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Area of the region bounded by the curve $y=\sqrt{49-x^2}$ and $X$-axis is
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Verified Answer
The correct answer is:
$\frac{49 \pi}{2}$ sq. units
$\begin{aligned}
\text { Required area } & =2 \int_0^7 \sqrt{49-x^2} \mathrm{~d} x \\
& =2\left[\frac{x}{2} \sqrt{49-x^2}+\frac{49}{2} \sin ^{-1}\left(\frac{x}{7}\right)\right]_0^7 \\
& =2\left[\frac{7}{2} \sqrt{49-49}+\frac{49}{2} \sin ^{-1}\left(\frac{7}{7}\right)\right]-0 \\
& =2 \times \frac{49}{2} \times \frac{\pi}{2} \\
& =\frac{49 \pi}{2} \text { sq. units }
\end{aligned}$
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