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

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

Evaluate

$$-\frac{5}{6}\approx -0.833333333$$

Solution Steps

Apply the distributive property by multiplying each term of 2x-1 by each term of x+1.

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

Combine 2x and -x to get x.

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

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

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$-\frac{5}{6}$$