Solve ∫ (from 02 to 12) of (-800V+1160)dV

$$\int _ { 02 } ^ { 12 } ( - 800 V + 1160 ) d V$$

Evaluate

$$-43680$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -800V+1160\mathrm{d}V$$

Integrate the sum term by term.

$$\int -800V\mathrm{d}V+\int 1160\mathrm{d}V$$

Factor out the constant in each of the terms.

$$-800\int V\mathrm{d}V+\int 1160\mathrm{d}V$$

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}.

$$-400V^{2}+\int 1160\mathrm{d}V$$

Find the integral of 1160 using the table of common integrals rule \int a\mathrm{d}V=aV.

$$-400V^{2}+1160V$$

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.

$$-400\times 12^{2}+1160\times 12-\left(-400\times \left(0\times 2\right)^{2}+1160\times 0\times 2\right)$$

Simplify.

$$-43680$$