PROBLEM 5 Consider the region bounded by the graphs of x=(y-1) x = 3 - y x=2/y As shown below. As it's given (x = f(y))i

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Problem 5 Consider The Region Bounded By The Graphs Of X Y 1 X 3 Y X 2 Y As Shown Below As It S Given X F Y I 1 (237.52 KiB) Viewed 23 times

PROBLEM 5 Consider the region bounded by the graphs of x=(y-1) x = 3 - y x=2/y As shown below. As it's given (x = f(y))it suggests using a horizontal slicing setup, and that's where we're ( headed first. Here's a graph of the region - start by labeling all the functions on the boundaries (you need to know which thing is which!) and start by slicing it up with some horizontal slices. Notice that to set up the integrals that give area, it will take two different integrals - there's a place where the region has to be divided. It's OK for this one to use the Desmos intersection points - I won't make you hand solve. (1,2) (2, 1) (0.743, 0.138) Part one: Find the area - set up and evaluate (show the by hand evaluation steps) the integrals needed to give the area of that region, using a horizontal f(y)dy setup. Round to three places. Part two: Next, you're going to re-think the region as a vertically sliced region. This will make things worse from a Calculus perspective, since it will now take three integrals to get the area. Your first job is to get all the functions switched over to y = f(x) form (include on this page), and use that information to replicate the graph below in Desmos. The switching over isn't strictly need to graph the boundaries, but it is needed to set up the inequalities for the shading, and that makes a nice check on it. If you can get the shading right, then you are lined up to get the correct integrals set up. Recreate it, and be sure to screenshot and include. As part of your screenshot, I want to see the side bar where you entered in the equations in Desmos to create the shading. (1,2) (2.1) (0.743, 0.138) Part three: Finally, set up the three integrals needed to give the area using vertical slices (an f(x) dx setup). Integrate them using Symbolab or Wolfram - you don't have to show the integration by hand. What this is doing is checking your setup - you should of course get the same answer for total area that you got for part one, and if you don't, there's an error. Show your integrals and their values here (and give the total area of the region again, showing the check):