Solve ∫ (from 0 to 1 div 3) of (9x^2+9) wrt x

$$\int_{ 0 }^{ 1 \div 3 } (9 { x }^{ 2 } +9) d x$$

Evaluate

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

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 9 times \frac{x^{3}}{3}.

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

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

$$3x^{3}+9x$$

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.

$$3\times \left(\frac{1}{3}\right)^{3}+9\times \frac{1}{3}-\left(3\times 0^{3}+9\times 0\right)$$

Simplify.

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