Definite Integration Question 3
Question: If $ I=\int_0^{100\pi }{\sqrt{(1-\cos 2x)}dx,} $ then the value of $ I $ is
Options:
A) $ 100\sqrt{2} $
B) $ 200\sqrt{2} $
C) $ 50\sqrt{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int_0^{\pi }{\sqrt{(1-\cos 2x)}dx+\int _{\pi }^{2\pi }{\sqrt{(1-\cos 2x)}}dx}+…. $
$ …..+\int _{(r-1)\pi }^{r\pi }{\sqrt{(1-\cos 2x)}dx+…..+\int _{99\pi }^{100\pi }{\sqrt{(1-\cos 2x)}}dx} $
$ \because \int_0^{na}{f(x)dx=n\int_0^{a}{f(x)dx}} $ , if $ f(a+x)=f(x) $
$ \therefore $ $ I=100\int_0^{\pi }{\sqrt{(1-\cos 2x)}}dx $
$ I=100\sqrt{2}\int_0^{\pi }{\sin xdx=200\sqrt{2}\int_0^{\pi /2}{\sin xdx}} $
$ =200\sqrt{2}[-\cos x]_0^{\pi /2}=200\sqrt{2} $ .
BETA
