一 Z 1
−1
(2 − 2x2− (1 − x4))dx = (1 5x5−
2
3x3+ x)
1
−1
= 16 15
二 利用殼層法可得,
Z 1 0
2π · x · ((3 − 2x2) − x2)dx = 6π · (1 2x2−
1 4x4)
1 0
= 3 2π 三 Let f (x) = ln(1 − x2).
f′(x) = 1
1 − x2 · (−2x) =
2x
(x + 1)(x − 1) = 1
x+ 1 + 1 x − 1
f(n)(x) = (−1)n−1· (n − 1)! ·(x + 1)−n+ (x − 1)−n
Then f(2k)(0) = −2 · (2k − 1)! and f(2k+1)(0) = 0.
So Pn(x) =
n
X
i=1
f(i)(0) i! xi =
n 2
X
i=1
−2 · (2i − 1)!
(2i)! x2i =
n 2
X
i=1
−x2i
i ,if n is even,
Pn(x) =
n
X
i=1
f(i)(0) i! xi =
n−1 2
X
i=1
−2 · (2i − 1)!
(2i)! x2i =
n−1 2
X
i=1
−x2i
i ,if n is odd.
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