Search any question & find its solution
Question:
Answered & Verified by Expert
If $I_1=\int_e^{e^2} \frac{d x}{\log x}$ and $I_2=\int_1^2 \frac{e^x}{x} d x_1$ then
Options:
Solution:
1990 Upvotes
Verified Answer
The correct answer is:
$l_1-l_2$
Put $\log x=u$ in $l_1$, so that $d x=x d u=e^u d u$ Also as $x=e$ to $e^2, u=1$ to 2
Thus, $I_1=\int_1^2 \frac{e^u}{u} d u=\int_1^2 \frac{e^x}{x} d x$. Hence, $I_1=I_2$.
Thus, $I_1=\int_1^2 \frac{e^u}{u} d u=\int_1^2 \frac{e^x}{x} d x$. Hence, $I_1=I_2$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.