Question

Let $I=\underset{0}{\overset{100\pi }{\int }}$ $\sqrt{\left(1-\cos 2x\right)}dx,$ then

• A. $I=0$
• B. $I=200\sqrt{2}$
• C. $I=\pi \sqrt{2}$
• D. $I=100$

Answer 1

Answer: (B)

$I=\underset{0}{\overset{100\pi }{\int }}\sqrt{1-\cos 2x}dx$
$=\underset{0}{\overset{100\pi }{\int }}\sqrt{2{\sin }^{2}x}dx$
$=\sqrt{2}\underset{0}{\overset{100\pi }{\int }}\sin x\mid dx$
$=\sqrt{2}\times 100\underset{0}{\overset{\pi }{\int }}\sin x\mid dx$
[ $\sin x\mid$ has period of $\pi$ ]
$=100\sqrt{2}\underset{0}{\overset{\pi }{\int }}\sin xdx$
$=100\sqrt{2}[-\cos x{]}_0^{\pi }$
$=100\sqrt{2}[(-\cos \pi )-(-\cos 0)]$
$=100\sqrt{2}[-(-1)-(-1)]$
$=100\sqrt{2}\times 2$
$=200\sqrt{2}$