Video: The Fundamental Theorem of Calculus: Functions Defined by Integrals

The Fundamental Theorem of Calculus: Functions Defined by Integrals

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Video Transcript

The Fundamental Theorem of Calculus: Functions Defined by Integrals

In this lesson, we’ll learn how to apply the fundamental theorem of calculus to find the derivative of a function defined by an integral. The fundamental theorem of calculus has such a big, important name because it relates the two branches of calculus. At this point, we should be familiar with the fact that differential calculus gives us a way of calculating the slope of the tangent to a curve at a point. And integral calculus gives us a way of calculating area under a curve between bounds or limits. However in the 1600s, Isaac Barrow, teacher of Isaac Newton, realized that differentiation and integration, two seemingly unrelated processes, were in fact the inverse of each other. Shortly after this, Newton himself, alongside likeness, would go on to complete the development of the theory and formalize much of the notation that we’re familiar with today.

Okay, so the first part of the fundamental theorem of calculus, which we’ll be abbreviating to FTC to save some space, is as follows. If the function lower case 𝑓 is continuous on the closed interval between π‘Ž and 𝑏 and the function capital 𝐹 of π‘₯ is equal to the integral between π‘Ž and 𝑏 of 𝑓 of 𝑑 with respect to 𝑑, then the following is true. Capital 𝐹 is continuous on the closed interval between π‘Ž and 𝑏. It is also differentiable on the open interval between π‘Ž and 𝑏. And crucially, capital 𝐹 Prime of π‘₯ is equal to lower case 𝑓 of π‘₯ for all values of π‘₯ in the open interval between π‘Ž and 𝑏. A few points on the notation here, it might seem a little strange at first, but we have to find our function capital 𝐹 of π‘₯, using the definite integral of another function 𝑓 of 𝑑. And the limits of integration are π‘Ž and π‘₯.

It’s worth noting that here π‘Ž is representing a constant which has no π‘₯ dependence. Now, one of the implications of this theorem is that any continuous function, here lower case 𝑓 of π‘₯, has an antiderivative, uppercase 𝐹 of π‘₯. And of course, the important rule that we stated earlier, this theorem gives us the connection between differentiation and integration. We can perhaps see this connection more clearly by representing the derivative portion as follows. Of course we defined capital 𝐹 of π‘₯ in the following way, taking capital 𝐹 prime of π‘₯ when differentiating with respect to π‘₯. And of course, this is equal to lowercase 𝑓 of π‘₯ as per the previous line. Now that we see the result of a differential operator acting upon an integral, we might start to get a sense that these are inverse processes. We’ll come back to our theory later to try and develop an intuitive understanding. But for now, let’s look at an example.

Use the fundamental theorem of calculus to find the derivative of the function 𝑔 of π‘₯, which is equal to the integral between three and π‘₯ of the natural log of one plus 𝑑 to the power of five with respect to 𝑑.

For this question, know that we’ve been given a function 𝑔 of π‘₯, which is defined by an integral. We then been asked to find the derivative of this function. Now, our first thought might be to try and differentiate the integral with standard techniques and then to differentiate with respect to π‘₯. Here, this would be a mistake, since the integral we’ve been given would probably be messy and difficult to tackle. Instead, the question gives us a hint that we should be using the fundamental theorem of calculus, which we’ll be abbreviating to FTC. Specifically, the first part of the theorem tells us that if 𝑓 is a continuous function on the closed interval between π‘Ž and 𝑏 and capital 𝐹 of π‘₯ is defined by the integral between π‘Ž and π‘₯ of 𝑓 of 𝑑 with respect to 𝑑. Then 𝐹 prime of π‘₯ is equal to 𝑓 of π‘₯ for all values of π‘₯ on the open interval between π‘Ž and 𝑏.

This is an incredibly powerful theorem and we can understand its meaning by applying it to our question. Indeed, we know that the function we’ve been given in the question does match the form of the fundamental theorem of calculus with 𝑔 of π‘₯ representing capital 𝐹 of π‘₯, the natural log of one plus 𝑑 to the power of five representing lowercase 𝑓 of 𝑑, the lower limit of our integration three being the constant π‘Ž, and of course, the upper limit being π‘₯. Given the forms match, we can directly use the fundamental theorem of calculus to reach a result for 𝑔 prime of π‘₯, which here represents capital 𝐹 prime of π‘₯. We know the function lowercase 𝑓 of 𝑑 and so to find lowercase 𝑓 of π‘₯, we simply replace the 𝑑s by π‘₯s. This means lowercase 𝑓 of π‘₯ is equal to the natural log of one plus π‘₯ to the power of five. And in fact, we’ve already reached our answer for 𝑔 prime of π‘₯.

