Definite Integration Question 73
Question: The integral $ \int _{-1/2}^{1/2}{{ [x]+\log ( \frac{1+x}{1-x} ) }}dx $ equal (where [.] is the greatest integer function)
[IIT Screening 2002]
Options:
A) $ -\frac{1}{2} $
B) 0
C) 1
D) $ 2\log \frac{1}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int _{-1/2}^{1/2}{[x]dx+\int _{-1/2}^{1/2}{\log ( \frac{1+x}{1-x} )dx}} $
If $ f(x)=\log ( \frac{1+x}{1-x} ) $ , then $ f(-x)=\log ( \frac{1-x}{1+x} )=-\log ( \frac{1+x}{1-x} )=-f(x) $
$ \therefore I=\int _{-1/2}^{1/2}{\text{ }[x]}dx+0 $ (being integral of odd function) $ =\int _{-1/2}^{0}{-1dx+\int_0^{1/2}{0dx}}=-(x) _{-1/2}^{0}=\frac{-1}{2}. $
BETA
