Solve ∫ (from 0 to 3) of (y-3)(2y+3) wrt y

$$\int _ { 0 } ^ { 3 } ( y - 3 ) ( 2 y + 3 ) d y$$

Evaluate

$$-\frac{45}{2}=-22.5$$

Solution Steps

Apply the distributive property by multiplying each term of y-3 by each term of 2y+3.

$$\int _{0}^{3}2y^{2}+3y-6y-9\mathrm{d}y$$

Combine 3y and -6y to get -3y.

$$\int _{0}^{3}2y^{2}-3y-9\mathrm{d}y$$

Evaluate the indefinite integral first.

$$\int 2y^{2}-3y-9\mathrm{d}y$$

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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 -3 times \frac{y^{2}}{2}.

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

Find the integral of -9 using the table of common integrals rule \int a\mathrm{d}y=ay.

$$\frac{2y^{3}}{3}-\frac{3y^{2}}{2}-9y$$

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.

$$\frac{2}{3}\times 3^{3}-\frac{3}{2}\times 3^{2}-9\times 3-\left(\frac{2}{3}\times 0^{3}-\frac{3}{2}\times 0^{2}-9\times 0\right)$$

Simplify.

$$-\frac{45}{2}$$