Evaluate
\frac{490}{3}\approx 163.333333333
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\int 4x^{2}-2x+1\mathrm{d}x
Evaluate the indefinite integral first.
\int 4x^{2}\mathrm{d}x+\int -2x\mathrm{d}x+\int 1\mathrm{d}x
Integrate the sum term by term.
4\int x^{2}\mathrm{d}x-2\int x\mathrm{d}x+\int 1\mathrm{d}x
Factor out the constant in each of the terms.
\frac{4x^{3}}{3}-2\int x\mathrm{d}x+\int 1\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 4 times \frac{x^{3}}{3}.
\frac{4x^{3}}{3}-x^{2}+\int 1\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 -2 times \frac{x^{2}}{2}.
\frac{4x^{3}}{3}-x^{2}+x
Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.
\frac{4}{3}\times 5^{3}-5^{2}+5-\left(\frac{4}{3}\left(-2\right)^{3}-\left(-2\right)^{2}-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{490}{3}
Simplify.
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