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