Solve ∫ (from -1 to 2) of v^2-vdv

$$\int_{ -1 }^{ 2 } { v }^{ 2 } -v d v$$

Evaluate

$$\frac{3}{2}=1.5$$

Solution Steps

Evaluate the indefinite integral first.

$$\int v^{2}-v\mathrm{d}v$$

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

$$\int v^{2}\mathrm{d}v-\int v\mathrm{d}v$$

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}.

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

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 -1 times \frac{v^{2}}{2}.

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

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.

$$\frac{2^{3}}{3}-\frac{2^{2}}{2}-\left(\frac{\left(-1\right)^{3}}{3}-\frac{\left(-1\right)^{2}}{2}\right)$$

Simplify.

$$\frac{3}{2}$$