Definite Integration Question 239
Question: The value of the integral $ \int _{-1}^{1}{\frac{d}{dx}( {{\tan }^{-1}}\frac{1}{x} )}dx $ is
Options:
A) $ \frac{\pi }{2} $
B) $ \frac{\pi }{4} $
C) $ -\frac{\pi }{2} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int _{-1}^{1}{\frac{d}{dx}}( {{\tan }^{-1}}\frac{1}{x} )dx=-2[{{\tan }^{-1}}(x)]_0^{1}=-\frac{\pi }{2} $ .
BETA
