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

$$\int _ { - 2 } ^ { 2 } ( 16 - x ^ { 4 } ) d x$$

Evaluate

$$\frac{256}{5}=51.2$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}.

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

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.

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

Simplify.

$$\frac{256}{5}$$