Question

If the value of the integral ${\int }_1^2{e}^{x^2}dx$ is $a$, then the value of ${\int }_e^{e^4}\sqrt{lnx}dx$ is

• A. $e^4-e-\alpha$
• B. $2e^4-e-\alpha$
• C. $2(e^4-e)-\alpha$
• D. $2e^4-1-\alpha$

Answer 1

The correct option is B $2e^4-e-\alpha$

Given
${\int }_1^2{e}^{2x}dx=\alpha$
Now $l={\int }_e^{e4}1\cdot \sqrt{lnx}dx=[{\left(x\sqrt{lnx}\right)}_0^{e4}-{\int }_e^{e4}\frac{x}{2x\sqrt{lnx}}dx]$
$l_1={\int }_e^{e4}\frac{dx}{\sqrt[2]{2lnx}}[x=e^{t4}\Rightarrow dx=e^{t2}2tdt]$
$={\int }_1^2\frac{e^{t2}}{2t}\cdot 2tdt={\int }_1^2{e}^{r2}dt=\alpha \therefore l=2e^4-e-\alpha$