Video: Properties of Definite Integrals

In this video, we will learn how to use properties of definite integration, such as the order of integration limits, zero-width limits, sums, and differences.

15:42

Video Transcript

In this video, we’ll learn how to use properties of definite integration, such as the order of integration limits, zero with limits, sums, and differences. We’ll also learn how these properties can help us to simplify problems involving definite integrals.

On defining the definite integral, the integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯, we implicitly assumed that π‘Ž was less than 𝑏. But thinking about the definition of a definite integral as the limit of Reimann sums, we see that this still holds if π‘Ž is greater than 𝑏. Notice also that if we reverse π‘Ž and 𝑏, then Ξ”π‘₯ changes to π‘Ž minus 𝑏 over 𝑛. With that, we find that the definite integral between 𝑏 and π‘Ž of 𝑓 of π‘₯ with respect to π‘₯ is equal to the negative definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. Now, if π‘Ž is equal to 𝑏, then Ξ”π‘₯ becomes zero. And so, the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ is itself equal to zero. We’ll now look at some other important properties of integrals that we’re able to recall.

Now, it’s outside of the scope of this video to fully derive these. But we will have a think about where they might come from. Let’s assume we have some functions, 𝑓 and 𝑔, which are continuous. The second property we’re interested in is that the definite integral between π‘Ž and 𝑐 of 𝑓 of π‘₯ with respect to π‘₯ is equal to the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ plus the definite integral between 𝑏 and 𝑐 of 𝑓 of π‘₯ with respect to π‘₯. And geometrically, this is really to be expected. We think about the definite integral as the area between the curve and the π‘₯-axis. And so, the total area between the curve and the π‘₯-axis, bounded by the lines π‘₯ equals π‘Ž and π‘₯ equals 𝑐, must be equal to the area bounded by the lines between π‘₯ equals π‘Ž and π‘₯ equals 𝑏 and the area bounded between the lines π‘₯ equals 𝑏 and π‘₯ equals 𝑐.

For constant value 𝑐, the definite integral between π‘Ž and 𝑏 of 𝑐 with respect to π‘₯ is 𝑐 times 𝑏 minus π‘Ž. In other words, the integral of a constant function 𝑐 is that constant times the length of the integral. Using 𝑐 greater than zero and π‘Ž less than 𝑏, then this is to be expected. Since 𝑐 times 𝑏 minus π‘Ž is the area of the rectangle shown. The next property is that the integral of a sum of two functions is equal to the sum of the integrals of those functions. So for positive functions, the area under 𝑓 plus 𝑔 is equal to the area under 𝑓 plus the area under 𝑔.

Similarly, the integral of a constant times a function is equal to that constant times the integral of the function. Then, by combining the previous two properties, we find that the integral of the difference of two functions is equal to the integral of 𝑓 of π‘₯ with respect to π‘₯ plus negative one times the integral of 𝑔 of π‘₯ with respect to π‘₯. But, of course, that must be equal to the difference of the integrals of these two functions. So let’s now have a look at an example of some of these properties.

The function 𝑓 is continuous on the closed interval negative four to four and satisfies the definite integral between zero and four of 𝑓 of π‘₯ with respect to π‘₯ is equal to nine. Determine the definite integral between zero and four of 𝑓 of π‘₯ minus six with respect to π‘₯.

To answer this question, we’ll begin by recalling some basic properties of definite integrals. Firstly, we know that the integral of the sum of two functions is equal to the sum of the integral of each respective function. Similarly, the integral of the difference of two functions is equal to the difference of the integrals of those functions. This means we can begin by rewriting our integral as the difference between the integral between zero and four of 𝑓 of π‘₯ with respect to π‘₯ and the integral between zero and four of six with respect to π‘₯. We also recall that for constant function 𝑐, the definite integral between π‘Ž and 𝑏 of 𝑐 with respect to π‘₯ is equal to 𝑐 times 𝑏 minus π‘Ž.

Now, our 𝑐 is equal to six, and π‘Ž is zero, 𝑏 is four. So the definite integral between zero and four of six with respect to π‘₯ is simply six times four minus zero, which is equal to 24. As per the question, we now replace the integral between zero and four of 𝑓 of π‘₯ with respect π‘₯ with nine. And we find that the integral of 𝑓 of π‘₯ minus six between the limits of zero and four with respect to π‘₯ is nine minus 24. That’s equal to negative 15. So, despite not knowing the function 𝑓, we see that the definite integral between zero and four of 𝑓 of π‘₯ minus six with respect to π‘₯ is negative 15.

Let’s now have a look at another example involving simple properties of definite integrals.

If the definite integral between negative seven and eight of 𝑔 of π‘₯ with respect to π‘₯ is equal to 10, determine the value of the definite integral between eight and negative seven of seven 𝑔 of π‘₯ with respect to π‘₯.

