Integral equation with fixed boundaries

Question

How to solve integral equation $$ f(x) = 1 + \int_0^1 f(x-t) dt $$ for $x\geq 1$ and we also know that $f(x) = e^x-1$ for $x \in (0,1) $? I would like to obtain solution for $x \in [2,3]$ without integral, or is there any way how can I plot it without having exact formula?

Answer 1

$f(x)=1+\int_{0}^{1} f(x-t) dt$ by changing variable $$ f(x)=1-\int_{x-1}^{x} f(y) dy \Rightarrow f(x)=1-\int_{0}^{x} f(y) dy+\int_{0}^{x-1} f(y) dy \Rightarrow \\ f'(x)=f(x-1)-f(x),$$ so you get the following delay differential equation \begin{align*} \begin{cases} f'(x)=f(x-1)-f(x), \\ f(x)=e^x-1, x \in (0,1). \end{cases} \end{align*} you can solve it by method of steps.