Question

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

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

Answer 1

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

Given ${\int }_1^2{e}^{x^2}dx=a$.

Let $I={\int }_e^{e^4}\sqrt{{\log }_{e}x}dx$

Substitute $\ln x=t^2,\therefore x=e^{t^2},\Rightarrow dx=2t{e}^{t^2}$

then $I={\int }_1^{2}t.2e^{t^2}dt=2{\int }_1^2{t}^2{e}^{t^2}dt$

$={\int }_1^{2}t.\left(2t{e}^{t^2}\right)dt=2{\int }_1^2{t}^2{e}^{t^2}dt={\int }_1^{2}t.\left(2t{e}^{t^2}\right)dt$

$={[t.e^{t^2}]}_1^2-{\int }_1^2{e}^{t^2}dt(\because {\int }_1^2{e}^{x^2}dx=a)$

$=2e^4-e^1-a=2e^4-e-a$