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\int 3x+1\mathrm{d}x
Evaluate the indefinite integral first.
\int 3x\mathrm{d}x+\int 1\mathrm{d}x
Integrate the sum term by term.
3\int x\mathrm{d}x+\int 1\mathrm{d}x
Factor out the constant in each of the terms.
\frac{3x^{2}}{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 3 times \frac{x^{2}}{2}.
\frac{3x^{2}}{2}+x
Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.
\frac{3}{2}\times 2^{2}+2-\left(\frac{3}{2}\times 0^{2}+0\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.
8
Simplify.
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