Lesson Video: Area between Curves Mathematics • Higher Education

In this video, we will learn how to apply integration to find the area bounded by the curves of two or more functions.

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Video Transcript

In this video, we’ll learn how to apply integration to find the area bounded by the curves of two or more functions. By this stage, you should feel confident in applying processes for integration to evaluate definite and indefinite integrals. We’ll now look at how integration can help us to find the area of the regions that lie between the graphs of two or more functions.

Consider the region that lies between the curve with equation 𝑦 equals 𝑓 of π‘₯ and the π‘₯-axis which is bounded by the vertical lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏. I’ve shaded this region pink. If 𝑓 is a continuous function, we know that we can evaluate the area of this region by integrating the function 𝑓 of π‘₯ with respect to π‘₯ between the limits of π‘Ž and 𝑏.

Now, let’s add another curve to our diagram. This time, the curve has the equation 𝑦 equals 𝑔 of π‘₯, where 𝑔 is continuous and 𝑔 of π‘₯ is less than or equal to 𝑓 of π‘₯ in the closed interval π‘Ž to 𝑏. Once again, we can find the area of the region between the curve 𝑦 equals 𝑔 of π‘₯ and the π‘₯-axis and these two vertical lines, now I’ve shaded this region in yellow, by evaluating the integral of 𝑔 of π‘₯ between the limits of π‘Ž and 𝑏.

We can now see that if we subtract the area between the curve 𝑔 of π‘₯ and the π‘₯-axis from the area between the curve of 𝑓 of π‘₯ and the π‘₯-axis, that we’ll be left with this region 𝐴 three. This is the region between the two curves 𝑦 equals 𝑓 of π‘₯ and 𝑦 equals 𝑔 of π‘₯. We can, therefore, say that the area 𝐴 three of the region bounded by 𝑦 equals 𝑓 of π‘₯ and 𝑦 equals 𝑔 of π‘₯ is the integral of 𝑓 of π‘₯ evaluated between π‘Ž and 𝑏 minus the integral of 𝑔 of π‘₯ evaluated between π‘Ž and 𝑏.

But we also know that the sum or difference of the integral of two functions is equal to the integral of the sum or difference of these functions. So, we can say that the area is equal to the integral of 𝑓 of π‘₯ minus 𝑔 of π‘₯ with respect to π‘₯ evaluated between π‘₯ equals π‘Ž and π‘₯ equals 𝑏. This brings us to our first definition. The area 𝐴 of the region bounded by the curves 𝑦 equals 𝑓 of π‘₯ and 𝑦 equals 𝑔 of π‘₯, and the lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏, where 𝑓 and 𝑔 are continuous and 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ in the closed interval π‘Ž to 𝑏, is the definite integral of 𝑓 of π‘₯ minus 𝑔 of π‘₯ evaluated between the limits of π‘Ž and 𝑏.

Notice here that 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ for all π‘₯ in between and including π‘Ž and 𝑏. We’ll need to watch carefully for situations where this is not the case and apply some extra logic. For now though, we’ll look at the application of this formula.

Find the area of the region bounded by the curves 𝑦 equals three π‘₯ squared minus five π‘₯ and 𝑦 equals negative five π‘₯ squared.

We’ll recall that the area of the region bounded by the curves 𝑦 equals 𝑓 of π‘₯, 𝑦 equals 𝑔 of π‘₯, and the lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏 for continuous functions 𝑓 and 𝑔, such that 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ for all π‘₯ in the closed interval π‘Ž to 𝑏, is the definite integral between the limits of π‘Ž and 𝑏 of 𝑓 of π‘₯ minus 𝑔 of π‘₯. We’re, therefore, going to need to define the functions 𝑓 of π‘₯ and 𝑔 of π‘₯ really carefully and, of course, the values for π‘Ž and 𝑏, ensuring that 𝑓 of π‘₯ is greater than 𝑔 of π‘₯ in the closed interval π‘Ž to 𝑏.

The lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏 will mark the beginning and end of the region we’re interested in. So, what are the equations of these lines? They’re the π‘₯-coordinates at the points where the two graphs intercept. So, we can set the equations three π‘₯ squared minus five π‘₯ and negative five π‘₯ squared equal to each other and solve for π‘₯.

We begin by adding five π‘₯ squared to both sides. And then, we factor the expression on the left-hand side by taking out that factor of π‘₯. And we obtain π‘₯ times eight π‘₯ minus five to be equal to zero. We know that for this statement to be true, either π‘₯ itself must be equal to zero or eight π‘₯ minus five must be equal to zero. To solve this equation on the right, we add five and then divide through by eight. And we obtain π‘₯ to be equal to five-eighths.

So, we can see that the π‘₯-coordinates of the points of intersection of our two curves are zero and five-eighths. So, π‘Ž is equal to zero and 𝑏 is equal to five-eighths. Now, we’re going to need to decide which function is 𝑓 of π‘₯ and which function is 𝑔 of π‘₯. What we do next is sketch out the graphs of 𝑦 equals three π‘₯ squared minus five π‘₯ and 𝑦 equals negative five π‘₯ squared. We’re looking to establish which of the curves is essentially on top.

We know that the graph of 𝑦 equals three π‘₯ squared minus five π‘₯ is a U-shaped parabola. We can even factor the expression three π‘₯ squared minus five π‘₯, set it equal to zero, and solve for π‘₯. And we see that it passes through the π‘₯-axis at zero and five-thirds. So, it will look a little something like this. The graph of 𝑦 equals negative five π‘₯ squared is an inverted parabola which passes through the origin like this. And so, we obtain the region shaded.

We can now see that in the closed interval of zero to five-eighths, the function that’s on top, if you will, is the function defined by 𝑦 equals negative five π‘₯ squared. So, we can say that 𝑓 of π‘₯ is equal to negative five π‘₯ squared. Meaning, 𝑔 of π‘₯ is three π‘₯ squared minus five π‘₯. The area that we’re interested in must, therefore, be given by the definite integral evaluated between zero and five-eighths of negative five π‘₯ squared minus three π‘₯ squared minus five π‘₯ with respect to π‘₯.

Distributing the parentheses, and our integrand becomes negative eight π‘₯ squared plus five π‘₯. But wait a minute, we know that when we evaluate areas below the π‘₯-axis, we end up with a funny result. We get a negative value. You might wish to pause the video for a moment and consider what that means in this example. Did you work it out? We can see that our entire region sits below the π‘₯-axis and we’re just working out the difference between the areas. So, the negative results that we would obtain from integrating each function individually will simply cancel each other out. So, all that’s left is to evaluate this integral.

The integral of negative eight π‘₯ squared is negative eight π‘₯ cubed over three. And the integral of five π‘₯ is five π‘₯ squared over two. We need to evaluate this between zero and five-eighths, which is negative eight-thirds of five-eighths cubed plus five over two times five-eighths squared minus zero. That’s 125 over 384 square units. This question was fairly straightforward as the curve of 𝑦 equals negative five π‘₯ squared was greater than or equal to the curve of 𝑦 equals three π‘₯ squared minus five π‘₯ in the interval we’re interested in.

Let’s now look at what we might do if this wasn’t the case.

The curves shown are 𝑦 equals one over π‘₯ and 𝑦 equals one over π‘₯ squared. What is the area of the shaded region? Give an exact answer.

Remember, the area of a region bounded by the curves 𝑦 equals 𝑓 of π‘₯, 𝑦 equals 𝑔 of π‘₯, and the lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏 for continuous functions 𝑓 of 𝑔, when 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ for π‘₯ in the closed interval π‘Ž to 𝑏, is given by the definite integral evaluated between π‘Ž and 𝑏 of 𝑓 of π‘₯ minus 𝑔 of π‘₯.

