Solve ∫ (from - 2 to 2) of x^4-3x^2 wrt x

$$\int _ { - 2 } ^ { 2 } x ^ { 4 } - 3 x ^ { 2 } d x$$

Evaluate

$$-\frac{16}{5}=-3.2$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{4}-3x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int -3x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{4}\mathrm{d}x-3\int x^{2}\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}-3\int x^{2}\mathrm{d}x$$

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

$$\frac{x^{5}}{5}-x^{3}$$

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.

$$\frac{2^{5}}{5}-2^{3}-\left(\frac{\left(-2\right)^{5}}{5}-\left(-2\right)^{3}\right)$$

Simplify.

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