Question

The value of ${\int }_0^1\sqrt{x}{e}^{\sqrt{x}}dx$ is equal to

• A. $\frac{\left(e-2\right)}{2}$
• B. $2(e-2)$
• C. $2e-1$
• D. $2(e-1)$
• E. $\frac{e-1}{2}$

Answer 1

Answer: (B)

. Let $I={\int }_0^1\sqrt{x}{e}^{\sqrt{x}}dx$
Put $\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}dx=dt$
$\therefore I={\int }_0^{1}t{e}^t(2tdt)$
$=2{\int }_0^1{t}^2{e}^{t}dt$
$=2{\left[t^2{e}^t-\int 2t{e}^{t}dt\right]}_0^1$
$=2{\left[t^2{e}^t-2t{e}^t+2e^t\right]}_0^1$
$=2\left[1e^1-2e^1+2e^1-\left(0-0+2e^0\right)\right]$
$=2[e-2]$