Solve ∫ (from 1 div 3 to 2 div 3) of (9x^2-12x+13)) wrt x

$$\int_{ 1 \div 3 }^{ 2 \div 3 } (9 { x }^{ 2 } -12x+13)) d x$$

Evaluate

$$\frac{28}{9}\approx 3.111111111$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 9x^{2}-12x+13\mathrm{d}x$$

Integrate the sum term by term.

$$\int 9x^{2}\mathrm{d}x+\int -12x\mathrm{d}x+\int 13\mathrm{d}x$$

Factor out the constant in each of the terms.

$$9\int x^{2}\mathrm{d}x-12\int x\mathrm{d}x+\int 13\mathrm{d}x$$

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 9 times \frac{x^{3}}{3}.

$$3x^{3}-12\int x\mathrm{d}x+\int 13\mathrm{d}x$$

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 -12 times \frac{x^{2}}{2}.

$$3x^{3}-6x^{2}+\int 13\mathrm{d}x$$

Find the integral of 13 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$3x^{3}-6x^{2}+13x$$

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.

$$3\times \left(\frac{2}{3}\right)^{3}-6\times \left(\frac{2}{3}\right)^{2}+13\times \frac{2}{3}-\left(3\times \left(\frac{1}{3}\right)^{3}-6\times \left(\frac{1}{3}\right)^{2}+13\times \frac{1}{3}\right)$$

Simplify.

$$\frac{28}{9}$$