怎样使∫tf(x^2-t^2)dt(上限x,下限0)=1/2∫f(u)du（上限x^2,下限0)

Question

怎样使∫tf(x^2-t^2)dt(上限x,下限0)=1/2∫f(u)du（上限x^2,下限0)

Answer 1

令u=x^2-t^2
du=-dt^2
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)dt^2
=1/2∫[x^2,0] f(u)*(-du)
=1/2∫[0,x^2] f(u)du