int_0^4 frac{x+2}{sqrt{4x-x^2}} dx = | vedclass

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(B) Let $I = \int_0^4 \frac{x+2}{\sqrt{4x-x^2}} dx$. First,rewrite the numerator: $x+2 = \frac{1}{2}(2x-4) + 4$. So,$I = \int_0^4 \frac{\frac{1}{2}(2x

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$4\pi$

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(B) Let $I = \int_0^4 \frac{x+2}{\sqrt{4x-x^2}} dx$. First,rewrite the numerator: $x+2 = \frac{1}{2}(2x-4) + 4$. So,$I = \int_0^4 \frac{\frac{1}{2}(2x-4) + 4}{\sqrt{4x-x^2}} dx = \frac{1}{2} \int_0^4 \frac{2x-4}{\sqrt{4x-x^2}} dx + 4 \int_0^4 \frac{1}{\sqrt{4-(x-2)^2}} dx$. For the first integral,let $u = 4x-x^2$,then $du = (4-2x) dx$,so $\int \frac{2x-4}{\sqrt{4x-x^2}} dx = -2\sqrt{4x-x^2}$. Evaluating from $0$ to $4$: $[-2\sqrt{4x-x^2}]_0^4 = -2(0) - (-2(0)) = 0$. For the second integral,$\int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1}(\frac{x}{a})$. So,$4 \int_0^4 \frac{1}{\sqrt{2^2-(x-2)^2}} dx = 4 [\sin^{-1}(\frac{x-2}{2})]_0^4$. $= 4 [\sin^{-1}(1) - \sin^{-1}(-1)] = 4 [\frac{\pi}{2} - (-\frac{\pi}{2})] = 4 [\pi] = 4\pi$.

$\int_0^4 \frac{x+2}{\sqrt{4x-x^2}} dx =$

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