Evaluate
\frac{1}{3}\approx 0.333333333
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\int x^{2}\mathrm{d}x
Evaluate the indefinite integral first.
\frac{x^{3}}{3}
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}.
\frac{1^{3}}{3}-\frac{0^{3}}{3}
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{1}{3}
Simplify.
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