Answer
$$\int y^2(4-y^3)^{2/3}dy=-\frac{1}{5}\sqrt[3]{(4-y^3)^5}+C$$
Work Step by Step
$$A=\int y^2(4-y^3)^{2/3}dy$$
Let $u=4-y^3$
Then $du=-3y^2dy$. So $y^2dy=-\frac{1}{3}du$
Substitute into $A$, we have $$A=-\frac{1}{3}\int u^{2/3}du$$ $$A=-\frac{1}{3}\frac{u^{5/3}}{\frac{5}{3}}+C$$ $$A=-\frac{\sqrt[3]{u^5}}{5}+C$$ $$A=-\frac{1}{5}\sqrt[3]{(4-y^3)^5}+C$$
