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If the area of the region enclosed by the curve $x^2+y^2=16$ and then lines $x=2$ and $x=3$ is $\left(3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+k\right)$ sq units, then $k$ equals
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Verified Answer
The correct answer is:
$16 \sin ^{-1}\left(\frac{3}{4}\right)$
Area of region enclosed by the curve $x^2+y^2=16$ and line $x=2$ and $x=3$ is
$$
\begin{aligned}
& A=\int_2^3 \sqrt{16-x^2} d x \\
& A=\left[\frac{x}{2} \sqrt{16-x^2}+8 \sin ^{-1} \frac{x}{4}\right]_2^3 \\
& \Rightarrow 3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+k \\
& =\left(\frac{3}{2} \sqrt{7}+8 \sin ^{-1} \frac{3}{4}\right)-\left(2 \sqrt{3}+8 \sin ^{-1} \frac{2}{4}\right) \\
& \Rightarrow 3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+k \\
& =\frac{3}{2} \sqrt{7}+8 \sin ^{-1} \frac{3}{4}-2 \sqrt{3}-\frac{8 \pi}{6} \\
& =\frac{1}{2}\left(3 \sqrt{7}+16 \sin ^{-1} \frac{3}{4}-4 \sqrt{3}-\frac{8 \pi}{3}\right) \\
& \therefore \quad k=16 \sin ^{-1}\left(\frac{3}{4}\right) \\
&
\end{aligned}
$$
$$
\begin{aligned}
& A=\int_2^3 \sqrt{16-x^2} d x \\
& A=\left[\frac{x}{2} \sqrt{16-x^2}+8 \sin ^{-1} \frac{x}{4}\right]_2^3 \\
& \Rightarrow 3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+k \\
& =\left(\frac{3}{2} \sqrt{7}+8 \sin ^{-1} \frac{3}{4}\right)-\left(2 \sqrt{3}+8 \sin ^{-1} \frac{2}{4}\right) \\
& \Rightarrow 3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+k \\
& =\frac{3}{2} \sqrt{7}+8 \sin ^{-1} \frac{3}{4}-2 \sqrt{3}-\frac{8 \pi}{6} \\
& =\frac{1}{2}\left(3 \sqrt{7}+16 \sin ^{-1} \frac{3}{4}-4 \sqrt{3}-\frac{8 \pi}{3}\right) \\
& \therefore \quad k=16 \sin ^{-1}\left(\frac{3}{4}\right) \\
&
\end{aligned}
$$
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