![]() To find them, lets look at the graph again: Well, remember that our boundaries are simply the values of \(x\) that tell us where the region begins and ends. The area formula requires a definite integral, and that means we need limits of integration (aka "bounds"). Great! We're ready to integrate, right? Hold it there, partner. The piece to the right of the dotted line is bounded on the top by \(y=2\), and on the bottom by \(y=2x\). Ok, so the piece to the left of the dotted line is bounded on the top by \(y=3x\), and on the bottom by \(y=2x\). Now, we simply need to use the formula to find the area of each piece, and then add them together to get our final answer. If you look at each piece separately, you will see that now they each have a distinct "top" and "bottom" function. I've drawn a dotted line to show where I cut the region. The part to the right of the cut is bounded on top by \(y=2\). If we cut up this region at the point where \(y=3x\) and \(y=2\) intersect, the part that is to the left of the cut is bounded on top by \(y=3x\). Actually, it depends on which part of the triangle we're looking at. The answer is that \(y=3x\) is also on the top. Well, the line \(y=2\) looks the top function, and \(y=2x\) looks like the bottom function, but what about \(y=3x\)? Is it on top, on the bottom, or somewhere else entirely? Looking at this picture, its not clear which function is the "top" function and which is the "bottom" function. HINT: you can graph any line by picking two different values of \(x\), plugging them into the equation to find the corresponding values of \(y\), plotting those two points, and drawing a straight line through them. Try to do it yourself before looking at my graph. I also want you to label the lines with their equations. I'll repeat it here, since I've rambled quite a bit since I first stated it:įirst, I want you to graph the lines, all on the same axes. Calculus is a great way to learn more general problem-solving skills, math or otherwise. Learning how to break down problems, by the way, is the reason that you are required to take calculus for your major - even if you'll never see another integral again in your entire life. We're going to figure out a way to decompose, or break down, this difficult problem into several smaller, easier problems. How in the name of Bieber can we apply the above formula to a region enclosed by three functions? ![]() The two-function area problems are solved by integrating the difference between the "top" and "bottom" functions, like so:īut wait a minute! Our problem is giving us THREE functions, not two. Alright, well hopefully you've already seen problems that ask for the area between two functions. Strong words from the guy with the gradebook. You must use calculus or you will not receive any credit! The problem goes something like this:įind the area of the region enclosed by the lines \(y=2x\), \(y=3x\), and \(y=2\). Howdy folks, Alex here! I thought I'd start this blog off right, with one of the most popular (and head-spinning) problems that get thrown at my calculus students.
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