Since $\int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1}$ for $k\neq -1$, replace $\int x\mathrm{d}x$ with $\frac{x^{2}}{2}$. Multiply $7$ times $\frac{x^{2}}{2}$.
Find the integral of $-8$ using the table of common integrals rule $\int a\mathrm{d}x=ax$.
$$\frac{x^{3}}{3}+\frac{7x^{2}}{2}-8x$$
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.
