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\int \frac{x^{4}}{4}-x^{3}-3x^{2}+8x\mathrm{d}x
Evaluate the indefinite integral first.
\int \frac{x^{4}}{4}\mathrm{d}x+\int -x^{3}\mathrm{d}x+\int -3x^{2}\mathrm{d}x+\int 8x\mathrm{d}x
Integrate the sum term by term.
\frac{\int x^{4}\mathrm{d}x}{4}-\int x^{3}\mathrm{d}x-3\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{5}}{20}-\int x^{3}\mathrm{d}x-3\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply \frac{1}{4} times \frac{x^{5}}{5}.
\frac{x^{5}}{20}-\frac{x^{4}}{4}-3\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -1 times \frac{x^{4}}{4}.
\frac{x^{5}}{20}-\frac{x^{4}}{4}-x^{3}+8\int x\mathrm{d}x
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}. Multiply -3 times \frac{x^{3}}{3}.
\frac{x^{5}}{20}-\frac{x^{4}}{4}-x^{3}+4x^{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 8 times \frac{x^{2}}{2}.
\frac{1^{5}}{20}-\frac{1^{4}}{4}-1^{3}+4\times 1^{2}-\left(\frac{\left(-2\right)^{5}}{20}-\frac{\left(-2\right)^{4}}{4}-\left(-2\right)^{3}+4\left(-2\right)^{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{78}{5}
Simplify.