Evaluate
\frac{64}{3}\approx 21.333333333
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\int _{-1}^{3}\left(x^{2}-x\right)\left(x+2\right)\mathrm{d}x
Use the distributive property to multiply x by x-1.
\int _{-1}^{3}x^{3}+2x^{2}-x^{2}-2x\mathrm{d}x
Apply the distributive property by multiplying each term of x^{2}-x by each term of x+2.
\int _{-1}^{3}x^{3}+x^{2}-2x\mathrm{d}x
Combine 2x^{2} and -x^{2} to get x^{2}.
\int x^{3}+x^{2}-2x\mathrm{d}x
Evaluate the indefinite integral first.
\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x+\int -2x\mathrm{d}x
Integrate the sum term by term.
\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x-2\int x\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{4}}{4}+\int x^{2}\mathrm{d}x-2\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}.
\frac{x^{4}}{4}+\frac{x^{3}}{3}-2\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}.
\frac{x^{4}}{4}+\frac{x^{3}}{3}-x^{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 -2 times \frac{x^{2}}{2}.
\frac{3^{4}}{4}+\frac{3^{3}}{3}-3^{2}-\left(\frac{\left(-1\right)^{4}}{4}+\frac{\left(-1\right)^{3}}{3}-\left(-1\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{64}{3}
Simplify.
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