Solve ∫ (from - 1 to 0) of (x+1)(x^2-2) wrt x

$$\int _ { - 1 } ^ { 0 } ( x + 1 ) ( x ^ { 2 } - 2 ) d x$$

Evaluate

$$-\frac{11}{12}\approx -0.916666667$$

Solution Steps

Use the distributive property to multiply x+1 by x^{2}-2.

$$\int _{-1}^{0}x^{3}-2x+x^{2}-2\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

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}.

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

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

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

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.

$$\frac{0^{4}}{4}-0^{2}+\frac{0^{3}}{3}-2\times 0-\left(\frac{\left(-1\right)^{4}}{4}-\left(-1\right)^{2}+\frac{\left(-1\right)^{3}}{3}-2\left(-1\right)\right)$$

Simplify.

$$-\frac{11}{12}$$