John Postovit

University of North Dakota
M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.

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Area Between Curves

John Postovit
John Postovit

University of North Dakota
M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.


For some reason, you decided to take on calculus, why? Well I know I did because it's fun and it's really actually useful. It has got lots of uses in engineering and science. Today we're going to be talk about about the area between two curves. Now the area between two curves is just what it sounds like. You can have two different shapes there is an area trapped in between them, and finding out the area of trap between them is really just a matter of finding two integrals and subtracting them.

Now say an engineer wants to find the volume of a really complex casting, or say a say a bridge support. If they can work out some equations for the shapes that define that, and one small extension they can find what the volume of that shape is. Simply by starting with the area between those two curves.

Let's have a look at a problem. This one asked us to find the area between f(x)= x²-4x+7, that's the parabola and g(x)=7. What they're asking for is this area. Now this one has a very common complication, that it doesn't tell you the limits of integration. What you need to know is that, the limits of integration have to be places where the curves intersect.

So if you were to do this problem, first thing you would do, will be to find the area underneath the upper curve. Remember curves don't have to actually curve of a calculus. That's the rectangle and then you would subtract away the area underneath the parabola. And what's left, that will be the area trapped between the two curves.

Now there is a formula that you use for this, you don't really have to have it because the essentials are just doing one integral and subtracting the other integral from it. But it is often very convenient to subtract the two functions before you integrate. Now let's go and do the practise problem. This practise problem is one where you're going to be given the limits of integration.

Let's start by sketching the graphs. That's a really important because you need to know what the upper function is. If you don't know what the upper function is, you don't always know which one to set up first. After you've identified that upper function and lower function, subtract the lower function from the upper function. And then you need to find the limits of integration if they aren't given to you, and then finally integrate it.

So in this example, I have a parabola and a straight line, and we're going to do that on the intervals of 0 to 5. So the interval is given for you. Here is the graph. Looks like the upper function is a straight line, the lower function is the parabola. Our limits are from 0 to 5. So we're going to find the area that's stuck between them. And you might be thinking, "Wait a minute, this is below the X axis, so the area trapped between that and the x axis is going to be negative, and I'm going to mess around with changing between them." No you don't have to. The integral takes care of all of it. All you need to do is take this function, the x+10 and subtract that function, -x²+6x-9. So we'll put them together as a single integral.

So this is the formula, but the formula again just essentially tells you subtract them. So you have x-10 minus all that. The reason this is convenient is because we can do fewer parts of the integral, if we subtract first. So we have our area from 0 to 5. Be really careful of the plus and minus signs of these problems. Messing one of them up, you have no luck you have the wrong answer. It's really easy to mess one of them up, so make sure you show a very clear work, so you can go and check that plus and minus signs.

Now one way you can show clear work and save a little bit of time, since you can write on the test, is take care of this distributing this negative in, just by writing on the problem. When you distribute a negative in there it becomes a plus, that becomes a minus, that becomes a plus. So we've got work that we can follow without taking a ton of time to write. I mean always do that.

You may think, "Oh yeah! Subtracting problem I can do that, no problems I'm in calculus now." You'd be amazed on how easy it is to make a mistake on that step. So I have x² and I've got -6x+1x, that's -5x and we have a +9+10, that's 19. Now we'll integrate, since it's got three parts to it. And if we integrate this, the integral of x² is 1/3x cubed. Integral of -5x is -5/2x².

Remember I'm building the correction factors in there as we speak. The derivative of x² is 2x. So this 2 on the denominator is necessary to get rid of the 2 that comes from the derivative. And plus 19x and we integrate this from 0 to 5.

Well then you go and substitute. And again these problems often have a lot of substitution and a lot of fractions too, which makes it easier to make mistakes if you don't show clear work. We have though 1/3 of 5 cubed which is a (125/3 minus 5/2 times 5² plus 19 times 5) minus, this is the first part of the integral if I substitute the 5 into it, minus what I substitute the 0 into it. Now you might cast your mark to be leaving this off. And if you left it off at this one, it wouldn't have mattered but be aware you have to always double check to make sure that if one of your limits is 0, the result of putting it in really 0. 1/3 times 0 cubed, yeah that's 0. 5/2 times 0 square yeah that's 0. And 19 times 0 that's also 0. If you have a sine and cosine functions though, those might not all have been zeros. I'm not going to show all the work for dealing all these fractions, I've worked out that in advance.

The area that's contained between here and here turns out to be 445/6. That's the area of our first complete example.

Time to try another, noch einmal as the Germans would say. This one is going to be a little bit harder but the last one, we didn't worry about the curves intersecting. We just said, "Hey here are some limits find the area that's trapped in between those limits." This time the curves are going to meet and that's going to have to tell us what the limits are.

So we need to find the area contained between the parabola f(x) and g(x) which is a straight line. That would be this area right in here. And remember you find the area trapped between two different curves, or between any two different curves, by finding two integrals and subtracting them. Before you can do that though, you need to make at a definite integral where you'll know the limits of integration. You have to have this exact location. On the graph it looks like 0, but that's not good enough, you have to be sure that's 0. This one looks like it's probably 5, but you have to be sure about that.

You may actually run across something like this on the non calculator portion of the AP test, because this one can be solved out nice and easy. If you're given around the non calculator portion, there is a pretty good chance they might give you the graph but again you can count on it. You might actually have to graph that by hand. Not likely but it's possible. Anyhow remember back to Algebra 2, you did a lot of this, maybe a little bit than Algebra 1. Where you were finding the location where graphs crossed. That happens when the x coordinate and the y coordinate of both curves match up in the same time. So in other words that's when f(x) equals g(x).

