Let the area of the region (x,y) : 2y x^23+3, y+|x|3, y |x-1| be A.

Question

Let the area of the region $\left(x,y\right) : 2y \le \dfrac{x^2}{3}+3,\ y+|x|\le3,\ y \ge |x-1|$ be $A$. Then $6A$ is equal to:

Accepted Answer

The region is bounded above by $y=3-|x|$ and below by $y=|x-1|$, further capped by $y = \dfrac{x^2}{6}+\dfrac32$. Carrying out the piecewise integration (as in the source) gives:

$$A = \frac{7}{3} \implies 6A = 14$$

Solutions · 1