Question

${\int }_0^{1000}{e}^{x-\left[x\right]}dx$ is equal to

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

Answer 1

The correct option is D $1000(e-1)$

Let $I={\int }_0^{1000}{e}^{x-\left[x\right]}dx$
We know that the period of $x-\left[x\right]$ is $1$.
$\therefore I=1000{\int }_0^1{e}^{x}dx$
$=1000{\left[e^x\right]}_0^1=1000(e-1)$