Solve ∫ (from 2 to 5) of (-3v+4)dv

$$\int _ { 2 } ^ { 5 } ( - 3 v + 4 ) d v$$

Evaluate

$$-\frac{39}{2}=-19.5$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -3v+4\mathrm{d}v$$

Integrate the sum term by term.

$$\int -3v\mathrm{d}v+\int 4\mathrm{d}v$$

Factor out the constant in each of the terms.

$$-3\int v\mathrm{d}v+\int 4\mathrm{d}v$$

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 -3 times \frac{v^{2}}{2}.

$$-\frac{3v^{2}}{2}+\int 4\mathrm{d}v$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}v=av.

$$-\frac{3v^{2}}{2}+4v$$

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.

$$-\frac{3}{2}\times 5^{2}+4\times 5-\left(-\frac{3}{2}\times 2^{2}+4\times 2\right)$$

Simplify.

$$-\frac{39}{2}$$