Definite Integration Question 291
Question: The value of $ \int_1^{e}{\log xdx} $ is
[Pb. CET 2001]
Options:
A) $ 0 $
B) 1
C) $ e-1 $
D) $ e+1 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int_1^{e}{\log xdx} $
therefore $ I=\int_e^{1}{1.\log xdx} $
therefore $ I=[x\log x-x]_1^{e}=(e\log e-e)-(0-1) $
therefore $ I=1 $ .