Question

关于积分的 等式证明

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Answer 1

设∫f(x)dx=F(x)
则：∫(0,x)[∫(0,u)f(t)dt]du
=∫(0,x)[F(u)-F(0)]du
=∫(0,x)F(u)du-∫(0,x)F(0)du
=∫(0,x)F(u)du-xF(0)
∫(0,x)(x-u)f(u)du
=∫(0,x)xf(u)du-∫(0,x)uf(u)du
=x∫(0,x)f(u)du-∫(0,x)udF(u)
=xF(x)-xF(0)-uF(u)|(0,x)+∫(0,x)F(u)du
=xF(x)-xF(0)-xF(x)+∫(0,x)F(u)du
=∫(0,x)F(u)du-xF(0)
所以：∫(0,x)[∫(0,u)f(t)dt]du=∫(0,x)(x-u)f(u)du