tam110064

Cho $f(x)$ liên tục trên $R$.Đặt $f_1(x)=\int_{0}^{x}f(t)dt;f_2(x)=\int_{0}^{x}f_1(t)dt..,f_{n+1}(x)=\int_{0}^{x}f_n(t)dt$.

chứng minh rằng : $f_{n+1}(x)=\frac{1}{n!}\int_{0}^{x}(x-t)^nf(t)dt;\forall n\geq 1$