Evaluate
-43896
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\int -800V+1160\mathrm{d}V
Evaluate the indefinite integral first.
\int -800V\mathrm{d}V+\int 1160\mathrm{d}V
Integrate the sum term by term.
-800\int V\mathrm{d}V+\int 1160\mathrm{d}V
Factor out the constant in each of the terms.
-400V^{2}+\int 1160\mathrm{d}V
Since \int V^{k}\mathrm{d}V=\frac{V^{k+1}}{k+1} for k\neq -1, replace \int V\mathrm{d}V with \frac{V^{2}}{2}. Multiply -800 times \frac{V^{2}}{2}.
-400V^{2}+1160V
Find the integral of 1160 using the table of common integrals rule \int a\mathrm{d}V=aV.
-400\times 12^{2}+1160\times 12-\left(-400\times 0.2^{2}+1160\times 0.2\right)
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.
-43896
Simplify.
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