Given $\int\limits_{1}^{2}e^{x^2} dx = a$, the value of $\int\limits_{e}^{e^4}\sqrt{\log_e x} dx$, is
Answer & explanation
Correct answer: option 4
We have,
$I=\int\limits_{e}^{e^4}\sqrt{\log_e x} dx=\int\limits_{1}^{2}\sqrt{t^2}e^{t^2} 2t\,dt$, where $\log_e x = t^2$
$⇒I=2\int\limits_{1}^{2}t^2\,e^{t^2}dt$
$⇒I=\int\limits_{1}^{2}\underset{I}{t}\,e^{t^2}\underset{II}{2t}dt=[t\,e^{t^2}]_{1}^{2}-\int\limits_{1}^{2}e^{t^2}dt=2e^4-e-a$