Solve ∫ (from -2 to 0) of (-x^3+4x) wrt x

$$\int_{ -2 }^{ 0 } (- { x }^{ 3 } +4x) d x$$

Evaluate

$$-4$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 -1 times \frac{x^{4}}{4}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 4 times \frac{x^{2}}{2}.

$$-\frac{x^{4}}{4}+2x^{2}$$

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.

$$-\frac{0^{4}}{4}+2\times 0^{2}-\left(-\frac{\left(-2\right)^{4}}{4}+2\left(-2\right)^{2}\right)$$

Simplify.

$$-4$$