What is the area of the region in the first quadrant bounded by $y^{2} = x$ and the line $x = 9$?

Explanation

1. 01

   In the first quadrant $y = \sqrt{x}$: $\int_0^{9} \sqrt{x}\,dx = \frac{2}{3}x^{3/2}\Big|_0^{9}$.

2. 02

   $\frac{2}{3} \times 27 = 18$ square units.