I am stuck in a concept of Definite Integration.
Let us say we have a function that goes like ${f(x) = \int_0^x e^{x-t} f(t) \,dt}$
Now I wanted to know that if I put $x=0$, will the limit range from $0$ to $0$ (hence value is also $0$). Also will the $x$ inside integral attain the value $0$. Also let $x=1$, then will the limits now range from $0$ to $1$ and the integral now becomes ${f(x) = \int_0^1 e^{1-t} f(t) \,dt}$ ? Please explain how this works.
The expression on the right works the way it's written. The value of the integral depends on the value of $x$. For each particular value it's a number. You can interpret that number as the area under the curve $e^{x-t}f(t)$ for $t$ in the interval $[0,x]$. It is indeed $0$ when $x=0$.
Writing $$ {f(x) = \int_0^1 e^{1-t} f(t) \,dt} $$ makes no sense. What substituting $1$ for $x$ calculates is the value of $f$ at $x=1$. That's the ordinary integral $$ {f(1) = \int_0^1 e^{1-t} f(t) \,dt} $$
If you really mean to have $f(x)$ on the left and also as part of the integrand then the equation will work for some functions $f$ and not others. To "solve" that equation for the unknown function $f$ you can differentiate the right side (correctly, using the fundamental theorem of calculus and taking into account the fact that there's an $x$ in the integrand). That will lead to a differential equation you can try to solve for $f$.