Question

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

• A. $2squnit$
• B. $4squnit$
• C. $8squnit$
• D. $Noneofthese$

Answer 1

The correct option is B $4squnit$

${\int }_0^{2\pi }\sin xdx$

$=2{\int }_0^{\pi }\sin xdx=2{\int }_0^{\pi }-\cos x$

$=-2[-1-p]=-2\times -2$

$=4sq.unit$