Now, the key insight that help us to solve this question was that we didn’t need to worry about integrating the natural log of one plus 𝑑 to the power of five. This would have led to some lengthy calculations. But instead the fundamental theorem of calculus allowed us to reach a result far more quickly. Remember to look out for trick questions of this form, where you need to find the derivative of an integral. And the function inside the integral seems in practically difficult to evaluate. The first part of the fundamental theorem of calculus will provide you with a useful alternative route.

Okay, let’s return back to our theory and take a look at a visual representation to increase our understanding. We have some function lower case 𝑓, which is continuous on the closed interval between π‘Ž and 𝑏. Now, the integral between π‘Ž and π‘₯ of 𝑓 of 𝑑 with respect to 𝑑 can be thought of as the area under the curve between the limits of integration π‘Ž and π‘₯. We can define a function capital 𝐹 as this area. So capital 𝐹 of π‘₯ is equal to the integral between π‘Ž and π‘₯ of 𝑓 of 𝑑 with respect to 𝑑. Now, it’s worth noting that π‘₯ is a variable and can take any value within the interval between π‘Ž and 𝑏, over which we’ve stated continuity.

To show what we mean by this, let’s add some labels. Our first diagram, we have an π‘₯ one. And on our second diagram, we have an π‘₯ two. For our first diagram, the area is therefore expressed by capital 𝐹 of π‘₯ one. And for our second diagram, the area is expressed by capital 𝐹 of π‘₯ two. In this example, the area, 𝐹 of π‘₯ two, is clearly bigger than 𝐹 of π‘₯ one. Okay, but the fundamental theorem of calculus actually concerns the derivative 𝑓 prime of π‘₯. Of course, this is capital 𝐹 of π‘₯ differentiated with respect to π‘₯, so d by dπ‘₯ of this integral.

Okay, but what does this actually mean? Well, if we think of our integral as the area under a curve and we think of a derivative as the rate of change, then capital 𝐹 prime of π‘₯, the derivative of our integral, is the rate of change of this area with respect to π‘₯. In other words, what’s the rate at which the area of this geometric shape changes due to a change in π‘₯? We won’t go too far into infinite decimals or limits here. But we can get a visual understanding again with our graphs. The fundamental theorem of calculus tells us that capital 𝐹 prime of π‘₯ is equal to lowercase 𝑓 of π‘₯. So the rate of change of the area is the height of the curve at this point. The rate of change of this area is the function lowercase 𝑓 evaluated at π‘₯ one. And the rate of change of this area is the function lowercase 𝑓 evaluated at π‘₯ two.

Visually, it should make sense that if the right-hand side of our geometric shape has a greater height, then its area changes by a larger amount as π‘₯ changes. In our example, we see this represented in the fact that 𝑓 of π‘₯ two is greater than 𝑓 of π‘₯ one. A final point to tie up this visual example, don’t get confused into thinking the rate of change of the area is the gradient of the curve. Remember, it’s the height. Again, we haven’t dived too far into the details of infinite decimals here, but hopefully this should strengthen our intuition when it comes to the underlying principles behind the fundamental theorem of calculus.

Let’s now take a look at another example to practice applying this.

Use the fundamental theorem of calculus to find the derivative of the function 𝑅 of 𝑦 is equal to the integral between 𝑦 and five of three 𝑑 squared sin two 𝑑 with respect to 𝑑.

For this question, we first recognize that we’ve been given a function defined by an integral capital 𝑅 of 𝑦. And we’ve been asked to find the derivative of this, capital 𝑅 prime of 𝑦. A tool that we can use to solve this problem, as the question tells us, is the fundamental theorem of calculus. The first part of this tells us that if lowercase 𝑓 is a continuous function on the closed interval between π‘Ž and 𝑏 and the function capital 𝐹 of π‘₯ is defined by the integral between π‘Ž and π‘₯ of lowercase 𝑓 of 𝑑 with respect to 𝑑. Then 𝐹 prime of π‘₯ is equal to 𝑓 of π‘₯ for all values of π‘₯ on the open interval between π‘Ž and 𝑏. Now, the equation given in our question does not have a capital 𝐹 of π‘₯, but instead we have a capital 𝑅 of 𝑦.

Before we start to think about 𝑅 prime of 𝑦, let’s dial things back to 𝑅 of 𝑦 to see if we have the correct form of equation to apply the fundamental theorem of calculus. We do indeed have a function defined by an integral and a continuous function as our integrand. However, the variable that we’re operating on, 𝑦, appears as the lower limit of integration, and we have a constant as the upper limit. This is the other way round to the fundamental theorem of calculus, where the variable is the upper limit and the constant is the lower limit. This means the positions of our limits are reversed. Luckily, one of the properties of integrals is that switching the positions of the limits and multiplying by negative one is equal to the original Integral. We can do this to the integral, which defines 𝑅 of 𝑦. And of course, it doesn’t matter whether this factor of negative one is inside or outside of our integral.

