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

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

Evaluate

$$\frac{512}{15}\approx 34.133333333$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-4\right)^{2}.

$$\int _{-2}^{2}\left(x^{2}\right)^{2}-8x^{2}+16\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

$$\frac{x^{5}}{5}-8\int x^{2}\mathrm{d}x+\int 16\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}-\frac{8x^{3}}{3}+\int 16\mathrm{d}x$$

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

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

Simplify.

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

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.

$$\frac{512}{15}$$