 Hi everyone! I have an example here for you of another area between two curves problem. So we're going to find the area of the region bounded by the graphs of a cubic equation and then also a parabola. So let's go ahead and get them graphed in our graphing calculator. And you can do zoom six to get yourself a graph started. I did shorten up my x axis though to how you see them here to negative 6 to positive 6. So obviously the blue curve is the cubic and the red curve is the quadratic. And what is different about this problem compared to others that we've looked at is that this is an example of which if you take a look at the functions they almost like flip-flop they turn change places in terms of which one is the one quote on quote on top. Because remember that we want to think of our representative rectangle. In this case if we want that rectangle to hit both curves at the same time it needs to be going vertically therefore it's going to be a dx problem. So if we think of doing that integrand as the top function minus the bottom function that's what's going to change places once you hit the origin going from the left to the right. So this is one we're going to have to set up in two integrals. So you're going to have to find those points of intersection. I shall leave that to you. Again I've already done it so I know what they are. That one down in the third quadrant on the lower left that x value should be negative 2 that you find. Then the curves each pass through the origin and then they cross each other again over at x equals positive 2. Alright so the first integral we're going to go left to right. The first integral is going to go from negative 2 to 0 and it's going to have to be the cubic equation minus the quadratic one. So let's set that up. You could even just call it f of x minus g of x if you'd like. If you don't feel like writing all this out because we'll probably just evaluate it in our calculators. Then for the second integral let's go back to our graph. We're going to go from 0 to 2 but now the quadratic is on top. So we're going to have to think of this as g of x which is the quadratic function minus f of x which was the cubic. So we have two different integrals to do on our calculator. Let's see I know this was y1 in my calculator and this one was y2. So I might be able to make use of that as I go ahead and set them up. So we can do them separately and then just add our answers together if you want. So let's go back to our quit screen and we'll go ahead and do math 9. So the first one was negative 2 to 0. This is the one remember the cubic was on top which is what I had under y1. So that was y1 minus y2. All right so that's my cubic as y1 minus my quadratic as y2 dx. So let's see what we get for that one. 12. Okay let's try the next one. So now we're going from 0 to 2. Now this is where they change places so now the top function was the quadratic which I had under y2. Oops I said y2 minus y1 which was my g of x function the cubic. I actually get 12 for that. So the combined total of course is 24. Now let me show you something with that area between program though because we've been talking about how we could use that to check our answers. So let's go back let's go ahead and run the program and what a lot of people would try to do is they would think okay well I'm just going to run it from negative 2 to positive 2 and just find the whole area in one lump sum. All right so let's try to do that. So negative 2 to positive 2 and look what you get. This is a case in which you actually have to be smarter than your calculator. All right that obviously is scientific notation your 9.174 times 10 to the negative 15th essentially 0. So the calculator does not realize that the functions have switched places that's really what happens. It's almost thinking of this as a positive area and a negative area offsetting each other. Now of course if you were to run this from negative 2 to 0 and then we could do the same from 0 to 2 it will give you the right answer then. So if you wanted to use this to check your answers I would I'd highly advise you to go ahead and do it in two separate integrals but I wanted to show you what happens if you were to run it from the far left end point to the far right end point. It doesn't always work you still have to be smarter than your calculator.