Solve ∫ (from - 2 to 2) of (8-2x^2)

$$\int _ { - 2 } ^ { 2 } ( 8 - 2 x ^ { 2 } )$$

Evaluate

$$\frac{64}{3}\approx 21.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 8-2x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8\mathrm{d}x+\int -2x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 8\mathrm{d}x-2\int x^{2}\mathrm{d}x$$

Find the integral of 8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$8x-2\int x^{2}\mathrm{d}x$$

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}.

$$8x-\frac{2x^{3}}{3}$$

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.

$$8\times 2-\frac{2}{3}\times 2^{3}-\left(8\left(-2\right)-\frac{2}{3}\left(-2\right)^{3}\right)$$

Simplify.

$$\frac{64}{3}$$