Area of the region bounded by the curve $y=\sqrt{49-x^2}$ and $X$-axis is

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Aicharya · Accepted Answer

$\begin{aligned} 	ext { Required area } & =2 \int_0^7 \sqrt{49-x^2} \mathrm{~d} x \ & =2\left[\frac{x}{2} \sqrt{49-x^2}+\frac{49}{2} \sin ^{-1}\left(\frac{x}{7}ight)ight]_0^7 \ & =2\left[\frac{7}{2} \sqrt{49-49}+\frac{49}{2} \sin ^{-1}\left(\frac{7}{7}ight)ight]-0 \ & =2 	imes \frac{49}{2} 	imes \frac{\pi}{2} \ & =\frac{49 \pi}{2} 	ext { sq. units } \end{aligned}$

Area of the region bounded by the curve $y=\sqrt{49-x^2}$ and $X$-axis is

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