Solve ∫ (from - 4 to 4) of (-1/8x^4+2x^2) wrt x

$$\int _ { - 4 } ^ { 4 } ( - \frac { 1 } { 8 } x ^ { 4 } + 2 x ^ { 2 } ) d x$$

Evaluate

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

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply -\frac{1}{8} times \frac{x^{5}}{5}.

$$-\frac{x^{5}}{40}+2\int x^{2}\mathrm{d}x$$

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 2 times \frac{x^{3}}{3}.

$$-\frac{x^{5}}{40}+\frac{2x^{3}}{3}$$

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.

$$-\frac{4^{5}}{40}+\frac{2}{3}\times 4^{3}-\left(-\frac{\left(-4\right)^{5}}{40}+\frac{2}{3}\left(-4\right)^{3}\right)$$

Simplify.

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