Evaluate
\frac{1645}{4}=411.25
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\int x^{3}+5x\mathrm{d}x
Evaluate the indefinite integral first.
\int x^{3}\mathrm{d}x+\int 5x\mathrm{d}x
Integrate the sum term by term.
\int x^{3}\mathrm{d}x+5\int x\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{4}}{4}+5\int x\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.
\frac{x^{4}}{4}+\frac{5x^{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}. Multiply 5 times \frac{x^{2}}{2}.
\frac{6^{4}}{4}+\frac{5}{2}\times 6^{2}-\left(\frac{1^{4}}{4}+\frac{5}{2}\times 1^{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{1645}{4}
Simplify.
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