Solve ∫ (from -4 to 4) of {left(sqrt{16-x^2}right)}^2 wrt x

$$\int_{ -4 }^{ 4 } { \left( \sqrt{ 16- { x }^{ 2 } } \right) }^{ 2 } d x$$

Evaluate

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

Solution Steps

Calculate \sqrt{16-x^{2}} to the power of 2 and get 16-x^{2}.

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

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.

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