推 june10th : H大 感謝您 05/11 19:16
※ 引述《june10th (囧天使)》之銘言：
: a,b,c皆實數
: 證明
: a
: ∫(x-a)(x-b)dx=(1/6)*(a-b)^3
: b
全部展開也不會花你多少時間
高中數學計算麻煩的多的是
利用平移將函數化簡成簡單的形式
x = (a + b)/2 + u
dx = du
b
∫(x-a)(x-b)dx
a
(b - a)/2
= ∫ [(b - a)/2 + u][(a - b)/2 + u] du
(a - b)/2
(b - a)/2
= ∫ [u^2 - (1/4)(b - a)^2] du
(a - b)/2
(b - a)/2
= (1/3)u^3 - (u/4)(b - a)^2 |
(a - b)/2
= (1/12)(b - a)^3 - (1/4)(b - a)^3
= (-1/6)(b - a)^3
= (1/6)(a - b)^3
: 證明
: c
: ∫ (x-a)(x-b)(x-c)dx=(1/12)*(a+c-2b)*(a-c)^3
: a
: 請板友們給個方向吧
c
∫ (x - a)(x - b)(x - c)dx
a
c c
= ∫(x - a) x (x - c) dx - b ∫(x - a)(x - c) dx
a a
= I_1 - b I_2
類似上面那一題
x = (a + c)/2 + u
(c - a)/2
I_1 = ∫ [(c - a)/2 + u][(a + c)/2 + u][(a - c)/2 + u] du
(a - c)/2
(c - a)/2
= ∫[u^2 - (1/4)(c - a)^2][u + (a + c)/2] du
(a - c)/2
(c - a)/2
= ∫ [u^3 +(1/2)(a+c)u^2 -(1/4)(c - a)^2 u -(1/8)(c+a)(c - a)^2]du
(a - c)/2
= (1/4)u^4 +(1/6)(c+a)u^3 -(1/8)(c - a)^2 u^2 -(1/8)(c+a)(c - a)^2 u
代上下限
= (1/24)(c+a)(c - a)^3 - (1/8)(c+a)(c - a)^3
= (-1/12)(c+a)(c - a)^3
I_2 = (-1/6)(c - a)^3
所以原積分 = (-1/12)(c+a)(c - a)^3 + (b/6)(c - a)^3
= (1/12)[c + a - 2b](a - c)^3
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