Solve ∫ (from 1 to 5) of x^2-6x+5

$$\int _ { 1 } ^ { 5 } x ^ { 2 } - 6 x + 5$$

Evaluate

$$-\frac{32}{3}\approx -10.666666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int x^{2}-6x+5\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int -6x\mathrm{d}x+\int 5\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 5\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}-6\int x\mathrm{d}x+\int 5\mathrm{d}x$$

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 -6 times \frac{x^{2}}{2}.

$$\frac{x^{3}}{3}-3x^{2}+\int 5\mathrm{d}x$$

Find the integral of 5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}-3x^{2}+5x$$

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.

$$\frac{5^{3}}{3}-3\times 5^{2}+5\times 5-\left(\frac{1^{3}}{3}-3\times 1^{2}+5\times 1\right)$$

Simplify.

$$-\frac{32}{3}$$