If the area of the region bounded by the curves $y=4-\frac{x^2}{4}$ and $y=\frac{x-4}{2}$ is equal to $\alpha$, then $6 \alpha$ equals

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Area }=\int_{-6}^4\left\{\left(4-\frac{x^2}{4}ight)-\left(\frac{x-4}{2}ight)ight\} d x \ & =\int_{-6}^4\left\{-\frac{x^2}{4}-\frac{x-6}{2}ight\} d x \ & \alpha=-\frac{x^3}{12}-\frac{x^2}{4}+\left.6 xight|_{-6} ^4=\frac{125}{3} \ & 	herefore 6 \alpha=250\end{aligned}$

If the area of the region bounded by the curves \(y=4-\frac{x^2}{4}\) and \(y=\frac{x-4}{2}\) is equal to…

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