Evaluate
\frac{15}{4}=3.75
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\int _{1}^{2}x^{3}\mathrm{d}x
To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.
\int x^{3}\mathrm{d}x
Evaluate the indefinite integral first.
\frac{x^{4}}{4}
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{2^{4}}{4}-\frac{1^{4}}{4}
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{15}{4}
Simplify.
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