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