Solve ∫ (from - 2 to 2) of 4x^3 wrt x

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

Evaluate

$$0$$

Solution Steps

Evaluate the indefinite integral first.

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

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 4 times \frac{x^{4}}{4}.

$$x^{4}$$

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.

$$2^{4}-\left(-2\right)^{4}$$

Simplify.

$$0$$