Thank you for reporting, we will resolve it shortly
Q.
$\int^{1000}_{0}e^{x-\left|x\right|}dx$ is
Integrals
Solution:
We know, $e^{x-\left|x\right|}$ is periodic function with period 1.
$\therefore I=I=\int^{1000}_{0}e^{x-\left[x\right]}dx$
$ I=\int^{1000}_{0}e^{x-\left[x\right]}dx $$\int\limits^{1000}_{{0}}$$e^{\left\{x\right\}}dx$
=$\int\limits^{1}_{{0}}e^x\,dx+$$\int\limits^{2}_{{0}}e^x\,dx+$$\int\limits^{3}_{{0}}e^x\,dx+.....+\int\limits_{999}^{1000}e^{x} dx$
$=\int\limits^{1}_{{0}}e^x\,dx+$$\int\limits^{1}_{{0}}e^x\,dx+$$\int\limits^{1}_{{0}}e^x\,dx+.....1000 \,\,terms$
$=1000\int\limits^{1000}_{{0}}e^x\,dx=1000[e^x]^1_0=1000[e^1-e^0]$
$=1000(e-1)$