The area enclosed by the curves \(y=x^3\) and \(y=\sqrt{x}\) is:

\(\mathrm{The}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{these}\mathrm{curves}\mathrm{will}\mathrm{be}\\ \sqrt{\mathrm{x}}={\mathrm{x}}^{3}\\ \Rightarrow \mathrm{x}=0,1\)

Then, the area is given by

\(A={\int }_{0}^{1}\left(\sqrt{x}-{x}^{3}\right)dx\\ ={\left[\frac{2}{3}{x}^{\frac{3}{2}}-\frac{{x}^{4}}{4}\right]}_{0}^{1}\\ =\frac{2}{3}(1)-\frac{1}{4}\\ =\frac{5}{12}\)