Definite Integration Question 392
Question: If $ I_1=\int_e^{e^{2}}{\frac{dx}{\log x}} $ and $ I_2=\int_1^{2}{\frac{e^{x}}{x}dx,} $ then
[Karnataka CET 2000]
Options:
A) $ I_1=I_2 $
B) $ I_1>I_2 $
C) $ I_1<I_2 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ \log x=u $ in $ I_1, $ so that $ dx=xdu=e^{u}du $
Also as $ x=e $ to $ e^{2},u=1 $ to 2 Thus, $ I_1=\int_1^{2}{\frac{e^{u}}{u}du=\int_1^{2}{\frac{e^{x}}{x}dx}} $ .
Hence, $ I_1=I_2 $ .
BETA
