Definite Integration Question 212
Question: $ \int _{-1}^{1}{\log \frac{2-x}{2+x}dx}= $
[Roorkee 1986; Kurukshetra CEE 1998]
Options:
A) 2
B) 1
C) $ -1 $
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ f(x)=\log ( \frac{2-x}{2+x} ) $
$ \Rightarrow f(-x)=\log {{( \frac{2-x}{2+x} )}^{-1}}=-\log ( \frac{2-x}{2+x} )=-f(x) $
$ \therefore $ $ \int _{-1}^{1}{\log ( \frac{2-x}{2+x} )dx=0} $ .
BETA
