Question

Let f be a derivable function satisfying the equation ${\int }_0^{x}f\left(t\right)dt+{\int }_0^{x}t.f(x-t)dt=e^{-x}-1$

${\int }_0^{1}f\left(x\right)dx$ is equal to

• A. $e^{-1}$
• B. -1
• C. 1
• D. $-e^{-1}$

Answer 1

The correct option is D

$-e^{-1}$

Let f be a derivable function satisfying the equation ${\int }_0^{x}f\left(t\right)dt+{\int }_0^{x}t.f(x-t)dt=e^{-x}-1$

Given ${\int }_0^{x}f\left(t\right)dt+{\int }_0^{x}t.f(x-t)dt=c^{-x}-1$
or, ${\int }_0^{x}f\left(t\right)dt+x{\int }_0^{x}f\left(t\right)dt-{\int }_0^{x}tf\left(t\right)dt=e^{-x}-1$
Diff. both sides with respect to x, we get.
$f\left(x\right)+x.f\left(x\right)+{\int }_0^{x}f\left(t\right)dt-xf\left(x\right)=-e^{-x}$
$\Rightarrow f\left(x\right)+{\int }_0^{x}f\left(t\right)dt=-e^{-x}$
again diff. w.r.t to $x\Rightarrow e^x\left(f^{\prime }\right(x\left)\right)+f\left(x\right)=1\dots \left(2\right)$
$e^x.f\left(x\right)=x+c,from\left(1\right),f\left(0\right)=-1$
$\therefore {\int }_0^{x}f\left(x\right)dx={\int }_0^1(x-1)e^{-x}dx=(-(x-1)e^{-x}{)}_0^1|{\int }_0^1{e}^{-x}dx=-1-(e^{-x}{)}_0^1=-1-(\frac{1}{e}-1)=\frac{1}{e}$