Question

Consider the integral $I={\int }_0^{10}\frac{\left\{\left[x\right]e^{\left[x\right]}\right\}}{\left[e^{x-1}\right]}dx$ where $\left[x\right]$ denotes the greatest integer less than or equal to $x$. Then the value of I is equal to :

• A. $45\left(e-1\right)$
• B. $45\left(e+1\right)$
• C. $9\left(e-1\right)$
• D. $9\left(e+1\right)$

Answer 1

The correct option is A

$45\left(e-1\right)$

Explanation for the correct options:

Step1. Simplifying the given integral

$I={\int }_0^{10}\frac{\left\{\left[x\right]e^{\left[x\right]}\right\}}{\left[e^{x-1}\right]}dx$

$$\begin{gathered} ={\int }_0^{10}\left[x\right]e^{\left[x\right]-\left(x-1\right)}dx \\ ={\int }_0^{10}\left[x\right]e^{\left[x\right]-x+1}dx \\ ={\int }_0^{10}\left[x\right]e^{\left[x\right]-x}.edx \\ =e{\int }_0^{10}\left[x\right]e^{\left[x\right]-x}dx \end{gathered}$$

Step2. Finding the value of greatest integer function.

As, $\left[x\right]$ represents greatest integer function therefore

$\left[x\right]=0\forall x\in [0,1)$

similarly,

$$\begin{gathered} \left[x\right]=1\forall x\in [1,2) \\ \left[x\right]=2\forall x\in [2,3) \\ . \\ . \\ . \\ \left[x\right]=9\forall x\in [9,10) \end{gathered}$$

Step3.Finding value of $I$

$\Rightarrow I=e{\int }_0^1\left[x\right]e^{\left[x\right]-x}dx+e{\int }_1^2\left[x\right]e^{\left[x\right]-x}dx+e{\int }_2^3\left[x\right]e^{\left[x\right]-x}dx+.....+e{\int }_9^{10}\left[x\right]e^{\left[x\right]-x}dx$

$$\begin{gathered} =e·0·\left(\frac{e-1}{e}\right)+e·1·\left(\frac{e-1}{e}\right)+e·2·\left(\frac{e-1}{e}\right)+e·3·\left(\frac{e-1}{e}\right)+......+e·10·\left(\frac{e-1}{e}\right) \\ =e·\left(\frac{e-1}{e}\right)+2e·\left(\frac{e-1}{e}\right)+3e·\left(\frac{e-1}{e}\right)++.......+10e·\left(\frac{e-1}{e}\right) \\ =\left(e-1\right)+2\left(e-1\right)+3\left(e-1\right)+......+10\left(e-1\right) \\ =\left(1+2+3+...+10\right)\left(e-1\right) \\ =45\left(e-1\right) \end{gathered}$$

Hence, option A is correct.