Question

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

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

Answer 1

Answer: (B)

$2(e-2)$

Let

$I={\int }_0^1\sqrt{x}{e}^{\sqrt{x}}dx$

Put $\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}dx=dt$

Therefore,

$I={\int }_0^{1}t{e}^t\left(2tdt\right)$

$=2{\int }_0^1\underset{I}{t^2}\underset{II}{e^t}dt$

$=2{[t^2{e}^t-\int \underset{I}{2t}+\underset{II}{e^t}dt]}_0^1$...............(applying by-parts formula for integration)

$=2[t^2{e}^t-2t{e}^t+2e^t{]}_0^1$

$=2[1.e^1-2e^1+2e^1-(0-0+2e^0\left)\right]$

$=2[e-2]$