Z 2
x
π( )dx = π
V2=
2
0
V = V 1 − V 2 = 2π − π = π
3. (6分) 求 ln(3 − 2x) 的 n-次泰勒多項式；須寫出一般項的表式。
sol:
x2 x3
xn
+
+ ... +
+ ...)
2
3
n
2x
2x
ln(3 − 2x) = ln[3(1 − )] = ln3 + ln(1 − )
3
3
2x 2
2x 3
2x n
( )
( )
2x ( )
= ln3 − ( + 3 + 3 + ... + 3
+ ...)
3
2
3
n
2
∞ ( )n xn
X
3
= ln3 −
n
n=1
ln(1 − x) = −(x +
1