The value of int_{3}^{4} sqrt{(4-x)(x-3)} d x | vedclass

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(B) Let $I = \int_{3}^{4} \sqrt{(4-x)(x-3)} d x$.We can rewrite the integrand as $\sqrt{-x^2 + 7x - 12}$.Completing the square: $-x^2 + 7x - 12 = -\le

Vedclass · Accepted Answer

$\frac{\pi}{8}$

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(B) Let $I = \int_{3}^{4} \sqrt{(4-x)(x-3)} d x$.We can rewrite the integrand as $\sqrt{-x^2 + 7x - 12}$.Completing the square: $-x^2 + 7x - 12 = -\left(x^2 - 7x + \frac{49}{4} - \frac{49}{4}ight) - 12 = -\left(x - \frac{7}{2}ight)^2 + \frac{49}{4} - 12 = \left(\frac{1}{2}ight)^2 - \left(x - \frac{7}{2}ight)^2$.So,$I = \int_{3}^{4} \sqrt{\left(\frac{1}{2}ight)^2 - \left(x - \frac{7}{2}ight)^2} d x$.Let $t = x - \frac{7}{2}$,then $dt = dx$. When $x=3, t=-\frac{1}{2}$ and when $x=4, t=\frac{1}{2}$.$I = \int_{-1/2}^{1/2} \sqrt{\left(\frac{1}{2}ight)^2 - t^2} dt$.Since the integrand is an even function,$I = 2 \int_{0}^{1/2} \sqrt{\left(\frac{1}{2}ight)^2 - t^2} dt$.Using the formula $\int \sqrt{a^2 - t^2} dt = \frac{t}{2}\sqrt{a^2 - t^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{t}{a}ight)$,where $a = \frac{1}{2}$:$I = 2 \left[ \frac{t}{2}\sqrt{\frac{1}{4} - t^2} + \frac{1}{8}\sin^{-1}(2t) ight]_{0}^{1/2}$.$I = 2 \left[ (0 + \frac{1}{8}\sin^{-1}(1)) - (0 + 0) ight] = 2 	imes \frac{1}{8} 	imes \frac{\pi}{2} = \frac{\pi}{8}$.

The value of $\int_{3}^{4} \sqrt{(4-x)(x-3)} d x$ is

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