Question

Given ${\int }_1^2{e}^{x^2}dx=a$ the the value of ${\int }_e^{e^4}\sqrt{lnx}dx$ is

• A. $2e^4-e-a$
• B. $e^4-a$
• C. $2e^4-a$
• D. $e^4-e$

Answer 1

The correct option is A $2e^4-e-a$

${\int }_1^2{e}^{x^2}dx=a$ (given)
Let $I={\int }_e^{e^4}\sqrt{\log x}dx$ Putting $\log x=t^{2}dx=2t{e}^{t^2}dt$
$={\int }_1^{2}t\left(2t\right)e^{t^2}dt={\int }_1^{2}t\left(2t{e}^{t^2}\right)dt$
$=[t{e}^{t^2}{]}_1^2-{\int }_1^2{e}^{t2}dt=2e^4-e-a$
$\{As{\int }_1^2{e}^{t^2}dt={\int }_{1}62e^{x^2}dx=a)$
Hence (a) is correct choice.