What is the area of the region bounded by $x^{2} = 4y$ and the line $y = 1$?

Explanation

1. 01

   They meet at $x = \pm 2$; the line lies above $y = \frac{x^{2}}{4}$: $\int_{-2}^{2} \left(1 - \frac{x^{2}}{4}\right)dx$.

2. 02

   $4 - \frac{16}{12} = \frac{8}{3}$ square units.