一道微积分题目设f(x)=∫(0到x)e^(-y^2+2y)dy求∫(0到1)((x-1)^2)f(x)dx要过程不要只

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/07/09 04:32:49

一道微积分题目
设f(x)=∫(0到x)e^(-y^2+2y)dy
求∫(0到1)((x-1)^2)f(x)dx
要过程不要只是一个结果,

∫(0/1) ((x-1)^2)f(x)dx
=(1/3)∫(0/1) f(x) d(x-1)^3
=(1/3)f(x)(x-1)^3 (0/1) -(1/3)∫(0/1)(x-1)^3df(x).
其中：
(1/3)f(x)(x-1)^3 (0/1)
=（1/3）f(1)(1-1)^3-(1/3)f(0)(x-1)^3
=0-(1/3)(x-1)^3∫(0/0)e^(-y^2+2y)dy
=0-0
=0
f(x)=∫(0/x)e^(-y^2+2y)dy
f'(x)=e^(-x^2+2x)
所以：
df(x)=e^(-x^2+2x)dx
则：
-(1/3)∫(0/1)(x-1)^3df(x).
=-（1/3）∫(0/1) （x-1)^3e^(-x^2+2x)dx
=-(e/3)∫(0/1) (x-1)^3e^[-(x-1)^2]dx
=-(e/6)∫(0/1)(x-1)^2e^[-(x-1)^2d(x-1)^2
=(e/6)∫(0/1)(x-1)^2 d e^[-(x-1)^2 应用分部积分可得到：
=(e/6)(x-1)^2e^[-(x-1)^2] (0/1)+(e/6)e^[-(x-1)^2](0/1)
=(e/6)e^[-(x-1)^2](x^2-2x+2) (0/1)
=(e-2)/6.

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