The given curve is $y^2 = x$, which represents a parabola. We need to find the area enclosed by the curve between $x = 0$ and $x = 1$. First, solve for $y$: \[ y = \sqrt{x}. \] To find the area, we integrate the function from $x = 0$ to $x = 1$: \[ \text{Area} = \int_0^1 \sqrt{x} \, dx. \] The integral of $\sqrt{x}$ is: \[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2}. \] Evaluating this from 0 to 1: \[ \text{Area} = \left[\frac{2}{3} x^{3/2} \right]_0^1 = \frac{2}{3} (1) - 0 = \frac{2}{3}. \] Thus, the area is 2 square units.