Solve ∫ (from 0 to 5) of 4(y-5/frac{-10{3}}) wrt y

$$\int_{ 0 }^{ 5 } 4( \frac{ y-5 }{ \frac{ -10 }{ 3 } } ) d y$$

Evaluate

$$15$$

Solution Steps

Divide y-5 by \frac{-10}{3} by multiplying y-5 by the reciprocal of \frac{-10}{3}.

$$\int _{0}^{5}4\times \frac{\left(y-5\right)\times 3}{-10}\mathrm{d}y$$

Express 4\times \frac{\left(y-5\right)\times 3}{-10} as a single fraction.

$$\int _{0}^{5}\frac{4\left(y-5\right)\times 3}{-10}\mathrm{d}y$$

Multiply 4 and 3 to get 12.

$$\int _{0}^{5}\frac{12\left(y-5\right)}{-10}\mathrm{d}y$$

Divide 12\left(y-5\right) by -10 to get -\frac{6}{5}\left(y-5\right).

$$\int _{0}^{5}-\frac{6}{5}\left(y-5\right)\mathrm{d}y$$

Use the distributive property to multiply -\frac{6}{5} by y-5.

$$\int _{0}^{5}-\frac{6}{5}y-\frac{6}{5}\left(-5\right)\mathrm{d}y$$

Express -\frac{6}{5}\left(-5\right) as a single fraction.

$$\int _{0}^{5}-\frac{6}{5}y+\frac{-6\left(-5\right)}{5}\mathrm{d}y$$

Multiply -6 and -5 to get 30.

$$\int _{0}^{5}-\frac{6}{5}y+\frac{30}{5}\mathrm{d}y$$

Divide 30 by 5 to get 6.

$$\int _{0}^{5}-\frac{6}{5}y+6\mathrm{d}y$$

Evaluate the indefinite integral first.

$$\int -\frac{6y}{5}+6\mathrm{d}y$$

Integrate the sum term by term.

$$\int -\frac{6y}{5}\mathrm{d}y+\int 6\mathrm{d}y$$

Factor out the constant in each of the terms.

$$-\frac{6\int y\mathrm{d}y}{5}+\int 6\mathrm{d}y$$

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{6}{5} times \frac{y^{2}}{2}.

$$-\frac{3y^{2}}{5}+\int 6\mathrm{d}y$$

Find the integral of 6 using the table of common integrals rule \int a\mathrm{d}y=ay.

$$-\frac{3y^{2}}{5}+6y$$

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.

$$-\frac{3}{5}\times 5^{2}+6\times 5-\left(-\frac{3}{5}\times 0^{2}+6\times 0\right)$$

Simplify.

$$15$$