Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x^{2}\mathrm{d}x$ with $\frac{x^{3}}{3}$.
$$\frac{x^{3}}{3}+\int 4\mathrm{d}x$$
Find the integral of $4$ using the table of common integrals rule $\int a\mathrm{d}x=ax$.
$$\frac{x^{3}}{3}+4x$$
The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.
