Step-by-Step Solution

The area of the region bounded by the curve \(y^2 = x\) between \(x = 0\) and \(x = 1\) can be found by integrating the function \(y = \sqrt{x}\) with respect to \(x\) from 0 to 1 and then multiplying by 2 because the curve is symmetric about the x-axis.

Step 1: Find the integral of \(y = \sqrt{x}\) with respect to \(x\).

\(\int \sqrt{x} \, dx = \int x^{\frac{1}{2}} \, dx = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{2}{3}x^{\frac{3}{2}} + C\)

Step 2: Evaluate the definite integral from \(x = 0\) to \(x = 1\).

\(\int_{0}^{1} \sqrt{x} \, dx = \left[ \frac{2}{3}x^{\frac{3}{2}} \right]_{0}^{1} = \frac{2}{3}(1^{\frac{3}{2}}) - \frac{2}{3}(0^{\frac{3}{2}}) = \frac{2}{3}(1) - \frac{2}{3}(0) = \frac{2}{3}\)

Step 3: Multiply the result by 2 to account for the area below the x-axis (due to the symmetry of \(y^2 = x\)).

Area = \(2 \times \frac{2}{3} = \frac{4}{3}\)