Solve ∫ (from -2 to -1) of 2x^2-2 wrt x

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

Evaluate

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

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

Find the integral of -2 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

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