Since $\int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1}$ for $k\neq -1$, replace $\int t^{2}\mathrm{d}t$ with $\frac{t^{3}}{3}$.
$$\frac{t^{3}}{3}+\int 1\mathrm{d}t$$
Find the integral of $1$ using the table of common integrals rule $\int a\mathrm{d}t=at$.
$$\frac{t^{3}}{3}+t$$
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.
