Question

The value of ${\int }_0^2{e}^{x+\left[x\right]}d(x-[x\left]\right),$ where $\left[x\right]=$ greatest integer $\le x,$ is equal to ?

• A. $(e^2-1)(e+1)$
• B. $e-1$
• C. $(e^2+1)(e-1)$
• D. $e^2+1$

Answer 1

The correct option is C $(e^2+1)(e-1)$

$I={\int }_0^2{e}^{x+\left[x\right]}d(x-[x\left]\right)$

$I={\int }_0^1{e}^{x}dx+{\int }_1^2{e}^{1+x}d(x-1)$

Let $I_1={\int }_0^1{e}^{x}dx$

$I_1=e-1$

Let $I_2={\int }_1^2{e}^{1+x}d(x-1)$

Let $x-1=t$

$\Rightarrow dx=dt$

$\Rightarrow I_2={\int }_0^1{e}^{t+2}dt$

$\Rightarrow I_2=e^2({\int }_0^1{e}^{t}dt)$

$\Rightarrow I_2=e^2(e-1)$

$I=I_1+I_2$

$I=e^2(e-1)+(e-1)$

$I=(e^2+1)(e-1)$

Hence, option C is correct.