MCQ
$\int_0^{10}\left(5-\sqrt{10 x-x^2}\right) d x=$
- A$50-25 \pi$
- B$(100-25 \pi)$
- ✓$\frac{1}{2}(100-25 \pi)$
- D$\frac{1}{4}(100-25 \pi)$
✓
Answer
Correct option: C.
$\frac{1}{2}(100-25 \pi)$
(C)
$\int_0^{10}\left(5-\sqrt{10 x-x^2}\right) d x$
$\begin{array}{l}=\int_0^{10} 5 d x-\int_0^{10} \sqrt{5^2-(x-5)^2} d x \\ =5[x]_0^{10}-\left[\frac{x-5}{2} \sqrt{5^2-(x-5)^2}+\frac{5^2}{2} \sin ^{-1} \frac{(x-5)}{5}\right]_0^{10} \\ =50-\left[\frac{5^2}{2} \cdot \frac{\pi}{2}-\frac{5^2}{2}\left(-\frac{\pi}{2}\right)\right] \\ =50-\frac{25 \pi}{2}=\frac{1}{2}(100-25 \pi)\end{array}$
$\int_0^{10}\left(5-\sqrt{10 x-x^2}\right) d x$
$\begin{array}{l}=\int_0^{10} 5 d x-\int_0^{10} \sqrt{5^2-(x-5)^2} d x \\ =5[x]_0^{10}-\left[\frac{x-5}{2} \sqrt{5^2-(x-5)^2}+\frac{5^2}{2} \sin ^{-1} \frac{(x-5)}{5}\right]_0^{10} \\ =50-\left[\frac{5^2}{2} \cdot \frac{\pi}{2}-\frac{5^2}{2}\left(-\frac{\pi}{2}\right)\right] \\ =50-\frac{25 \pi}{2}=\frac{1}{2}(100-25 \pi)\end{array}$
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