int_{0}^{3} sqrt{9 - x^2} dx = | vedclass

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(B) Let $I = \int_{0}^{3} \sqrt{3^2 - x^2} dx$.Using the standard integral formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^

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$\frac{9\pi}{4}$

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(B) Let $I = \int_{0}^{3} \sqrt{3^2 - x^2} dx$.Using the standard integral formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a}) + C$.Applying the limits from $0$ to $3$:$I = [\frac{x}{2} \sqrt{9 - x^2} + \frac{9}{2} \sin^{-1}(\frac{x}{3})]_{0}^{3}$.Substitute the upper limit $x = 3$:$(\frac{3}{2} \sqrt{9 - 9} + \frac{9}{2} \sin^{-1}(\frac{3}{3})) = (0 + \frac{9}{2} \sin^{-1}(1)) = \frac{9}{2} 	imes \frac{\pi}{2} = \frac{9\pi}{4}$.Substitute the lower limit $x = 0$:$(\frac{0}{2} \sqrt{9 - 0} + \frac{9}{2} \sin^{-1}(0)) = (0 + 0) = 0$.Therefore, $I = \frac{9\pi}{4} - 0 = \frac{9\pi}{4}$.

$\int_{0}^{3} \sqrt{9 - x^2} dx =$

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