令F(x)=∫(0→x)(x^2-t^2)f(t)dt=(x^2)∫(0→x)f(t)dt-∫(0→x)(t^2)f(t)dt
则F'(x)=[2x∫(0→x)f(t)dt+(x^2)f(x)]-(x^2)f(x)=2x∫(0→x)f(t)dt