Evaluate
\frac{16}{3}\approx 5.333333333
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\int x^{3}-x^{2}-6x\mathrm{d}x
Evaluate the indefinite integral first.
\int x^{3}\mathrm{d}x+\int -x^{2}\mathrm{d}x+\int -6x\mathrm{d}x
Integrate the sum term by term.
\int x^{3}\mathrm{d}x-\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{4}}{4}-\int x^{2}\mathrm{d}x-6\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}-6\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 -1 times \frac{x^{3}}{3}.
\frac{x^{4}}{4}-\frac{x^{3}}{3}-3x^{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 -6 times \frac{x^{2}}{2}.
\frac{0^{4}}{4}-\frac{0^{3}}{3}-3\times 0^{2}-\left(\frac{\left(-2\right)^{4}}{4}-\frac{\left(-2\right)^{3}}{3}-3\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{16}{3}
Simplify.
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