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 $ .