INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More

A nice problem from ISI 10+2

Compute I = $(\int_e^{e^4}\sqrt{\log(x)}dx)$ if it is given that $(\int _1^2 e^{t^2} dt = \alpha )$

I = $([x \sqrt{\log(x)}]_e^{e^4} - \int_e^{e^4} x \frac{1}{2 \sqrt{log(x)}} \frac {1}{x} dx )$
= $([e^4 \sqrt {\log_e e^4} - e \sqrt {\log _e e}] - \frac{1}{2} \int_e^{e^4}\frac{1}{\sqrt{log(x)}} dx )$
= $(2 e^4 - e - \frac{1}{2} \int_e^{e^4}\frac{1}{\sqrt{\log(x)}} dx )$

let $log(x) = (t^2)$

x =$(e^{t^2})$

dx = 2t $(e^{t^2})$ dt

Thus I = $(2 e^4 - e - \frac{1}{2} \int_e^{e^4}\frac{1}{\sqrt{\log(x)}} dx )$

=  $(2 e^4 - e - \frac{1}{2} \int _1^2 \frac {1}{t} 2 t e^{t^2} dt )$
=  $(2 e^4 - e - \int _1^2 e^{t^2} dt )$
=  $(2 e^4 - e - \alpha )$

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
CAREERTEAM
support@cheenta.com