Solve ∫ (from 1 to 5) of (-x^2+6x-5) wrt x

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

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}. Multiply -1 times \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.

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

Simplify.

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