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\int 3+x\mathrm{d}x
Evaluate the indefinite integral first.
\int 3\mathrm{d}x+\int x\mathrm{d}x
Integrate the sum term by term.
3x+\int x\mathrm{d}x
Find the integral of 3 using the table of common integrals rule \int a\mathrm{d}x=ax.
3x+\frac{x^{2}}{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}.
3\times 4+\frac{4^{2}}{2}-\left(3\times 3+\frac{3^{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{13}{2}
Simplify.