Question

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

• A. $1$ sq unit
• B. $2$ sq unit
• C. $4$ sq unit
• D. $8$ sq unit

Answer 1

Answer: (C)

$4$ sq unit

Given curve is $y=\sin x$ between $x=0$ and $x=2\pi$

$\therefore$ Required area

$={\int }_0^{\pi }|\sin x|dx+{\int }_{\pi }^{2\pi }|\sin x|dx$

$={\int }_0^{\pi }\sin xdx+{\int }_{\pi }^{2\pi }(-\sin x)dx$

$(\because \sin x\ge 0forx\in [0,\pi ]and\sin x\le 0forx\in [\pi ,2\pi \left]\right)$

$=[-\cos x{]}_0^{\pi }+[\cos x{]}_{\pi }^{2\pi }$

$=-\cos \pi +\cos 0+\cos 2\pi -\cos \pi$

$=-(-1)+1+1-(-1)$

$=4$ sq unit