Definite Integration Question 35
Question: $ \int _{\pi }^{10\pi }{|\sin x|dx} $ is
[AIEEE 2002]
Options:
A) 20
B) 8
C) 10
D) 18
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int _{\pi }^{10\pi }{|\sin x|dx=\int_0^{\pi }{|\sin x|dx+\int _{\pi }^{10\pi }{|\sin x|dx}}}-\int_0^{\pi }{|\sin x|dx} $
$ =\int_0^{10\pi }{|\sin x|dx-\int_0^{\pi }{|\sin x|dx}} $
$ =10\int _{0}^{\pi }{|\sin x|dx-\int _{0}^{\pi }{|\sin x|dx}} $
$ =9\int _{0}^{\pi }{\sin xdx} $
$ [\because |\sin x| $ is periodic with period $ \pi $ and in $ [0,\pi ],\sin x\ge 0] $
$ =9[-\cos x]_0^{\pi }=9(-\cos \pi +\cos 0) $
$ =9(1+1)=18 $ .
BETA
