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\int _{0}^{8}\left(-16-8x+\frac{20}{3}\times 4+x\right)\times 1\mathrm{d}x
Reduce the fraction \frac{80}{12} to lowest terms by extracting and canceling out 4.
\int _{0}^{8}\left(-16-8x+\frac{20\times 4}{3}+x\right)\times 1\mathrm{d}x
Express \frac{20}{3}\times 4 as a single fraction.
\int _{0}^{8}\left(-16-8x+\frac{80}{3}+x\right)\times 1\mathrm{d}x
Multiply 20 and 4 to get 80.
\int _{0}^{8}\left(-\frac{48}{3}-8x+\frac{80}{3}+x\right)\times 1\mathrm{d}x
Convert -16 to fraction -\frac{48}{3}.
\int _{0}^{8}\left(\frac{-48+80}{3}-8x+x\right)\times 1\mathrm{d}x
Since -\frac{48}{3} and \frac{80}{3} have the same denominator, add them by adding their numerators.
\int _{0}^{8}\left(\frac{32}{3}-8x+x\right)\times 1\mathrm{d}x
Add -48 and 80 to get 32.
\int _{0}^{8}\left(\frac{32}{3}-7x\right)\times 1\mathrm{d}x
Combine -8x and x to get -7x.
\int _{0}^{8}\frac{32}{3}-7x\mathrm{d}x
Use the distributive property to multiply \frac{32}{3}-7x by 1.
\int \frac{32}{3}-7x\mathrm{d}x
Evaluate the indefinite integral first.
\int \frac{32}{3}\mathrm{d}x+\int -7x\mathrm{d}x
Integrate the sum term by term.
\int \frac{32}{3}\mathrm{d}x-7\int x\mathrm{d}x
Factor out the constant in each of the terms.
\frac{32x}{3}-7\int x\mathrm{d}x
Find the integral of \frac{32}{3} using the table of common integrals rule \int a\mathrm{d}x=ax.
\frac{32x}{3}-\frac{7x^{2}}{2}
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}.
\frac{32}{3}\times 8-\frac{7}{2}\times 8^{2}-\left(\frac{32}{3}\times 0-\frac{7}{2}\times 0^{2}\right)
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.
-\frac{416}{3}
Simplify.