Solve ∫ (from -2 to 0) of (-4x-2x^2) wrt x

$$\int_{ -2 }^{ 0 } (-4x-2 { x }^{ 2 } ) d x$$

Evaluate

$$\frac{8}{3}\approx 2.666666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -4x-2x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -4x\mathrm{d}x+\int -2x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-4\int x\mathrm{d}x-2\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -4 times \frac{x^{2}}{2}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -2 times \frac{x^{3}}{3}.

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

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.

$$-2\times 0^{2}-\frac{2}{3}\times 0^{3}-\left(-2\left(-2\right)^{2}-\frac{2}{3}\left(-2\right)^{3}\right)$$

Simplify.

$$\frac{8}{3}$$