Question

设f(x)为连续函数，求

Answer 1

如图

Answer 2

∫<0,x>tf(x²-t²)dt
=(-1/2)∫<0,x>f(x²-t²)d(x²-t²)
令x²-t²=u，则上式=(-1/2)∫<x²,0>f(u)du=(1/2)∫<0,x²>f(u)du
所以，原式=[(1/2)∫<0,x²>f(u)du]'=(1/2)[f(x²)·2x-0]
=xf(x²)

Answer 3

let
u=x^2-t^2
du =-2tdt
t=0, u=x^2
t=x, u=0
∫(0->x) tf(x^2-t^2) dt
=-(1/2)∫(0->x) f(x^2-t^2) (-2tdt)
=-(1/2)∫(x^2->0) f(u) du
=(1/2)∫(0->x^2) f(u) du
d/dx (∫(0->x) tf(x^2-t^2) dt)
=d/dx [(1/2)∫(0->x^2) f(u) du]
=(1/2)(2x) f(x^2)
=xf(x^2)