Question

Let $f$ be a continuous function satisfying the equation $\underset{0}{\overset{x}{\int }}f\left(t\right)dt+\underset{0}{\overset{x}{\int }}tf(x-t)dt=e^{-x}-1.$ Then the value of $e^{10}f\left(10\right)$ is

Answer 1

$\underset{0}{\overset{x}{\int }}f\left(t\right)dt+\underset{0}{\overset{x}{\int }}tf(x-t)dt=e^{-x}-1$
$\Rightarrow \underset{0}{\overset{x}{\int }}f\left(t\right)dt+\underset{0}{\overset{x}{\int }}(x-t)f\left(t\right)dt=e^{-x}-1$
$\Rightarrow x\underset{0}{\overset{x}{\int }}f\left(t\right)dt+\underset{0}{\overset{x}{\int }}f\left(t\right)(1-t)dt=e^{-x}-1$
Differentiating w.r.t. $x,$
$x\left(f\right(x)\cdot 1-0)+\underset{0}{\overset{x}{\int }}f\left(t\right)dt+f\left(x\right)(1-x)\cdot 1=-e^{-x}$
$\Rightarrow f\left(x\right)+\underset{0}{\overset{x}{\int }}f\left(t\right)dt=-e^{-x}\cdots \left(1\right)$
Again differentiating w.r.t. $x,$
$f^{\prime }\left(x\right)+f\left(x\right)=e^{-x}$
$\Rightarrow \int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=\int dx$
$\Rightarrow e^{x}f\left(x\right)=x+C\cdots \left(2\right)$
Put $x=0$ in $\left(1\right),$
$f\left(0\right)+0=-1\Rightarrow f\left(0\right)=-1$
Put $x=0$ in $\left(2\right),$
$f\left(0\right)=0+C\Rightarrow C=f\left(0\right)=-1$
$\therefore e^{x}f\left(x\right)=x-1$
At $x=10,$ we have $e^{x}f\left(x\right)=9$