91Ó°ÊÓ

To find the points of intersection of the lines \(y=2x+3\) and \(y=3x-4\), we need to set \(y=0\) and solve for \(x\) for each given line: For the line \(y=2x+3\), when \(y=0\): \(0 = 2x + 3 \Rightarrow x =-\dfrac{3}{2}\) For the line \(y=3x-4\), when \(y=0\): \(0 = 3x - 4 \Rightarrow x = \dfrac{4}{3}\) Now, let's find the point of intersection of the two lines by setting them equal to each other: \(2x+3 = 3x-4 \Rightarrow x = 7\) and \(y = 3(7)-4 = 17\).

Now, we need to determine the limits of integration for our double integral. To do that, we will first draw the region bounded by the given lines and the x-axis. From the intersection points found in step 1, the region is bounded below by the x-axis, on the left by the line \(y=2x+3\), on the right by the line \(y=3x-4\), and between the x-values of \(-\dfrac{3}{2}\) and \(\dfrac{4}{3}\). Analyzing this region, we can set up the limits of integration in two different orders: 1. Integrate with respect to y first: In this case, \(y\) goes from the x-axis (i.e., \(y=0\)) to the line connecting the points of intersection of the given lines (i.e., \(y=3x-4\)). The x-values go from the left intersection point to the right intersection point (i.e., \(x\) goes from \(-\dfrac{3}{2}\) to \(\dfrac{4}{3}\)). So, the integral would look like: $$ \int_{-\frac{3}{2}}^{\frac{4}{3}}\int_{0}^{3x-4} dA $$ 2. Integrate with respect to x first: In this case, \(x\) goes from the line connecting the points of intersection on the left to the respective x-axis points (i.e., \(x\) goes from \(\dfrac{y-3}{2}\) to \(\dfrac{y+4}{3}\)). The y-values go from 0 (at the x-axis) to the maximum y-value at the intersection point of the two lines (i.e., \(y\) goes from 0 to 17). So, the integral would look like: $$ \int_{0}^{17}\int_{\frac{y-3}{2}}^{\frac{y+4}{3}} dA $$

Upon analyzing the limits of integration for both possible orders, the first order (integrating with respect to y first) appears to be the more convenient option. This is because the limits of integration are in a simpler form than the second order, involving less complex fractions. So, the recommended order of integration to find the area of the region bounded by the x-axis and the lines \(y=2x+3\) and \(y=3x-4\) using a double integral would be to integrate with respect to y first, and then with respect to x. Thus, the appropriate double integral to use is: $$ \int_{-\frac{3}{2}}^{\frac{4}{3}}\int_{0}^{3x-4} dA $$