# If f is a continuous function such that int_0^xf(t)dt=xe^(2x) + int_0^xe^-tf(t)dt for all x, fi nd an explicit formula for f(x) ?

## If $f$ is a continuous function such that ${\int}_{0}^{x} f \left(t\right) \mathrm{dt} = x {e}^{2 x}$ + ${\int}_{0}^{x} {e}^{-} t f \left(t\right) \mathrm{dt}$ for all $x$, fi nd an explicit formula for $f \left(x\right)$ ?

Jun 23, 2018

$f \left(x\right) = \frac{{e}^{3 x} \left(2 x + 1\right)}{{e}^{x} - 1}$

#### Explanation:

${\int}_{0}^{x} f \left(t\right) \mathrm{dt} = x {e}^{2 x} + {\int}_{0}^{x} {e}^{- t} f \left(t\right) \mathrm{dt}$

Differentiate both sides with respect to $x$ (hint: use the fundamental rule of calculus and the product rule):

$f \left(x\right) = 2 x {e}^{2 x} + {e}^{2 x} + {e}^{- x} f \left(x\right)$

Now, arrange both sides to get
$f \left(x\right) = \frac{{e}^{2 x} \left(2 x + 1\right)}{1 - {e}^{- x}}$
$\textcolor{w h i t e}{f \left(x\right)} = \frac{{e}^{3 x} \left(2 x + 1\right)}{{e}^{x} - 1}$