Question

${\int }_0^{1000}{e}^{x-|x|}dx$ is

• A. $e^{1000}-1$
• B. $\frac{e^{1000}-1}{e-1}$
• C. $1000(e-1)$
• D. $\frac{e-1}{1000}$

Answer 1

Answer: (C)

We know, $e^{x-|x|}$ is periodic function with period 1.
$\therefore I=I={\int }_0^{1000}{e}^{x-\left[x\right]}dx$
$I={\int }_0^{1000}{e}^{x-\left[x\right]}dx$$\underset{0}{\overset{1000}{\int }}$$e^{\left\{x\right\}}dx$
=$\underset{0}{\overset{1}{\int }}{e}^{x}dx+$$\underset{0}{\overset{2}{\int }}{e}^{x}dx+$$\underset{0}{\overset{3}{\int }}{e}^{x}dx+.....+\underset{999}{\overset{1000}{\int }}{e}^{x}dx$
$=\underset{0}{\overset{1}{\int }}{e}^{x}dx+$$\underset{0}{\overset{1}{\int }}{e}^{x}dx+$$\underset{0}{\overset{1}{\int }}{e}^{x}dx+.....1000terms$
$=1000\underset{0}{\overset{1000}{\int }}{e}^{x}dx=1000[e^x{]}_0^1=1000[e^1-e^0]$
$=1000(e-1)$