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$\int_0^{10}\left(5-\sqrt{10 x-x^2}\right) d x=$
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2234 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}(100-25 \pi)$
$$
\begin{aligned}
& \text { Let } \\
& I=\int_0^{10} 5-\sqrt{\left.10 x-x^2\right)} d x, I=\int_0^{10}\left(5-\sqrt{(5)^2-(x-5)^2} d x\right. \\
& I=\left[5 x-\frac{x-5}{2} \sqrt{10 x-x^2}-\frac{25}{2} \sin ^{-1} \frac{x-5}{5}\right]_0^{10} \\
& I=\left(50-0-\frac{25}{2} \sin ^{-1} 1\right)-\left(0-0-\frac{25}{2} \sin ^{-1}(-1)\right) \\
& I=50-\frac{25}{2} \times \frac{\pi}{2}-\frac{25}{2} \times \frac{\pi}{2} \\
& I=50-\frac{25 \pi}{2} \Rightarrow I=\frac{1}{2}(100-25 \pi)
\end{aligned}
$$
\begin{aligned}
& \text { Let } \\
& I=\int_0^{10} 5-\sqrt{\left.10 x-x^2\right)} d x, I=\int_0^{10}\left(5-\sqrt{(5)^2-(x-5)^2} d x\right. \\
& I=\left[5 x-\frac{x-5}{2} \sqrt{10 x-x^2}-\frac{25}{2} \sin ^{-1} \frac{x-5}{5}\right]_0^{10} \\
& I=\left(50-0-\frac{25}{2} \sin ^{-1} 1\right)-\left(0-0-\frac{25}{2} \sin ^{-1}(-1)\right) \\
& I=50-\frac{25}{2} \times \frac{\pi}{2}-\frac{25}{2} \times \frac{\pi}{2} \\
& I=50-\frac{25 \pi}{2} \Rightarrow I=\frac{1}{2}(100-25 \pi)
\end{aligned}
$$
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