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\int x^{2}+9x+10\mathrm{d}x
Evaluate the indefinite integral first.
\int x^{2}\mathrm{d}x+\int 9x\mathrm{d}x+\int 10\mathrm{d}x
Integrate the sum term by term.
\int x^{2}\mathrm{d}x+9\int x\mathrm{d}x+\int 10\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{3}}{3}+9\int x\mathrm{d}x+\int 10\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^{3}}{3}+\frac{9x^{2}}{2}+\int 10\mathrm{d}x
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 9 times \frac{x^{2}}{2}.
\frac{x^{3}}{3}+\frac{9x^{2}}{2}+10x
Find the integral of 10 using the table of common integrals rule \int a\mathrm{d}x=ax.
\frac{\left(-1\right)^{3}}{3}+\frac{9}{2}\left(-1\right)^{2}+10\left(-1\right)-\left(\frac{\left(-4\right)^{3}}{3}+\frac{9}{2}\left(-4\right)^{2}+10\left(-4\right)\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{33}{2}
Simplify.