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(A) To find the area $\alpha$ bounded by the curves $y=4-\frac{x^2}{4}$ and $y=\frac{x-4}{2}$,we first find the points of intersection by setting the

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$250$

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(A) To find the area $\alpha$ bounded by the curves $y=4-\frac{x^2}{4}$ and $y=\frac{x-4}{2}$,we first find the points of intersection by setting the equations equal:$4-\frac{x^2}{4} = \frac{x-4}{2}$$16-x^2 = 2x-8$$x^2+2x-24 = 0$$(x+6)(x-4) = 0$So,the intersection points are $x=-6$ and $x=4$.The area $\alpha$ is given by the integral:$\alpha = \int_{-6}^4 \left\{ \left(4-\frac{x^2}{4}ight) - \left(\frac{x-4}{2}ight) ight\} dx$$\alpha = \int_{-6}^4 \left( 4 - \frac{x^2}{4} - \frac{x}{2} + 2 ight) dx = \int_{-6}^4 \left( 6 - \frac{x}{2} - \frac{x^2}{4} ight) dx$$\alpha = \left[ 6x - \frac{x^2}{4} - \frac{x^3}{12} ight]_{-6}^4$$\alpha = \left( 6(4) - \frac{16}{4} - \frac{64}{12} ight) - \left( 6(-6) - \frac{36}{4} - \frac{-216}{12} ight)$$\alpha = \left( 24 - 4 - \frac{16}{3} ight) - \left( -36 - 9 + 18 ight)$$\alpha = \left( 20 - \frac{16}{3} ight) - (-27) = \frac{44}{3} + 27 = \frac{44+81}{3} = \frac{125}{3}$Therefore,$6 \alpha = 6 	imes \frac{125}{3} = 2 	imes 125 = 250$.

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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