试用分部积分法证明：若f(x)连续，则∫0x[∫0tf(x)dx]dt=∫x0f(t)(x-1)dt

试用分部积分法证明：若f(x)连续，则
∫₀^x[∫₀^tf(x)dx]dt=∫_x⁰f(t)(x-1)dt
【正确答案】：证明：∫₀^x[∫₀^tf(x)dx]dt =t∫₀^tf(x)dx|₀^x-∫₀^xtd∫₀^xf(x)dx =x∫₀^xf(y)dy-∫₀^xtf(t)dt =∫₀^xf(t)dt-∫₀^xtf(t)dt =∫₀^xf(t)(x-t)dt