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

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

Evaluate

$$\frac{20}{3}\approx 6.666666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}.

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

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}.

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

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.

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

Simplify.

$$\frac{20}{3}$$