Question

If ${\int }_0^{25}{e}^{x-\left[x\right]}dx=k(e-1)$, then the value of $k$ is

• A. $12$
• B. $25$
• C. $23$
• D. $24$

Answer 1

The correct option is B

$25$

Explanation for the correct option.

Find the value of $k$:

Given,

${\int }_0^{25}{e}^{x-\left[x\right]}dx=k(e-1)$.

Concept: $x-\left[x\right]$ is a periodic function with $1$.

So, $e^{x-\left[x\right]}$ has period $1$.

Now, we know that if $f\left(n\right)$ is periodic with period $t$, then,

${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{nt}}\mathbf{f}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{dx}\mathbf{=}\mathbf{n}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}\mathbf{f}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{dx}$.

$\begin{array}{rcl}{\int }_0^{25\times 1}{e}^{x-\left[x\right]}dx& =& k(e-1)\\ & \Rightarrow & 25{\int }_0^1{e}^{x-0}dx=k(e-1)\\ & \Rightarrow & 25{\int }_0^1{e}^{x}dx=k(e-1)\\ & \Rightarrow & 25{\left(e^x\right)}_0^1=k(e-1)\\ & \Rightarrow & 25\left(e^1-e^0\right)=k(e-1)\\ & \Rightarrow & 25(e-1)=k(e-1)\end{array}$

On comparing the coefficient of both side $k=25$.

Hence, the correct option is B.