Solve ∫ (from -1 to 1) of x^3+2x^2-x-2 wrt x

$$\int_{ -1 }^{ 1 } { x }^{ 3 } +2 { x }^{ 2 } -x-2 d x$$

Evaluate

$$-\frac{8}{3}\approx -2.666666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

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

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 -1 times \frac{x^{2}}{2}.

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

Find the integral of -2 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$\frac{1^{4}}{4}+\frac{2}{3}\times 1^{3}-\frac{1^{2}}{2}-2-\left(\frac{\left(-1\right)^{4}}{4}+\frac{2}{3}\left(-1\right)^{3}-\frac{\left(-1\right)^{2}}{2}-2\left(-1\right)\right)$$

Simplify.

$$-\frac{8}{3}$$