Now, we do have a little bit of a problem here. We can see quite clearly that the region is bounded by the vertical lines π‘₯ equals 0.5 and π‘₯ equals two. But in the closed interval π‘₯ from 0.5 to two, we can see that one of our functions is not always greater than or equal to the other. So, we can’t actually apply this definition. We can see, however, that if we split our region up a little more, we do achieve that requirement. I’ve added a third line at the point where the two curves intersect. This has the equation π‘₯ equals one.

In the closed interval 0.5 to one, the values on the red line are always greater than or equal to those on the green line. And in the closed interval π‘₯ between one and two, the reverse is true. So, all we do is split our region up and then add the values at the end. Let’s find the area of our first region, 𝑅 one. To do so, we’re going to need to double check which line is which. We can probably deduce that the red line is more likely to be one over π‘₯ squared. But let’s choose a coordinate pair and substitute these values in just to be safe.

We can see that the curve passes through the point with coordinates 0.5, 4. So, let’s substitute π‘₯ equals 0.5 into the equation 𝑦 equals one over π‘₯ squared. When we do, we get 𝑦 equals one over 0.5 squared, which is one over 0.25, which is four as required. So, the red line has the equation 𝑦 equals one over π‘₯ squared and the green line has equation 𝑦 equals one over π‘₯. And when evaluating the area of 𝑅 one, 𝑓 of π‘₯ is therefore one over π‘₯ squared and 𝑔 of π‘₯ is equal to one over π‘₯.

The area is, therefore, given by the definite integral between the limits of 0.5 and one of one over π‘₯ squared minus one over π‘₯. So, all that’s left here is to evaluate this integral. This is much easier to do if we rewrite one over π‘₯ squared as π‘₯ to the power of negative two and then recall some standard results. To integrate π‘₯ to the power of negative two, we add one to the power and then divide by this new number. That gives us π‘₯ to the power of negative one over negative one, which is negative one over π‘₯. The integral of one over π‘₯, however, is the natural log of the absolute value of π‘₯. So, our integral is negative one over π‘₯ minus the natural log of the absolute value of π‘₯.

We’re going to now evaluate this between π‘₯ equals 0.5 and π‘₯ equals one. That’s negative one over one minus the natural log of one minus negative one over 0.5 minus the natural log of 0.5. And notice, I’ve lost the symbol for the absolute value because one and 0.5 are already positive. The natural log of zero is one. Negative one over one is negative one. And negative one over 0.5 is two. I’ve also rewritten the natural log of 0.5 as the natural log of a half and distributed the parentheses. And this simplifies to one plus the natural log of one-half.

A really important skill, though, is to be able to spot when we can further simplify a logarithmic term. If we rewrite the natural log of a half as the natural log of two to the power of negative one. And then, use the fact that the natural log of π‘Ž to the 𝑏th power is equal to 𝑏 times the natural log of π‘Ž. We see that the exact area of the first region 𝑅 one is one minus the natural log of two.

Let’s clear some space and repeat this process for region two. This time, the green line is above the red line, so we’re going to let 𝑓 of π‘₯ be equal to one over π‘₯ and 𝑔 of π‘₯ be equal to one over π‘₯ squared. Our area is the definite integral between one and two of one over π‘₯ minus one over π‘₯ squared which, when we integrate, gives us the natural log of the absolute value of π‘₯ plus one over π‘₯. Evaluating between the limits of one and two, and we get the natural log of two plus a half minus the natural log of one plus one, which is equal to the natural log of two minus a half.

We want to find the area of the whole region, so we add these two values. It’s one minus the natural log of two plus the natural log of two minus one-half, which simplifies to one-half. The area of the shaded region is a half square units. In this example, we saw that the area formula can be applied to find the area between two curves where one curve is above the other for part of the integration interval and the opposite in the second part of the interval, as long as we remember to split the region up at the point where the curves intersect.

