Question

The area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi $ given by

• A. $4\, sq.\, units$
• B. $3\, sq.\, units$
• C. $0\, sq.\, units$
• D. $2\, sq. \, units$

Answer 1

Answer: (A)

We have, $y = \sin x, x = 0$ and $x = 2\pi$
$ \therefore $ Required area = Area of shaded region
$\int\limits_{0}^{2\pi} \left|\sin x\right|dx = \int\limits_{0}^{\pi} \sin x dx + \int\limits_{\pi}^{2\pi} \left(-\sin x\right)dx $
$=- \left[\cos x\right]^{\pi}_{0} + \left[\cos x\right]^{2\pi}_{\pi} $
$= -\left[\cos \pi -\cos 0\right]+\left[\cos 2\pi -\cos \pi\right] $
$=- \left[-1 -1 \right]+\left[1+1\right]=4$