Evaluate
2\sqrt{3}\approx 3.464101615
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\int \sqrt{3}x\mathrm{d}x
Evaluate the indefinite integral first.
\sqrt{3}\int x\mathrm{d}x
Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.
\sqrt{3}\times \frac{x^{2}}{2}
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}.
\frac{\sqrt{3}x^{2}}{2}
Simplify.
\frac{1}{2}\times 3^{\frac{1}{2}}\times 2^{2}-\frac{1}{2}\times 3^{\frac{1}{2}}\times 0^{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.
2\sqrt{3}
Simplify.
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