Question

The value of${\int }_1^4{e}^{\sqrt{x}}dx$ is

• A. $e^2$
• B. $2e^2$
• C. $4e^2$
• D. $3e^2$

Answer 1

The correct option is B

$2e^2$

Compute the required value:

Given: ${\int }_1^4{e}^{\sqrt{x}}dx$

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

$\frac{1}{2\sqrt{x}}·dx=dt$

when $$\begin{gathered} x=4,t=2 \\ x=1,t=1 \end{gathered}$$

$\Rightarrow {\int }_1^2{e}^t·t·dt$

using integration by parts $\int \left(f\left(x\right)·g\left(x\right)\right)dx=f\left(x\right)·g\text{'}\left(x\right)-\int f\text{'}\left(x\right)·\left(\int g\left(x\right)·dx\right)·dx$

putting $f\left(x\right)=t,g\left(x\right)=e^t$

$$\begin{gathered} \Rightarrow 2{\int }_1^2{e}^t·t·dt=2\left[\left(t\right)\left(e^t\right)-\int e^t·dt\right] \\ \Rightarrow 2{\int }_1^2{e}^t·t·dt=2\left|\left(t\right)\left(e^t\right)-e^t\right| \\ \Rightarrow 2{\int }_1^2{e}^t·t·dt=2{\left|\left(t\right)\left(e^t\right)-e^t\right|}_1^2 \\ \Rightarrow 2{\int }_1^2{e}^t·t·dt=2\left|2\left(e^2\right)-e-e^2+e\right| \\ \Rightarrow 2{\int }_1^2{e}^t·t·dt=2e^2 \end{gathered}$$

Hence, option (B) is the correct answer.