Solve ∫ (from 1 to 6) of -(x^2-7x+6) wrt x

$$\int _ { 1 } ^ { 6 } - ( x ^ { 2 } - 7 x + 6 ) d x$$

Evaluate

$$\frac{125}{6}\approx 20.833333333$$

Solution Steps

To find the opposite of x^{2}-7x+6, find the opposite of each term.

$$\int _{1}^{6}-x^{2}+7x-6\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -x^{2}+7x-6\mathrm{d}x$$

Integrate the sum term by term.

$$\int -x^{2}\mathrm{d}x+\int 7x\mathrm{d}x+\int -6\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\int x^{2}\mathrm{d}x+7\int x\mathrm{d}x+\int -6\mathrm{d}x$$

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}.

$$-\frac{x^{3}}{3}+7\int x\mathrm{d}x+\int -6\mathrm{d}x$$

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 7 times \frac{x^{2}}{2}.

$$-\frac{x^{3}}{3}+\frac{7x^{2}}{2}+\int -6\mathrm{d}x$$

Find the integral of -6 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{x^{3}}{3}+\frac{7x^{2}}{2}-6x$$

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.

$$-\frac{6^{3}}{3}+\frac{7}{2}\times 6^{2}-6\times 6-\left(-\frac{1^{3}}{3}+\frac{7}{2}\times 1^{2}-6\right)$$

Simplify.

$$\frac{125}{6}$$