$\int_{0}^{2}{({{x}^{3}}+{{x}^{2}})dx}=\left[ \tfrac{{{x}^{4}}}{4}+\tfrac{{{x}^{3}}}{3} \right]_{0}^{2}=(\tfrac{16}{4}+\tfrac{8}{3})-(0+0)=\tfrac{20}{3}=6\tfrac{2}{3}$