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\int \sqrt{x^{3}}\mathrm{d}x
Evaluate the indefinite integral first.
\frac{2x^{\frac{5}{2}}}{5}
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{3}{2}}\mathrm{d}x with \frac{2x^{\frac{5}{2}}}{5}.
\frac{2}{5}\times 4^{\frac{5}{2}}-\frac{2}{5}\times 0^{\frac{5}{2}}
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{64}{5}
Simplify.