设f(x)在[0,1]上连续,并设∫(0~1)f(x)dx=A,求∫(0~1)dx∫(x~1)f(x)f(y)dy.

设其原函数是F（x)
∫(0~1)f(x)dx=A＝F(1)-F(0)
∫(0~1)dx∫(x~1)f(x)f(y)dy
=∫(0~1)f(x)dx∫(x~1)f(y)dy
=∫(0~1)[F(1)-F(x)]f(x)dx
=∫(0~1)[F(1)-F(x)]dF(x)
=[F(1)F(x)-1/2F^2(x)](0~1)
=F^2(1)-1/2F^2(1)-F(1)F(0)+1/2F^2(0)
=1/2F^2(1)-F(1)F(0)+1/2F^2(0)
=1/2[F(1)-F(0)]^2
=1/2A^2