Find the area enclosed by the curve

Question:

Find the area enclosed by the curve $y=\sin x$ and the $X$-axis between $x=0$ and $x=\pi$.

Solution:

$y=\sin (x)$

Area under the curve from $x=0 t x=\pi$ is calculated by the method of integration.

Area $=\int_{x=0}^{x-\pi} y d x=\int_{x=0}^{x-\pi} \sin x d x=-[\cos \pi-\cos 0]=2$

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