To answer this question, we’re going to need to recall two specific properties of definite integrals. The first is that the definite integral between 𝑏 and π‘Ž of 𝑓 of π‘₯ with respect to π‘₯ is equal to the negative definite integral between π‘Ž and 𝑏 of that same function with respect to π‘₯. We also know that the definite integral between π‘Ž and 𝑏 of some constant 𝑐 times 𝑓 of π‘₯ with respect to π‘₯ is equal to 𝑐 times the definite integral of 𝑓 of π‘₯ with respect to π‘₯. And this second one’s great because it means we can simply take the constant outside of the integral and focus on integrating the function itself.

Let’s begin by taking that constant factor of seven outside the integral. When we do, we see that the definite integral we’re looking to evaluate is equal to seven times the definite integral between eight and negative seven of 𝑔 of π‘₯ with respect to π‘₯. Now, in the question, we’ve been told that the definite integral between negative seven and eight of our function is equal to 10. So we use the first property we recall to rewrite the definite integral between eight and negative seven of 𝑔 of π‘₯ with respect to π‘₯ as negative the definite integral between negative seven and eight of 𝑔 of π‘₯ with respect to π‘₯.

This can, of course, be simplified. And we can say that our integral is equal to negative seven times the definite integral between negative seven and eight of 𝑔 of π‘₯ with respect to π‘₯. But, of course, we know that that definite integral is equal to 10. So our definite integral is negative seven times 10, which is simply negative 70.

Now, what’s really nice about the properties we’ve considered so far is that they hold for values of π‘Ž less than 𝑏, values of π‘Ž greater than 𝑏, and values of π‘Ž equal to 𝑏. Now, there are some properties that we can consider if, and only if, π‘Ž is less than or equal to 𝑏. We call these comparison properties of the integral. And this is because they allow us to compare the general size of definite integrals.

The first says that if 𝑓 of π‘₯ is greater than or equal to zero, where π‘₯ is in the closed interval π‘Ž to 𝑏, then the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ will also be greater than zero. And, of course, this makes a lot of sense if we think about the definite integral as the area between the graph of 𝑓 and the π‘₯-axis. The geometric interpretation is simply that the area here is positive. It sits above the π‘₯-axis. The second simply says that a bigger function has a bigger integral. This also makes sense geometrically but can also be inferred from the first property because 𝑓 minus 𝑔 must be greater than or equal to zero.

The third takes a little more thinking about. If 𝑓 is continuous, we can take lowercase π‘š and capital 𝑀 to be the absolute minimum and maximum values of 𝑓 on the closed interval π‘Ž to 𝑏, respectively. This property tells us that the area under the graph of 𝑓 is greater than the area of the rectangle with a height of lowercase π‘š but less than the area of a rectangle with a height of capital 𝑀.

Let’s now have a look at a question that involves some of these properties.

Suppose that on the closed interval negative two to five, the values of 𝑓 lie in the closed interval lowercase π‘š to capital 𝑀. Between which bounds does the definite integral between negative two and five of 𝑓 of π‘₯ with respect to π‘₯ lie?

We recall one of the comparison properties of integrals. It tells us that if 𝑓 of π‘₯ is greater than or equal to lowercase π‘š and less than or equal to capital 𝑀 for values of π‘₯ greater than or equal to π‘Ž and less than or equal to 𝑏. Then 𝑀 times 𝑏 minus π‘Ž is less than or equal to the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯, which in turn is less than or equal to capital 𝑀 times 𝑏 minus π‘Ž. In other words, given that lower case π‘š is the absolute minimum of 𝑓 and uppercase 𝑀 is the absolute maximum of 𝑓, the area under the graph of 𝑓 is greater than the area of the rectangle with a height lower case π‘š. But it’s less than the area of the rectangle with a height uppercase 𝑀.

In this example, we’re going to let π‘Ž be equal to negative two and 𝑏 be equal to five. Then, we see that 𝑀 times five minus negative two is less than or equal to the definite integral between negative two and five of 𝑓 of π‘₯ with respect to π‘₯ which, in turn, is less than or equal to capital 𝑀 times five minus negative two. Five minus negative two is seven. And so, we see that our definite integral must be greater than or equal to seven π‘š and less than or equal to seven uppercase 𝑀.

The final properties we’re interested in involve integration of odd and even functions. We recall that if a function is odd, 𝑓 of negative π‘₯ is equal to negative 𝑓 of π‘₯. And a function is said to be even if 𝑓 of negative π‘₯ equals 𝑓 of π‘₯. Geometrically, an odd function has rotational symmetry about the origin. For example, 𝑦 equals sin of π‘₯. Whereas an even function has reflectional symmetry about the 𝑦-axis, the graph of 𝑦 equals cos of π‘₯, for example. For the closed interval negative π‘Ž to π‘Ž, when the function is odd, we say that the definite integral between negative π‘Ž and π‘Ž of 𝑓 of π‘₯ with respect to π‘₯ is equal to zero. And when it’s even, we find that it’s equal to two times the definite integral between zero and π‘Ž of 𝑓 of π‘₯ with respect to π‘₯.

Once again, geometrically, this makes a lot of sense. We saw that an even function has reflectional symmetry about the 𝑦-axis. So integrating between negative π‘Ž and π‘Ž would provide an area double the size of that between zero and π‘Ž. For an odd function, the areas would be equal in size but on either side of the π‘₯-axis, thereby cancelling one another out.

