Answer
\[\frac{5}{12}\]
Work Step by Step
\[\begin{align}
& \int_{0}^{1}{\int_{{{y}^{2}}}^{\sqrt[3]{y}}{dx}dy} \\
& \text{Find the integral} \\
& \int_{0}^{1}{\int_{{{y}^{2}}}^{\sqrt[3]{y}}{dx}dy}=\int_{0}^{1}{\left( \sqrt[3]{y}-{{y}^{2}} \right)dy} \\
& =\left[ \frac{3}{4}{{y}^{4/3}}-\frac{1}{3}{{y}^{3}} \right]_{0}^{1} \\
& =\left[ \frac{3}{4}{{\left( 1 \right)}^{4/3}}-\frac{1}{3}{{\left( 1 \right)}^{3}} \right]-\left[ \frac{3}{4}{{\left( 0 \right)}^{4/3}}-\frac{1}{3}{{\left( 0 \right)}^{3}} \right] \\
& =\frac{3}{4}-\frac{1}{3} \\
& =\frac{5}{12} \\
& \text{Using the graph to switch the order of integration} \\
& \int_{0}^{1}{\int_{{{y}^{2}}}^{\sqrt[3]{y}}{dx}dy}=\int_{0}^{1}{\int_{{{x}^{3}}}^{\sqrt{x}}{dy}dx} \\
& =\int_{0}^{1}{\left( \sqrt{x}-{{x}^{3}} \right)dx} \\
& =\left[ \frac{2}{3}{{x}^{3/2}}-\frac{1}{4}{{x}^{4}} \right]_{0}^{1} \\
& =\left[ \frac{2}{3}{{\left( 1 \right)}^{3/2}}-\frac{1}{4}{{\left( 1 \right)}^{4}} \right]-\left[ \frac{2}{3}{{\left( 0 \right)}^{3/2}}-\frac{1}{4}{{\left( 0 \right)}^{4}} \right] \\
& =\frac{2}{3}-\frac{1}{4} \\
& =\frac{5}{12} \\
\end{align}\]
