Problem
Let $f(x)$ be a continuous function, and $$\int_0^x f(x-t){\rm d}t=e^{-2x}(x+1)-1.$$ Find $\int_0^1 f(x){\rm d}x$ and $f(x).$
Solution
Differentiate the both sides of the assumption equality with respect to $x$. For the left hand side, we make a substitution $x-t=:u$ and obtain \begin{align*} \frac{{\rm d}}{{\rm d}x}\int_0^xf(x-t){\rm d}t&=\frac{{\rm d}}{{\rm d}(u+t)}\int_x^0f(u){\rm d}(x-u)\\ &=\frac{{\rm d}}{{\rm d}u}\int_0^xf(u){\rm d}u\\ &=f(x). \end{align*}
As for the right hand side, we may directly obtain $$\frac{{\rm d}}{{\rm d}x}[e^{-2x}(x+1)-1]=-e^{-2x}(2x+1).$$
Thus $$f(x)=-e^{-2x}(2x+1),$$ $$\int_0^1f(x){\rm d}x=[e^{-2x}(x+1)]_0^1=\frac{2}{e^2}-1.$$