Question

The value of $\sum _{n=1}^{100}\underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx,$ where $\left[x\right]$ is the greatest integer $\le x,$ is :

• A. $100(e-1)$
• B. $100e$
• C. $100(1-e)$
• D. $100(1+e)$

Answer 1

The correct option is A $100(e-1)$

$\sum _{n=1}^{100}\underset{n-1}{\overset{n}{\int }}{e}^{x-\left[x\right]}dx$
$=\underset{0}{\overset{1}{\int }}{e}^{\left\{x\right\}}dx+\underset{1}{\overset{2}{\int }}{e}^{\left\{x\right\}}dx+\underset{2}{\overset{3}{\int }}{e}^{\left\{x\right\}}dx+\cdots +\underset{99}{\overset{100}{\int }}{e}^{\left\{x\right\}}dx(\because \{x\}=x-[x\left]\right)$
$=e^x{|}_0^1+e^{(x-1)}{|}_1^2+e^{(x-2)}{|}_2^3+\cdots +e^{(x-99)}{|}_{99}^{100}=(e-1)+(e-1)+(e-1)+\cdots +(e-1)=100(e-1)$