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\int \sqrt{x}\mathrm{d}x
Evaluate the indefinite integral first.
\frac{2x^{\frac{3}{2}}}{3}
Rewrite \sqrt{x} as x^{\frac{1}{2}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{2}}\mathrm{d}x with \frac{x^{\frac{3}{2}}}{\frac{3}{2}}. Simplify.
\frac{2}{3}\times 8^{\frac{3}{2}}-\frac{2}{3}\times 1^{\frac{3}{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{32\sqrt{2}-2}{3}
Simplify.