Question

Which order of integration would you use to find the area of the region bounded by the \(x\) -axis and the lines \(y=2 x+3\) and \(y=3 x-4\) using a double integral?

Short answer

Based on the given solution, describe the steps to find the intersection points of the lines and the x-axis, and the order of integration to calculate the area of the region bounded by the given lines and the x-axis using a double integral.

Step-by-step solution

Finding the intersection points of the given lines and the x-axis

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 \(y = 3(7)-4 = 17\).

Setting up the limits of integration

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: $$ \int_{0}^{17}\int_{\frac{y-3}{2}}^{\frac{y+4}{3}} dA $$

Deciding on the order of integration

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: $$ \int_{-\frac{3}{2}}^{\frac{4}{3}}\int_{0}^{3x-4} dA $$