Question

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

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

Answer 1

Answer: (B)

$I=\int\limits_{0}^{100 \pi} \sqrt{1-\cos 2 x}\, d x$
$=\int\limits_{0}^{100 \pi} \sqrt{2 \sin ^{2} x} \,d x$
$=\sqrt{2} \int\limits_{0}^{100 \pi} \sin x \mid d x$
$=\sqrt{2} \times 100 \int\limits_{0}^{\pi} \sin x \mid d x$
[ $\sin x \mid$ has period of $\pi$ ]
$=100 \sqrt{2} \int\limits_{0}^{\pi} \sin x \,d x$
$=100 \sqrt{2}[-\cos x]_{0}^{\pi}$
$=100 \sqrt{2}[(-\cos \pi)-(-\cos 0)]$
$=100 \sqrt{2}[-(-1)-(-1)]$
$=100 \sqrt{2} \times 2$
$=200 \sqrt{2}$