Find the area bounded by the curve
Question:

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

Solution:

The graph of y = sin x can be drawn as

$\therefore$ Required area = Area OABO + Area BCDB

$=\int_{0}^{\pi} \sin x d x+\left|\int_{\pi}^{2 \pi} \sin x d x\right|$

$=[-\cos x]_{0}^{\pi}+\left|[-\cos x]_{\pi}^{2 \pi}\right|$

$=[-\cos \pi+\cos 0]+|-\cos 2 \pi+\cos \pi|$

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

$=2+|-2|$

$=2+2=4$ units

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