Solve ∫ (from -2 to 2) of (4-x^2)-(8x-2x^2) wrt x

$$\int_{ -2 }^{ 2 } (4- { x }^{ 2 } )-(8x-2 { x }^{ 2 } ) d x$$

Evaluate

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

Solution Steps

To find the opposite of 8x-2x^{2}, find the opposite of each term.

$$\int _{-2}^{2}4-x^{2}-8x+2x^{2}\mathrm{d}x$$

Combine -x^{2} and 2x^{2} to get x^{2}.

$$\int _{-2}^{2}4+x^{2}-8x\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4+x^{2}-8x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4\mathrm{d}x+\int x^{2}\mathrm{d}x+\int -8x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 4\mathrm{d}x+\int x^{2}\mathrm{d}x-8\int x\mathrm{d}x$$

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

$$4x+\int x^{2}\mathrm{d}x-8\int x\mathrm{d}x$$

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}.

$$4x+\frac{x^{3}}{3}-8\int x\mathrm{d}x$$

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 -8 times \frac{x^{2}}{2}.

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

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.

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

Simplify.

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