Solve ∫ (from 2 to 6) of (3x-9) wrt x

$$\int _ { 2 } ^ { 6 } ( 3 x - 9 ) d x$$

Evaluate

$$12$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 3x-9\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x\mathrm{d}x+\int -9\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x\mathrm{d}x+\int -9\mathrm{d}x$$

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}.

$$\frac{3x^{2}}{2}+\int -9\mathrm{d}x$$

Find the integral of -9 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{3x^{2}}{2}-9x$$

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.

$$\frac{3}{2}\times 6^{2}-9\times 6-\left(\frac{3}{2}\times 2^{2}-9\times 2\right)$$

Simplify.

$$12$$