## Limit for Double Integrals When Boundaries are Functions

\[0\leq x\leq 1, 2 \leq y\leq 3\]

. It is not so simple when one or more boundaries of the region of integration is given by an equation in the variables.Suppose we are to evaluate

\[\int_R xy dR\]

over the region \[R\]

satisfying \[x+2y \leq 4, x-3y \geq6, x \geq0\]

. We can evaluate this integral by considering the limits. For the region \[R\]

\[x/3 -2 \leq y\leq2-x/2\]

and \[0 \leq x\leq 24/5\]

Not that

\[\frac{24}{5}\]

is the \[x\]

intecpt of the top and bottom boundaries of \[R\]

.We can write the integral as

\[\begin{equation} \begin{aligned} \int^{24/5}_0 \int^{2-x/2}_{x/3-2} xy dy dx &= \int^{24/5}_0 [xy^2/2]^{2-x/2}_{x/3-2} dx \\ &= \int^{24/5}_0 ((x(2-x/2)^2 - x(x/3-2)^2) dx \\ &= \int^{24/5}_0 (-2x^2/3+5x^3/36) dx \\ &= [-2x^3/9 5x^4/144]^{24/5}_0 \\ &= 663/125 \end{aligned} \end{equation} \]

Notice that on the left hand boundary

\[x=0 \]

, a constant, and on the right hand boundary, \[x=24/5\]

.This is why we integrated with respect to

\[y\]

first, then \[x\]

.