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\text{.}
\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\text{.}
\end{align*}
As expected, we get the same answer both ways.