Solve ∫ (y+3)(3-y)2.3 wrt y

$$\int{ (y+3)(3-y)2.3 }d y$$

Evaluate

$$-\frac{23y^{3}}{30}+\frac{207y}{10}+С$$

Solution Steps

Apply the distributive property by multiplying each term of y+3 by each term of 3-y.

$$\int \left(3y-y^{2}+9-3y\right)\times 2.3\mathrm{d}y$$

Combine 3y and -3y to get 0.

$$\int \left(-y^{2}+9\right)\times 2.3\mathrm{d}y$$

Use the distributive property to multiply -y^{2}+9 by 2.3.

$$\int -2.3y^{2}+20.7\mathrm{d}y$$

Integrate the sum term by term.

$$\int -\frac{23y^{2}}{10}\mathrm{d}y+\int 20.7\mathrm{d}y$$

Factor out the constant in each of the terms.

$$-\frac{23\int y^{2}\mathrm{d}y}{10}+\int 20.7\mathrm{d}y$$

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 -2.3 times \frac{y^{3}}{3}.

$$-\frac{23y^{3}}{30}+\int 20.7\mathrm{d}y$$

Find the integral of 20.7 using the table of common integrals rule \int a\mathrm{d}y=ay.

$$-\frac{23y^{3}}{30}+\frac{207y}{10}$$

If F\left(y\right) is an antiderivative of f\left(y\right), then the set of all antiderivatives of f\left(y\right) is given by F\left(y\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$-\frac{23y^{3}}{30}+\frac{207y}{10}+С$$

Differentiate w.r.t. y

$$\frac{207-23y^{2}}{10}$$