Solve ∫ (from 0 to 5) of {left(2*(-10/3y+5)right)}^2 wrt y

$$\int_{ 0 }^{ 5 } { \left(2 \times (- \frac{ 10 }{ 3 } y+5) \right) }^{ 2 } d y$$

Evaluate

$$\frac{18500}{27}\approx 685.185185185$$

Solution Steps

Use the distributive property to multiply 2 by -\frac{10}{3}y+5.

$$\int _{0}^{5}\left(-\frac{20}{3}y+10\right)^{2}\mathrm{d}y$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{20}{3}y+10\right)^{2}.

$$\int _{0}^{5}\frac{400}{9}y^{2}-\frac{400}{3}y+100\mathrm{d}y$$

Evaluate the indefinite integral first.

$$\int \frac{400y^{2}}{9}-\frac{400y}{3}+100\mathrm{d}y$$

Integrate the sum term by term.

$$\int \frac{400y^{2}}{9}\mathrm{d}y+\int -\frac{400y}{3}\mathrm{d}y+\int 100\mathrm{d}y$$

Factor out the constant in each of the terms.

$$\frac{400\int y^{2}\mathrm{d}y}{9}-\frac{400\int y\mathrm{d}y}{3}+\int 100\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y^{2}\mathrm{d}y with \frac{y^{3}}{3}. Multiply \frac{400}{9} times \frac{y^{3}}{3}.

$$\frac{400y^{3}}{27}-\frac{400\int y\mathrm{d}y}{3}+\int 100\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y\mathrm{d}y with \frac{y^{2}}{2}. Multiply -\frac{400}{3} times \frac{y^{2}}{2}.

$$\frac{400y^{3}}{27}-\frac{200y^{2}}{3}+\int 100\mathrm{d}y$$

Find the integral of 100 using the table of common integrals rule \int a\mathrm{d}y=ay.

$$\frac{400y^{3}}{27}-\frac{200y^{2}}{3}+100y$$

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.

$$\frac{400}{27}\times 5^{3}-\frac{200}{3}\times 5^{2}+100\times 5-\left(\frac{400}{27}\times 0^{3}-\frac{200}{3}\times 0^{2}+100\times 0\right)$$

Simplify.

$$\frac{18500}{27}$$