Evaluate
\frac{3}{5}=0.6
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\int \left(\sqrt[3]{x}\right)^{2}\mathrm{d}x
Evaluate the indefinite integral first.
\frac{3x^{\frac{5}{3}}}{5}
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{2}{3}}\mathrm{d}x with \frac{3x^{\frac{5}{3}}}{5}.
\frac{3}{5}\times 1^{\frac{5}{3}}-\frac{3}{5}\times 0^{\frac{5}{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{3}{5}
Simplify.
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