Given curves

$y_1=4-\frac{x^2}{4}, \qquad y_2=\frac{x-4}{2}$

We need the area enclosed between these two curves.

Find points of intersection

Set $4-\frac{x^2}{4}=\frac{x-4}{2}$

Multiply by $4$: $16-x^2=2x-8$

Rearranging, $x^2+2x-24=0$

Factorize: $(x+6)(x-4)=0$

So the curves intersect at $x=-6 \quad \text{and} \quad x=4.$

Determine which curve is above

Consider $y_1-y_2=\left(4-\frac{x^2}{4}\right)-\frac{x-4}{2}$

Simplify: $y_1-y_2=4-\frac{x^2}{4}-\frac{x}{2}+2=6-\frac{x}{2}-\frac{x^2}{4}.$

At $x=0$, $y_1-y_2=6>0,$ so $y_1$ is above $y_2$ on $[-6,4]$.

Set up the area integral

$\alpha=\int_{-6}^{4}\left[\left(4-\frac{x^2}{4}\right)-\frac{x-4}{2}\right]dx$

$\alpha=\int_{-6}^{4}\left(6-\frac{x}{2}-\frac{x^2}{4}\right)dx$

Integrate

$\int \left(6-\frac{x}{2}-\frac{x^2}{4}\right)dx=6x-\frac{x^2}{4}-\frac{x^3}{12}$

Therefore, $\alpha=\left[6x-\frac{x^2}{4}-\frac{x^3}{12}\right]_{-6}^{4}.$

Evaluate at limits

At $x=4$: $6(4)-\frac{4^2}{4}-\frac{4^3}{12}=24-4-\frac{64}{12}=20-\frac{16}{3}=\frac{44}{3}.$

At $x=-6$: $6(-6)-\frac{(-6)^2}{4}-\frac{(-6)^3}{12}=-36-9+18=-27.$

Hence, $\alpha=\frac{44}{3}-(-27)=\frac{44}{3}+\frac{81}{3}=\frac{125}{3}.$

Compute $6\alpha$

$6\alpha=6\cdot \frac{125}{3}=250.$

Match with options

$6\alpha=250$

So the correct option is B.