设f(x)=积分(0->x)e^(-y^2+2y)dy,求积分(0->1)[(x-1)^2]f(x)dx

Question

如题所述

Answer 1

∫(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).
(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)
∴df(x)=e^(-x^2+2x)dx
∴-(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 =(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.