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

$$\int _ { 1 } ^ { 2 } ( x - 1 ) ( x + 3 ) d x =$$

Evaluate

$$\frac{7}{3}\approx 2.333333333$$

Solution Steps

Apply the distributive property by multiplying each term of x-1 by each term of x+3.

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

Combine 3x and -x to get 2x.

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

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

Find the integral of -3 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$\frac{7}{3}$$