Question

# If ${\int }_{0}^{25}{e}^{x-\left[x\right]}dx=k\left(e-1\right)$, then the value of $k$ is

A

$12$

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B

$25$

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C

$23$

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D

$24$

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Solution

## The correct option is B $25$Explanation for the correct option.Find the value of $k$:Given, ${\int }_{0}^{25}{e}^{x-\left[x\right]}dx=k\left(e-1\right)$.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,${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{nt}}\mathbf{f}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{dx}\mathbf{=}\mathbf{n}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}\mathbf{f}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{dx}$.$\begin{array}{rcl}{\int }_{0}^{25×1}{e}^{x-\left[x\right]}dx& =& k\left(e-1\right)\\ & ⇒& 25{\int }_{0}^{1}{e}^{x-0}dx=k\left(e-1\right)\\ & ⇒& 25{\int }_{0}^{1}{e}^{x}dx=k\left(e-1\right)\\ & ⇒& 25{\left({e}^{x}\right)}_{0}^{1}=k\left(e-1\right)\\ & ⇒& 25\left({e}^{1}-{e}^{0}\right)=k\left(e-1\right)\\ & ⇒& 25\left(e-1\right)=k\left(e-1\right)\end{array}$On comparing the coefficient of both side $k=25$.Hence, the correct option is B.

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