Let’s have a look at a couple of examples of this.

The function 𝑓 is odd, continuous on the closed interval negative one to seven, and satisfies the definite integral between one and seven of 𝑓 of π‘₯ with respect to π‘₯ equals negative 17. Determine the definite integral between negative one and seven of 𝑓 of π‘₯ with respect to π‘₯.

We’re firstly told that the function 𝑓 is odd. So we recall the following property for integrating odd functions. The definite integral between negative π‘Ž and π‘Ž of 𝑓 of π‘₯ with respect to π‘₯ is equal to zero. We’re also told that the definite integral between one and seven of 𝑓 of π‘₯ with respect to π‘₯ is equal to negative 17. So we split the integral up. And we see that the integral that we’re looking for between negative one and seven of 𝑓 of π‘₯ with respect to π‘₯ is equal to the definite integral between negative one and one of 𝑓 of π‘₯ plus the definite integral between one and seven of 𝑓 of π‘₯.

Now, the function 𝑓 is odd. So by the first property, we see that the definite integral between negative one and one of 𝑓 of π‘₯ with respect to π‘₯ must be equal to zero. Then, we simply take the definite integral between one and seven of 𝑓 of π‘₯ from the question. It’s negative 17. This means the definite integral we’re looking for is equal to zero plus negative 17 which is negative 17.

We’ll now have a look at an example that involves an even function.

The function 𝑓 is even, continuous on the closed interval negative eight to eight, and satisfies the definite integral between negative eight and eight of 𝑓 of π‘₯ with respect to π‘₯ is equal to 19 and the definite integral between zero and four of 𝑓 of π‘₯ with respect to π‘₯ is equal to 13. Determine the definite integral between negative eight and negative four of 𝑓 of π‘₯ with respect to π‘₯.

We begin by recalling the property of the integral of an even function That is, the definite integral between negative π‘Ž and π‘Ž of that even function is equal to two times the definite integral between zero and π‘Ž of 𝑓 of π‘₯ with respect to π‘₯. Now, in fact, we’re looking to find the definite integral between negative eight and negative four of our even function. So we’re going to do this in two parts. Firstly, we’re going to split it up and say that the definite integral must be equal to the integral between negative eight and zero minus the integral between negative four and zero. Now, actually, we’ll form an equation using the first part of this integral and the fact that the function is even.

The definite integral between negative eight and eight of 𝑓 of π‘₯ with respect to π‘₯ must be two times the definite integral between zero and eight of 𝑓of π‘₯ with respect to π‘₯. Now, it also follows that this must also be equal then to two times the definite integral between negative eight and zero of 𝑓 of π‘₯ with respect to π‘₯. Of course, in the question, we were told that the definite integral between negative eight and eight of 𝑓 of π‘₯ is 19. So we set 19 equal to two times the definite integral that we’re looking for. And then, we divide both sides of our equation by two. And we see that this is equal to 19 over two. The integral we’re looking for then is equal to 19 over two minus the definite integral between negative four and zero of 𝑓 of π‘₯ with respect to π‘₯.

Now, once again, the function is even. So this must, in turn, be equal to 19 over two minus the definite integral between zero and four of 𝑓 of π‘₯ with respect to π‘₯. Remember, this is because even functions have reflection or symmetry about the 𝑦-axis. Now, we’re told in the question that this definite integral is equal to 13. Then to evaluate 19 over two minus 13, we write 13 as 26 over two. So we’re looking to find 19 over two minus 26 over two which is negative seven over two. And so, we found the definite integral between negative eight and negative four of 𝑓 of π‘₯ with respect to π‘₯. It’s negative seven over two.

In this video, we’ve developed some basic properties of integrals that help us to evaluate integrals in a simple manner. For a continuous function 𝑓, the definite integral between 𝑏 and π‘Ž of 𝑓 of π‘₯ with respect to π‘₯ is equal to the negative definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. We saw that the integral of a constant function 𝑐 is the constant times the length of the interval. And that the integral of the sum or difference of two continuous functions is equal to this sum or difference of the integrals of each of those functions. We saw that we can take constant factors outside of the integral and focus on integrating the function itself.

We learned that comparison properties of the integral can be used when 𝑏 is greater than or equal to π‘Ž. The first of these says that if 𝑓 of π‘₯ is greater than or equal to zero, then the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ is also greater or equal than zero. A bigger function has a bigger integral. And if π‘š and capital 𝑀 are absolute minimum and maximum values of 𝑓 on some closed interval π‘Ž to 𝑏, then the area under the graph of 𝑓 must be greater than the area of the rectangle with height π‘š and less than the area of the rectangle with height capital 𝑀.

Finally, we learned that if a function is even or odd over some closed interval negative π‘Ž to π‘Ž, then we can apply the following rules. If it’s odd, the definite integral between negative π‘Ž and π‘Ž of 𝑓 of π‘₯ with respect to π‘₯ is zero. And if it’s even, it’s two times the definite integral between zero and π‘Ž of 𝑓 of π‘₯ with respect to π‘₯.

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