Solve ∫ (from -2 to 2) of (16-8x^2-x^4) wrt x

$$\int_{ -2 }^{ 2 } (16-8 { x }^{ 2 } - { x }^{ 4 } ) d x$$

Evaluate

$$\frac{128}{15}\approx 8.533333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 16-8x^{2}-x^{4}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 16\mathrm{d}x+\int -8x^{2}\mathrm{d}x+\int -x^{4}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 16\mathrm{d}x-8\int x^{2}\mathrm{d}x-\int x^{4}\mathrm{d}x$$

Find the integral of 16 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$16x-8\int x^{2}\mathrm{d}x-\int x^{4}\mathrm{d}x$$

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 -8 times \frac{x^{3}}{3}.

$$16x-\frac{8x^{3}}{3}-\int x^{4}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply -1 times \frac{x^{5}}{5}.

$$16x-\frac{8x^{3}}{3}-\frac{x^{5}}{5}$$

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.

$$16\times 2-\frac{8}{3}\times 2^{3}-\frac{2^{5}}{5}-\left(16\left(-2\right)-\frac{8}{3}\left(-2\right)^{3}-\frac{\left(-2\right)^{5}}{5}\right)$$

Simplify.

$$\frac{128}{15}$$