int_0^{10} (5 - sqrt{10x - x^2}) , dx = | vedclass

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Question

(C) Let $I = \int_0^{10} (5 - \sqrt{10x - x^2}) \, dx$. First,complete the square inside the square root: $10x - x^2 = -(x^2 - 10x) = -(x^2 - 10x + 25

Vedclass · Accepted Answer

$\frac{1}{2}(100 - 25\pi)$

Vedclass · Answer

(C) Let $I = \int_0^{10} (5 - \sqrt{10x - x^2}) \, dx$. First,complete the square inside the square root: $10x - x^2 = -(x^2 - 10x) = -(x^2 - 10x + 25 - 25) = 25 - (x - 5)^2$. So,$I = \int_0^{10} 5 \, dx - \int_0^{10} \sqrt{5^2 - (x - 5)^2} \, dx$. The first part is $\int_0^{10} 5 \, dx = [5x]_0^{10} = 50$. For the second part,use the formula $\int \sqrt{a^2 - u^2} \, du = \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2}\sin^{-1}(\frac{u}{a})$. Here $u = x - 5$,so $du = dx$. When $x=0, u=-5$; when $x=10, u=5$. $\int_{-5}^5 \sqrt{5^2 - u^2} \, du = [\frac{u}{2}\sqrt{25 - u^2} + \frac{25}{2}\sin^{-1}(\frac{u}{5})]_{-5}^5$. $= (0 + \frac{25}{2}\sin^{-1}(1)) - (0 + \frac{25}{2}\sin^{-1}(-1)) = \frac{25}{2}(\frac{\pi}{2}) - \frac{25}{2}(-\frac{\pi}{2}) = \frac{25\pi}{4} + \frac{25\pi}{4} = \frac{25\pi}{2}$. Thus,$I = 50 - \frac{25\pi}{2} = \frac{100 - 25\pi}{2} = \frac{1}{2}(100 - 25\pi)$.

$\int_0^{10} (5 - \sqrt{10x - x^2}) \, dx = $

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