Solve ∫ (from - 4 to 4) of 16-x^2

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

Evaluate

$$\frac{256}{3}\approx 85.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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 -1 times \frac{x^{3}}{3}.

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

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.

$$16\times 4-\frac{4^{3}}{3}-\left(16\left(-4\right)-\frac{\left(-4\right)^{3}}{3}\right)$$

Simplify.

$$\frac{256}{3}$$