Definite Integration Question 120
Question: $ \int_0^{2n\pi }{( |\sin x|-. | \frac{1}{2}\sin x . | )}\ dx $ equals
[Orissa JEE 2005]
Options:
A) n
B) 2n
C) -2n
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_0^{2n\pi }{( |\sin x|-\frac{1}{2}|\sin x| )}\ dx $ = $ \frac{1}{2}\int_0^{2n\pi }{\|\sin x|dx} $
$ =\frac{2n}{2}\times 2\int_0^{\pi /2}{\sin x\ dx=2n}[-\cos x]_0^{\pi /2}=2n. $
BETA
