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

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

Evaluate

$$-\frac{56}{3}\approx -18.666666667$$

Solution Steps

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}.

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

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.

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

Simplify.

$$-\frac{56}{3}$$