We’ll now see how we can develop this formula further to help us find the region bounded by three curves.

Find the area of the region bounded by the curves 𝑦 equals four minus π‘₯ squared, 𝑦 equals negative π‘₯, and 𝑦 equals the square root of π‘₯. Give your answer correct to one decimal place.

Remember, for continuous functions 𝑓 and 𝑔, the area of the region bounded by the curves 𝑦 equals 𝑓 of π‘₯, 𝑦 equals 𝑔 of π‘₯, and the lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏, as long as 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ in the closed interval π‘Ž to 𝑏, is given by the integral evaluated between π‘Ž and 𝑏 of 𝑓 of π‘₯ minus 𝑔 of π‘₯. We are going to need to be a little bit careful here, as we have three curves. So, let’s begin by sketching this out and see what we’re dealing with.

The area enclosed between the three curves looks a little something like this. Now, if we’re really clever, we can actually use the definition we looked up before. We can split this region into the region above the π‘₯-axis and the region below the π‘₯-axis. We then can split this up a little bit further. We see that we have 𝑅 one, that’s the region between the π‘₯-axis and the curve 𝑦 equals root π‘₯ between π‘₯ equals zero and π‘₯ equals 𝑏. Where 𝑏 is the π‘₯-coordinate at the point of intersection of the curve 𝑦 equals root π‘₯ and 𝑦 equals four minus π‘₯ squared.

We then have 𝑅 two. That’s the region between 𝑦 equals four minus π‘₯ squared, π‘₯ equals 𝑏, and π‘₯ equals two. And the reason we’ve chosen π‘₯ equals two as our upper limit is that’s the π‘₯-value at the point at which the curve crosses the π‘₯-axis. We can even split our three up into two further regions to make life easier. But let’s deal first with the area of 𝑅 one and 𝑅 two. We need to work out the value of 𝑏.

We said it’s the π‘₯-coordinate at the point of intersection of the two curves four minus π‘₯ squared and root π‘₯. So, we set these equal to each other and solve for π‘₯. That gives us an π‘₯-value of 1.648, correct to three decimal places. So, 𝑏 is equal to 1.648. We can either do this by hand or use our graphical calculators to evaluate each of these integrals. The area of 𝑅 one becomes 1.4104 and so on. And the area of 𝑅 two is 0.23326 and so on.

Let’s now consider the area of 𝑅 three. It’s the integral between zero and two of negative π‘₯ evaluated with respect to π‘₯. We need to be a little bit careful here, since this is below the π‘₯-axis and therefore will yield a negative result on integration. In fact, it gives us negative two. So, we can say that the area is the absolute value of this. It’s two. And notice, we could have actually used the formula for area of a triangle to work this area out.

Now, the area of 𝑅 four is the definite integral between two and 𝑐 of four minus π‘₯ squared minus negative π‘₯. And here, 𝑐 is the π‘₯-coordinate of the point of intersection of the lines 𝑦 equals negative π‘₯ and 𝑦 equals four minus π‘₯ squared. We can once again set four minus π‘₯ squared equal to negative π‘₯ and solve for π‘₯. And we find, correct to three decimal places, that they intersect at the point where π‘₯ equals 2.562. We type this into our calculator and we find that the area of this region is 0.59106. We find the total of these four values, which gives us 4.2347, which is 4.2 square units, correct to one decimal place.

In this video, we’ve seen that we can use the formula area is equal to the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ minus 𝑔 of π‘₯ with respect to π‘₯ for continuous functions, 𝑓 and 𝑔 as long as 𝑓 of π‘₯ is greater than or equal to 𝑔 for all π‘₯ in the closed interval π‘Ž to 𝑏. We also saw that for more complicated regions such as those bounded by three or more curves, those which involve regions above and below the π‘₯-axis, or those where 𝑓 of π‘₯ and 𝑔 of π‘₯ switch, we might need to split this region up a little further.

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