# Find the value of the integral \iint_A (3(x-3) -y) \, dx \, dy where A is the region \{...

## Question:

Find the value of the integral

{eq}\iint_A (3(x-3) -y) \, dx \, dy {/eq}

where {eq}A {/eq} is the region {eq}\{ (x,y) : 0 \leq y \leq x-3, \quad 3 \leq x \leq 5 \} {/eq}

## Double Integration:

Double integral is a definite integral that includes two variable. This is used to find the two-dimensional region of the given problem. In this question, we will first consider the double integration as regards x and then as regards y.

Given that {eq}\iint_A (3(x-3) -y) \, dx \, dy {/eq}

where {eq}A {/eq} is the region {eq}\{ (x,y) : 0 \leq y \leq x-3, \quad 3 \leq x \leq 5 \} {/eq}

{eq}\int _3^5\:\int _0^{x-3}\:(3(x-3) -y) \:dydx\\ =\int _3^5\:\left [ \left[\left(-3+x\right)\cdot \:3y\right]^{x-3}_0-\left[\frac{y^2}{2}\right]^{x-3}_0 \right ]dx\\ =\int _3^5\left(-\frac{\left(x-3\right)^2}{2}+3\left(x-3\right)^2\right)dx\\ =\int _3^5\frac{5}{2}\left(x-3\right)^2dx\\ =\frac{5}{2}\cdot \int _3^5\left(x-3\right)^2dx\\ =\frac{5}{2}\cdot \int _3^5\left (x^2-6x+9 \right )dx\\ =\frac{5}{2}\left(\int _3^5x^2dx-\int _3^56xdx+\int _3^59dx\right)\\ =\frac{5}{2}\left(\frac{98}{3}-48+18\right)\\ =\frac{5}{2}\left(\frac{98}{3}-30\right)\\ =\frac{5}{2}\cdot \frac{8}{3}\\ =\frac{20}{3} {/eq}