Definite Integration Question 468
Question: $ \int _{1/e}^{e}{|\log x|dx=} $
[UPSEAT 2001]
Options:
A) $ 1-\frac{1}{e} $
B) $ 2( 1-\frac{1}{e} ) $
C) $ {e^{-1}}-1 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int _{1/e}^{e}{|\log x|dx=\int _{1/e}^{1}{-\log xdx+\int_1^{e}{\log xdx}}} $
$ =[x-x\log x] _{1/e}^{1}+[x\log x-x]_1^{e} $
$ =(1-0)-{ \frac{1}{e}-\frac{1}{e}(-1) }+e-e+1 $
$ =2-\frac{2}{e}=2( 1-\frac{1}{e} ) $ .
BETA
