etermine the area of the region enlosed by \( y=\sqrt{x} \) and \( y=x^{2} \).

To determine the area of the region enclosed by the curves \( y=\sqrt{x} \) and \( y=x^{2} \), we need to find the points of intersection of these two curves and then calculate the area between them. 1. Find the points of intersection: To find the points of intersection, we need to set the two equations equal to each other and solve for \( x \): \[ \sqrt{x} = x^{2} \] 2. Solve for \( x \): \[ x^{2} - \sqrt{x} = 0 \] \[ x(\sqrt{x} - x) = 0 \] This equation has two solutions: \( x = 0 \) and \( x = 1 \). 3. Calculate the area between the curves: The area between the curves can be calculated by integrating the difference between the two functions over the interval where they intersect. The interval of interest is from \( x = 0 \) to \( x = 1 \). The area \( A \) can be calculated as: \[ A = \int_{0}^{1} (\sqrt{x} - x^{2}) dx \] Let's calculate the integral to find the area. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{0}^{1} \sqrt{x}-x^{2} dx\) - step1: Evaluate the power: \(\int_{0}^{1} x^{\frac{1}{2}}-x^{2} dx\) - step2: Evaluate the integral: \(\int x^{\frac{1}{2}}-x^{2} dx\) - step3: Use properties of integrals: \(\int x^{\frac{1}{2}} dx-\int x^{2} dx\) - step4: Evaluate the integral: \(\frac{2}{3}x^{\frac{3}{2}}-\frac{1}{3}x^{3}\) - step5: Return the limits: \(\left(\frac{2}{3}x^{\frac{3}{2}}-\frac{1}{3}x^{3}\right)\bigg |_{0}^{1}\) - step6: Calculate the value: \(\frac{1}{3}\) The area of the region enclosed by the curves \( y=\sqrt{x} \) and \( y=x^{2} \) is \( \frac{1}{3} \) or approximately 0.333.