Question

Let $x>0$ be a fixed real number. Then the integral ${\int }_0^{\infty }{e}^{-t}|x-t|dt$ is equal to.

• A. $x+2e^{-x}-1$
• B. $x-2e^{-x}+1$
• C. $x+2e^{-x}+1$
• D. $-x-2e^{-x}+1$

Answer 1

Answer: (A)

$x+2e^{-x}-1$

${\int }_0^{\infty }{e}^{-r}|(x-t)|dt$
$={\int }_0^x{e}_{II}^{-t}\underset{I}{(x-t)}dt+{\int }_x^{\infty }{e}_{II}^{-t}\underset{I}{(t-x)}dt$
${\{-(x-t)e^{-t}-(-1\left)e^{-t}\right\}}_0^x-{\{-(x-t)e^{-t}+e^{-t}\}}_x^{\infty }$
$=x+2e^{-x}-1$.