A) \[100\sqrt{2}\]
B) \[200\sqrt{2}\]
C) \[50\sqrt{2}\]
D) None of these
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 \[f(a+x)=f(x)\] \[\therefore \]\[I=100\int_{0}^{\pi }{\sqrt{(1-\cos 2x)}}dx\] \[I=100\sqrt{2}\int_{0}^{\pi }{\sin xdx=200\sqrt{2}\int_{0}^{\pi /2}{\,\sin x\,dx}}\] \[=200\sqrt{2}[-\cos x]_{0}^{\pi /2}=200\sqrt{2}\].
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