So we're going to be finding the place where the y co-ordinates are equal. And since there is x on both sides of it, the x are going to be equal as well. f(x) was x²-4x+7, g(x) is x+7, and we've got a polynomial x² and x. So you're going to have to use old friend there, the zero product property get everything on one side. Take x from both sides and we have x²-5x, take 7 from both sides and there no constant on the left side any more. This is going to work out nice and neat. Just factoring the x out.

So binomial factor of x is taken out you get x times (x-5) equals 0. I'm going to do a little shortcut, you know how to do these. The values of x that will make this work are x=0 and x=5. We've got our limits of integration, 5 and 0, and now let's go give it a try.

Set this one up. Remember what you're going to be doing is to find the area is taking the integral of one function minus the other. I wrote (f-g), but that was because that's the way I remember the formula. In this case, look the upper function, it's g, it's the line. I should really put that one first. So bear in mind that when you look at the way that is written in the definition of how to do these, you can flip flop. And what matters is that the upper function has to be the one that you're subtracting from.

We found out a moment ago that the limits are 0 and 5, and g is (x+7). So we have (x+7),don't forget your parenthesis. Looks like the board here is forgetting the numbers, there I just remembered, minus write all that out. Take the time to write it out, it's not very long. And it can sure save you mistakes, especially when you have things where you have numbers that match up. The may or may not subtract away. In this case they will.

Remember on the last one, I distributed that negative in to reduce my chances of errors.

That changes into a plus, that becomes minus, that becomes plus and that becomes a minus. And now our integral from 0 to 5. I've got -x², I'm going to put that first, and I have 4x plus one more x, that's 5x. And I have got 7 plus -7 the constant is going away. I won't show you the substitutions. It's a matter of putting 5 and 0 and working at all those crazy fractions. If you take care of that, you'll find out that in this case the answer for this is 125/6, the area trapped between those two curves.

So we're going to try another now. This one is going to be the one where you're going to need to find the graph and calculator. You're very likely you're going to find one of these on the calculator part of the AP test. I'm supposed to find the area contained between cosine of x² and g(x) equals x².

Here is your x², here is that cosine related function. The area trapped between them is here and here. Now the reason you have to have the calculator on this one, is because if you set this equal to each other, you wouldn't be able to find out those location in exact form. There is no way. What you're expected to do is know how to put both of these functions into your calculator. You'd graph that function as separate functions and then you'd use your calculator's function that's called intercept to find out exactly where they cross.

I did that on my calculator in advance, and found out that the two places where they crossed were at -0.85970, that's this location and +0.85970. This one is symmetrical but don't count on they'll always be symmetrical. Well we'll go ahead and set up our integral now. It looks like our upper function for just this part, is cosine. If we were doing a different part, that might not be the upper function. Lower function is x², we just found out what our limits are. This is what the integral looks like. And now you're probably thinking, "How am I going to integrate that thing?" If you try substitution, you substitute u=x², there isn't the du that you need. That doesn't matter. This is on the calculator portion. Let your friend the calculator give you advise, just type it in again to find out the area.

Now you will find it's typically better to erase the two functions that you just had in there for finding the intersections, erase both. Type this in as a graph. It will look totally different to this but that doesn't matter because you don't care what it looks like at this point. You just want to use the integrate function on your calculator to integrate between here and here. And if you go about doing that, you will find that the area on this one is approximately 1.2042. I'd recommend you using three or four decimal places of accuracy when you put these down on the AP test.

Also, be really careful on the free response section when you're allowed to use the calculator, they won't expect you to show a lot of work, but they will expect you to demonstrate that you know the proper notation for it, and that your showing clearly how you set this up.

So you really do need to write out this integral. Even though you worked it out on your calculator. You could even write the steps as a next step. You could write out the words, "I found this integral on my calculator." And then write down the answer, but you have to show the setup. Part of the points that you get for a free response question, always are involved in the setup.

Another really important reason to make sure that you always draw the graph, is that the functions might intersect in more that one point. This one asks you to find the area contained between the sine function and this line function. And look, there is actually two different regions, the intersect here and the intersect here. In that section, the line is the upper function. In the next section where they intersect, the sine function is the upper function. The way you have to deal with this, is by separating into two different integrals.

If you find this area, the setup would look something like this. In the first region between here and here, you'd have to have the line minus the sine like that. Plus, for this region with different limits of integration, you would have to have the integral of sine minus line. Now while I can go through all the steps with you on the problem, I think it's important that you try it yourself. But remember what you'll be doing. You'll be finding intersections, setting up integrals. And this problem has a lot of similarities to the one that we just demonstrated with just one extra step.

Area between curves, you're guaranteed to see at least one of them on the AP test, so you need to be prepared. We started off talking about the basic concept, how to set them up and we only moved on through practise problems of increasing difficulty.

The first one we did was one where the limits were given to you, limits of integration were all set up and ready to go. Next, we moved on to where we had to find our own limits of integration because it was the area trapped between two curves. On that problem, we were able to find the limits manually, just by using a little bit of algebra. The third one involved a trigonometric function. We were finding the limits of integration, the places where the curves meet, required using your calculator. But no matter how these problems get, no matter how complex they are, they're essential still what you're doing area minus area.

You might be feeling now though like you're really pushing and things are getting harder, in other words they say you're pushing the envelope. But remember no matter how much you push the envelope, it's still stationary.

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