Question

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

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

Answer 1

The correct option is A

$100\left(e-1\right)$

Explanation for the correct answer:

We know that, $x-\left[x\right]=\left\{x\right\}$, where $x$ id known as the fractional part of

Hence, the given integral is reduced to

$\sum _{n=1}^{100}{\int }_{n-1}^n{e}^{\left\{x\right\}}dx$

The period of the function fractional part of $x$=$\left\{x\right\}$ is $1$

$\Rightarrow$ $\sum _{n=1}^{100}{\int }_{n-1}^n{e}^{\left\{x\right\}}dx={\int }_0^1{e}^{\left\{x\right\}}dx+{\int }_1^2{e}^{\left\{x\right\}}dx+{\int }_2^3{e}^{\left\{x\right\}}dx+.......+{\int }_{99}^{100}{e}^{\left\{x\right\}}dx$

Since all limit values are integral values

$\Rightarrow$ $\sum _{n=1}^{100}{\int }_{n-1}^n{e}^{\left\{x\right\}}dx={\int }_0^1{e}^{x}dx+{\int }_1^2{e}^{x-1}dx+{\int }_2^3{e}^{x-2}dx+.......+{\int }_{99}^{100}{e}^{x-99}dx$

$\Rightarrow$ $\sum _{n=1}^{100}{\int }_{n-1}^n{e}^{\left\{x\right\}}dx={\left[e^x\right]}_0^1+{\left[e^{x-1}\right]}_1^2+{\left[e^{x-2}\right]}_2^3+.......+{\left[e^{x-99}\right]}_{99}^{100}$

$\Rightarrow$ $\sum _{n=1}^{100}{\int }_{n-1}^n{e}^{\left\{x\right\}}dx=\left(e-1\right)+\left(e-1\right)+.......+\left(e-1\right)$

$\Rightarrow$ $\sum _{n=1}^{100}{\int }_{n-1}^n{e}^{\left\{x\right\}}dx=100\left(e-1\right)$

Hence, the value of $\sum _{n=1}^{100}{\int }_{n-1}^n{e}^{x-\left[x\right]}dx$ is $100\left(e-1\right)$.

Hence, option $\left(A\right)$ is the correct answer.