Now that our equation is in the correct form with a variable appearing as the upper limit and the constant appearing as the lower limit, we can use the fundamental theorem of calculus. This allows us to say that capital 𝑅 prime of 𝑦 is equal to 𝑓 of 𝑦. And remember the variable is 𝑦 not π‘₯. Looking back at our integral, we have to find lowercase 𝑓 of 𝑑 to be negative three 𝑑 squared sin two 𝑑. This means that for lowercase 𝑓 of 𝑦 we’ll replace all of our 𝑑s for 𝑦s. We, therefore, find that capital 𝑅 prime of 𝑦 is equal to negative three 𝑦 squared sin two 𝑦. And we have, therefore, answered our question. We have found the derivative capital 𝑅 prime of 𝑦 using the fundamental theorem of calculus. And this allowed us to avoid evaluating the definite integral given in the question which may have led to messy or lengthy calculations.

Okay, in the examples that we have seen so far, the limits of our integral have involved a constant π‘Ž and the variable on which our function capital 𝐹 is operating on which is, of course, π‘₯. What if, instead, this limit was no π‘₯ but π‘₯ squared. Or indeed, the limit was some other function of π‘₯. Let’s say 𝑒 of π‘₯. You may come across various equations of this form and we’ll see how to deal with them in the following example.

Use the fundamental theorem of calculus to find the derivative of the function 𝑦 of π‘₯ equals the integral between two and π‘₯ to the power of four of five cos squared five πœƒ with respect to πœƒ.

Here, we have a function 𝑦 of π‘₯, which is defined by an integral. To find this derivative, instead of evaluating the integral directly, we’re gonna be using the fundamental theorem of calculus. The first part of the theorem tells us that if lowercase 𝑓 is a continuous function on the closed interval between π‘Ž and 𝑏 and we have another function capital 𝐹 of π‘₯, which is defined by the integral between π‘Ž and π‘₯ of lowercase 𝑓 of 𝑑 with respect to 𝑑. Then 𝐹 prime of π‘₯ is equal to lowercase 𝑓 of π‘₯ for all π‘₯ on the open interval between π‘Ž and 𝑏. Let’s now check that we have the correct form to use.

The function defined by the integral is 𝑦 of π‘₯ instead of capital 𝐹 of π‘₯. Our integrand is a function which is in fact continuous over the entire set of real numbers. And this is lowercase 𝑓 of πœƒ instead of lowercase 𝑓 of 𝑑. The lower limit of our integral is two, which is a constant. However, here we run into the problem and that the upper limit is not π‘₯ but rather is π‘₯ to the power of four. And this is a function of π‘₯. To use the fundamental theorem of calculus, we’ll therefore need to come up with a modification. First, we express 𝑦 of π‘₯ in a slightly tidier form. Next, we’ll define our troublesome upper limit as something else, for example, the variable 𝑒. We then have that 𝑦 of π‘₯ is equal to the integral between two and 𝑒 of 𝑓 of πœƒ dπœƒ.

Now, we need to find 𝑦 prime of π‘₯, which is 𝑦 of π‘₯ differentiated with respect to π‘₯. We can write this as d by dπ‘₯ of the integral which we formed here. This is the step at which the problem would occur. We can’t directly use the fundamental theorem of calculus to evaluate, yet, since our upper limit does not match the variable π‘₯. One tool that we can use to move forward is the chain rule which tells us that d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯. Of course, here we have a d𝑦 by dπ‘₯. At this crucial step, the chain rule allows us to re express this as d𝑦 by d𝑒 times d𝑒 by dπ‘₯. Looking at this part of the equation, we now see that it can be evaluated using the fundamental theorem of calculus since we’re taking a derivative with respect to 𝑒 and our upper limit, is indeed 𝑒.

You may be able to see this more clearly by observing that the form now matches as shown here. Using the theorem we’re able to say that this is equal to 𝑓 of 𝑒. And hence, 𝑦 Prime of π‘₯ is equal to 𝑓 of 𝑒 d by dπ‘₯. In fact, this is an extremely useful generalization that we can use when the limits of our integral involve a function of π‘₯ rather than π‘₯ itself. To be forward with our question, we can now substitute back into the equation using our definition, which is that 𝑒 is equal to π‘₯ to the power of four. We now recall that 𝑓 of πœƒ is equal to five cos squared five πœƒ. Replacing our πœƒ with π‘₯ to the power of four, we get that 𝑓 of π‘₯ to the power of four is equal to five cos squared five π‘₯ to the power of four.

Next, we need to differentiate π‘₯ the power of four with respect to π‘₯. And of course, this is four π‘₯ cubed. Multiplying these two things together, we are left with 20π‘₯ cubed times cos squared five π‘₯ to the power of four. And finally, we have reached the answer to our question, since this is 𝑦 prime of π‘₯.

