Solve ∫ (from 1 to 3) of x^2-3x+4

$$\int _ { 1 } ^ { 3 } x ^ { 2 } - 3 x + 4$$

Evaluate

$$\frac{14}{3}\approx 4.666666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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 -3 times \frac{x^{2}}{2}.

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

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$\frac{14}{3}$$