Solve ∫ (from 0 to 3) of 5(2-2/3x)^2

$$\int _ { 0 } ^ { 3 } 5 ( 2 - \frac { 2 } { 3 } x ) ^ { 2 }$$

Evaluate

$$20$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-\frac{2}{3}x\right)^{2}.

$$\int _{0}^{3}5\left(4-\frac{8}{3}x+\frac{4}{9}x^{2}\right)\mathrm{d}x$$

Use the distributive property to multiply 5 by 4-\frac{8}{3}x+\frac{4}{9}x^{2}.

$$\int _{0}^{3}20-\frac{40}{3}x+\frac{20}{9}x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 20-\frac{40x}{3}+\frac{20x^{2}}{9}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 20\mathrm{d}x+\int -\frac{40x}{3}\mathrm{d}x+\int \frac{20x^{2}}{9}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 20\mathrm{d}x-\frac{40\int x\mathrm{d}x}{3}+\frac{20\int x^{2}\mathrm{d}x}{9}$$

Find the integral of 20 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$20x-\frac{40\int x\mathrm{d}x}{3}+\frac{20\int x^{2}\mathrm{d}x}{9}$$

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 -\frac{40}{3} times \frac{x^{2}}{2}.

$$20x-\frac{20x^{2}}{3}+\frac{20\int x^{2}\mathrm{d}x}{9}$$

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 \frac{20}{9} times \frac{x^{3}}{3}.

$$20x-\frac{20x^{2}}{3}+\frac{20x^{3}}{27}$$

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.

$$20\times 3-\frac{20}{3}\times 3^{2}+\frac{20}{27}\times 3^{3}-\left(20\times 0-\frac{20}{3}\times 0^{2}+\frac{20}{27}\times 0^{3}\right)$$

Simplify.

$$20$$