Solve ∫ (from - 2 to 0) of (x^3-x^2-6x)

$$\int _ { - 2 } ^ { 0 } ( x ^ { 3 } - x ^ { 2 } - 6 x )$$

Evaluate

$$\frac{16}{3}\approx 5.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{3}-x^{2}-6x\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -x^{2}\mathrm{d}x+\int -6x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x-\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}-\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -1 times \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}-\frac{x^{3}}{3}-6\int x\mathrm{d}x$$

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 -6 times \frac{x^{2}}{2}.

$$\frac{x^{4}}{4}-\frac{x^{3}}{3}-3x^{2}$$

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.

$$\frac{0^{4}}{4}-\frac{0^{3}}{3}-3\times 0^{2}-\left(\frac{\left(-2\right)^{4}}{4}-\frac{\left(-2\right)^{3}}{3}-3\left(-2\right)^{2}\right)$$

Simplify.

$$\frac{16}{3}$$