Question

∫（t-x）f(2x-t)dt=sinx+cosx（积分上下限是x到2x,打不上）,求∫f(x)dx（上下限0到π/2

Answer 1

∵∫(t-x)f(2x-t)dt=∫(x-y)f(y)dy (令y=2x-t)

=x∫f(y)dy-∫yf(y)dy

又∫(t-x)f(2x-t)dt=sinx+cosx

∴ x∫f(y)dy-∫yf(y)dy=sinx+cosx

==>∫f(y)dy+xf(x)-xf(x)=cosx-sinx (等式两端对x求导数)

==>∫f(y)dy=cosx-sinx

==>∫f(y)dy=cos(π/2)-sin(π/2)=-1

故 ∫f(x)dx=-1.