int_0^2 sqrt{(x+3)(2-x)} , dx = | vedclass

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(C) Let $I = \int_0^2 \sqrt{(x+3)(2-x)} \, dx$.
Expanding the term inside the square root: $(x+3)(2-x) = -x^2 - x + 6 = 6 - (x^2 + x)$.
Completing t

Vedclass · Accepted Answer

$\frac{25\pi}{16} - \frac{\sqrt{6}}{4} - \frac{25}{8} \sin^{-1}\left(\frac{1}{5}\right)$

Vedclass · Answer

(C) Let $I = \int_0^2 \sqrt{(x+3)(2-x)} \, dx$.
Expanding the term inside the square root: $(x+3)(2-x) = -x^2 - x + 6 = 6 - (x^2 + x)$.
Completing the square: $6 - (x^2 + x + \frac{1}{4} - \frac{1}{4}) = \frac{25}{4} - (x + \frac{1}{2})^2$.
So, $I = \int_0^2 \sqrt{(\frac{5}{2})^2 - (x + \frac{1}{2})^2} \, dx$.
Using the formula $\int \sqrt{a^2 - u^2} \, du = \frac{u}{2} \sqrt{a^2 - u^2} + \frac{a^2}{2} \sin^{-1}(\frac{u}{a}) + C$:
$I = \left[ \frac{2x + 1}{4} \sqrt{(x+3)(2-x)} + \frac{25}{8} \sin^{-1}\left(\frac{2x + 1}{5}\right) \right]_0^2$.
Evaluating at $x=2$: $\frac{5}{4} \sqrt{0} + \frac{25}{8} \sin^{-1}(1) = \frac{25}{8} \cdot \frac{\pi}{2} = \frac{25\pi}{16}$.
Evaluating at $x=0$: $\frac{1}{4} \sqrt{6} + \frac{25}{8} \sin^{-1}(\frac{1}{5})$.
Therefore, $I = \left( \frac{25\pi}{16} \right) - \left( \frac{\sqrt{6}}{4} + \frac{25}{8} \sin^{-1}(\frac{1}{5}) \right) = \frac{25\pi}{16} - \frac{\sqrt{6}}{4} - \frac{25}{8} \sin^{-1}(\frac{1}{5})$.

$\int_0^2 \sqrt{(x+3)(2-x)} \, dx =$

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