Definite Integrals

Question

Let $f: \mathbb{R} 	o \mathbb{R}$ be a differentiable function such that $f(x) = x^2 + \int_0^x e^{-(x-t)} f(t) dt$. Then $f(x)$ equals:

RankGuru · Accepted Answer

Note: The integral term can be simplified by factorization logic using $e^{-(x-t)} = e^{-x}e^t$. We use differentiation.

Method: Differentiate both sides.
given $f(x) = x^2 + \int_0^x e^{-(x-t)} f(t) dt$
$f(x) = x^2 + e^{-x} \int_0^x e^t f(t) dt$.
Multiply by $e^x$:
$e^x f(x) = x^2 e^x + \int_0^x e^t f(t) dt$.
Differentiate wrt $x$:
$e^x f'(x) + e^x f(x) = (x^2 e^x + 2x e^x) + e^x f(x)$.
Cancel $e^x f(x)$:
$e^x f'(x) = e^x (x^2 + 2x)$.
divide by $e^x$:
$f'(x) = x^2 + 2x$.
Integrate:
$f(x) = x^3/3 + x^2 + C$.
From original eq, $f(0) = 0$. So $C=0$.
$f(x) = x^3/3 + x^2$.

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