Multiple integration is obviously overkill in this situation, but we proceed to establish its use.

The region \(R\) is bounded by \(x=-1\text{,}\) \(x=3\text{,}\) \(y=1\) and \(y=3\text{.}\) Choosing to integrate with respect to \(y\) first, we have \begin{equation*} A = \int_{-1}^3\int_1^3 1\ dy\ dx = \int_{-1}^3 \left(y\ \Big|_1^3\right)\ dx = \int_{-1}^3 2\ dx = 2x\Big|_{-1}^3=8. \end{equation*}

We could also integrate with respect to \(x\) first, giving: \begin{equation*} A = \int_1^3\int_{-1}^3 1\ dx \ dy =\int_1^3 \left(x\ \Big|_{-1}^3\right)\ dy = \int_1^3 4\ dy = 4y\Big|_1^3 = 8. \end{equation*}

Clearly there are simpler ways to find this area, but it is interesting to note that this method works.

The triangle is bounded by the lines as shown in the figure. Choosing to integrate with respect to \(x\) first gives that \(x\) is bounded by \(x=y\) to \(x = \frac{y+5}2\text{,}\) while \(y\) is bounded by \(y=1\) to \(y=5\text{.}\) (Recall that since \(x\)-values increase from left to right, the leftmost curve, \(x=y\text{,}\) is the lower bound and the rightmost curve, \(x=(y+5)/2\text{,}\) is the upper bound.) The area is \begin{align*} A \amp = \int_1^5\int_{y}^{\frac{y+5}2}\ dx\ dy\\ \amp = \int_1^5\left(x\ \Big|_y^{\frac{y+5}2}\right)\ dy\\ \amp = \int_1^5 \left(-\frac12y+\frac52\right)\ dy\\ \amp = \left(-\frac14y^2+\frac52y\right)\Big|_1^5\\ \amp =4. \end{align*}

We can also find the area by integrating with respect to \(y\) first. In this situation, though, we have two functions that act as the lower bound for the region \(R\text{,}\) \(y=1\) and \(y=2x-5\text{.}\) This requires us to use two iterated integrals. Note how the \(x\)-bounds are different for each integral: \begin{align*} A \amp = \int_1^3\int_1^x 1\ dy \ dx \amp +\amp \amp \amp \int_3^5\int_{2x-5}^x1\ dy\ dx\\ \amp = \int_1^3\big(y\big)\Big|_1^x\ dx \amp + \amp \amp \amp \int_3^5\big(y\big)\Big|_{2x-5}^x\ dx\\ \amp = \int_1^3\big(x-1\big)\ dx \amp + \amp \amp \amp \int_3^5\big(-x+5\big)\ dx\\ \amp = 2 \amp + \amp \amp \amp 2\\ \amp =4. \end{align*}

As expected, we get the same answer both ways.

Once again we'll find the area of the region using both orders of integration.

Using \(dy\ dx\text{:}\) \begin{equation*} \int_0^2\int_{x^2}^{2x}1\ dy \ dx = \int_0^2(2x-x^2)\ dx = \big(x^2-\frac13x^3\big)\Big|_0^2 = \frac43. \end{equation*}

Using \(dx\ dy\text{:}\) \begin{equation*} \int_0^4\int_{y/2}^{\sqrt{y}} 1\ dx\ dy = \int_0^4 (\sqrt{y}-y/2)\ dy = \left(\frac23y^{3/2} - \frac14y^2\right)\Big|_0^4 = \frac43. \end{equation*}

We need to use the bounds of integration to determine the region we are integrating over.

The bounds tell us that \(y\) is bounded by \(0\) and \(x/3\text{;}\) \(x\) is bounded by 0 and 6. We plot these four curves: \(y=0\text{,}\) \(y=x/3\text{,}\) \(x=0\) and \(x=6\) to find the region described by the bounds. Figure 13.1.15 shows these curves, indicating that \(R\) is a triangle.

To change the order of integration, we need to consider the curves that bound the \(x\)-values. We see that the lower bound is \(x=3y\) and the upper bound is \(x=6\text{.}\) The bounds on \(y\) are \(0\) to \(2\text{.}\) Thus we can rewrite the integral as \(\ds \int_0^2\int_{3y}^6 1\ dx \ dy\text{.}\)

We sketch the region described by the bounds to help us change the integration order. \(x\) is bounded below and above (i.e., to the left and right) by \(x=y^2/4\) and \(x=(y+4)/2\) respectively, and \(y\) is bounded between 0 and 4. Graphing the previous curves, we find the region \(R\) to be that shown in Figure 13.1.17.

To change the order of integration, we need to establish curves that bound \(y\text{.}\) The figure makes it clear that there are two lower bounds for \(y\text{:}\) \(y=0\) on \(0\leq x\leq 2\text{,}\) and \(y=2x-4\) on \(2\leq x\leq 4\text{.}\) Thus we need two double integrals. The upper bound for each is \(y=2\sqrt{x}\text{.}\) Thus we have \begin{equation*} \int_0^4\int_{y^2/4}^{(y+4)/2}1\ dx\ dy = \int_0^2\int_0^{2\sqrt{x}} 1\ dy\ dx + \int_2^4\int_{2x-4}^{2\sqrt{x}}1\ dy\ dx. \end{equation*}