This example illustrates a very useful modification to the fundamental theorem of calculus. The method can be generalized to deal with integrals, which have limits taking other functions of π‘₯. Let’s now take a look at one final example.

Find the derivative of the function 𝑔 of π‘₯ is equal to the integral between one minus two π‘₯ and one plus π‘₯ of five 𝑑 sin 𝑑 with respect to 𝑑.

For this question, we’ve been given a function defined by an integral, 𝑔 of π‘₯. And we need to find its derivative, 𝑔 prime of π‘₯. To do so we’ll be using the first part of the fundamental theorem of calculus, which tells us if we have an equation in this form, we can directly find its derivative using the following rules. Here, we have rewritten our equation. Setting 𝑓 of 𝑑 equal to five 𝑑 sin 𝑑. Now, the first thing we will notice when looking at our integral is that not only other top and bottom limits functions of π‘₯ instead of π‘₯ itself. But that of them are a constant, which we need to use our theorem.

At this point, we can recall that an integral can be split in the following way. To conceptualize this, it can be helpful to interpret a definite integral is the area under a curve as shown. Using this method we can artificially introduce a constant which we’ll call π‘Ž into the integrals which now form our sum. Next, we noticed that our first integral has this constant as its upper limit instead of its lower limit. And we need to switch things round to use the fundamental theorem of calculus. Luckily, switching the bounds of an integral and multiplying by minus one evaluates to the same result. And so we can do this.

Okay, we’re a bit closer to the form we need now. However, remember, the upper limit of both of these integrals instead of being π‘₯ is a function of π‘₯. To move forward, we can use the following modified form of the fundamental theorem of calculus. If lowercase 𝑓 is a continuous function over the closed interval between π‘Ž and 𝑏, and uppercase 𝐹 of π‘₯ is defined by the integral between π‘Ž and some function of π‘₯, which we’ll call 𝑒 of π‘₯ of 𝑓of 𝑑 with respect to 𝑑. Then capital 𝐹 prime of π‘₯ is equal to lower case 𝑓 of 𝑒 of π‘₯ multiplied by d by dπ‘₯ of 𝑒 of π‘₯, where 𝑒 of π‘₯ is in the open interval between π‘Ž and 𝑏. Okay, so looking back at 𝑔 of π‘₯, it is now expressed is the sum of two separate integrals.

Given the rules of differentiation, we confined 𝑔 prime of π‘₯ simply by differentiating each of these terms individually. And therefore, our modified theorem can also be applied individually to both of the terms. For the first term our upper limit is the function one minus two π‘₯. Our modified theorem tells us that this is, therefore, equal to negative 𝑓 of one minus two π‘₯ times d by dπ‘₯ of one minus two π‘₯. Our second term takes the same form, but using the upper limit of one plus π‘₯. We now remember that 𝑓 of 𝑑 is equal to five 𝑑 sin 𝑑. For 𝑓 of one minus two π‘₯, we’ll replace all of the 𝑑 terms in this equation with one minus two π‘₯. We must then multiply this by the derivative of one minus two π‘₯.

For our second term, we follow exactly the same pattern, first substituting in to lowercase 𝑓 and then finding the derivative of one plus π‘₯ and multiplying. After a bit of simplification, we’re left with the following result. We have now completed our question and we have found the derivative 𝑔 prime of π‘₯. This example illustrates that we can use the modified version of the fundamental theorem of calculus even when the function that we have is defined by an integral involving two different functions of π‘₯ in the upper and lower limits.

Great to finish off, let’s go through some key points. The first part of the fundamental theorem of calculus tells us that if lowercase 𝑓 is a continuous function on the closed interval between π‘Ž and 𝑏, and the function capital 𝐹 of π‘₯ is defined by the integral between π‘Ž and π‘₯ of lowercase 𝑓 of 𝑑 with respect to 𝑑. Then capital 𝐹 is continuous on the closed interval between π‘Ž and 𝑏, is differentiable on the open interval between π‘Ž and 𝑏, and capital 𝐹 prime of π‘₯ is equal to lowercase 𝑓 of π‘₯ for all values of π‘₯ on the open interval between π‘Ž and 𝑏. If the variable appears as the lower limit of integration instead of the upper, you can swap the limits and multiply by negative one, which is equal to the original Integral.

The first part of the fundamental theorem of calculus can be modified when the limit or limits involve functions of π‘₯ instead of π‘₯ itself. And here this function we’ve called 𝑒 of π‘₯. When a function is defined by an integral in which both of the limits have an π‘₯ dependence, the interval could be split to artificially include constants in the limits. You can then swap the bounds as shown previously and then multiply it by negative one. The question can then be solved by using the modified fundamental theorem of calculus on each of these integrals individually.

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