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 𝑎 and 𝑏 and the function capital 𝐹 of
𝑥 is equal to the integral between 𝑎 and 𝑏 of 𝑓 of 𝑡 with respect to 𝑡, then
the following is true. Capital 𝐹 is continuous on the
closed interval between 𝑎 and 𝑏. It is also differentiable on the
open interval between 𝑎 and 𝑏. And crucially, capital 𝐹 Prime of
𝑥 is equal to lower case 𝑓 of 𝑥 for all values of 𝑥 in the open interval between
𝑎 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 𝑥, using the definite integral of another function 𝑓 of 𝑡. And the limits of integration are
𝑎 and 𝑥.
It’s worth noting that here 𝑎 is
representing a constant which has no 𝑥 dependence. Now, one of the implications of
this theorem is that any continuous function, here lower case 𝑓 of 𝑥, has an
antiderivative, uppercase 𝐹 of 𝑥. 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
𝑥 in the following way, taking capital 𝐹 prime of 𝑥 when differentiating with
respect to 𝑥. And of course, this is equal to
lowercase 𝑓 of 𝑥 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 𝑥, which is equal to the
integral between three and 𝑥 of the natural log of one plus 𝑡 to the power of five
with respect to 𝑡.
For this question, know that we’ve
been given a function 𝑔 of 𝑥, 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 𝑥. 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
𝑎 and 𝑏 and capital 𝐹 of 𝑥 is defined by the integral between 𝑎 and 𝑥 of 𝑓 of
𝑡 with respect to 𝑡. Then 𝐹 prime of 𝑥 is equal to 𝑓
of 𝑥 for all values of 𝑥 on the open interval between 𝑎 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 𝑥 representing capital 𝐹 of 𝑥, the natural log of one plus 𝑡
to the power of five representing lowercase 𝑓 of 𝑡, the lower limit of our
integration three being the constant 𝑎, and of course, the upper limit being
𝑥. Given the forms match, we can
directly use the fundamental theorem of calculus to reach a result for 𝑔 prime of
𝑥, which here represents capital 𝐹 prime of 𝑥. We know the function lowercase 𝑓
of 𝑡 and so to find lowercase 𝑓 of 𝑥, we simply replace the 𝑡s by 𝑥s. This means lowercase 𝑓 of 𝑥 is
equal to the natural log of one plus 𝑥 to the power of five. And in fact, we’ve already reached
our answer for 𝑔 prime of 𝑥.
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 𝑡 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 𝑎 and 𝑏. Now, the integral between 𝑎 and 𝑥
of 𝑓 of 𝑡 with respect to 𝑡 can be thought of as the area under the curve between
the limits of integration 𝑎 and 𝑥. We can define a function capital 𝐹
as this area. So capital 𝐹 of 𝑥 is equal to the
integral between 𝑎 and 𝑥 of 𝑓 of 𝑡 with respect to 𝑡. Now, it’s worth noting that 𝑥 is a
variable and can take any value within the interval between 𝑎 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 𝑥
one. And on our second diagram, we have
an 𝑥 two. For our first diagram, the area is
therefore expressed by capital 𝐹 of 𝑥 one. And for our second diagram, the
area is expressed by capital 𝐹 of 𝑥 two. In this example, the area, 𝐹 of 𝑥
two, is clearly bigger than 𝐹 of 𝑥 one. Okay, but the fundamental theorem
of calculus actually concerns the derivative 𝑓 prime of 𝑥. Of course, this is capital 𝐹 of 𝑥
differentiated with respect to 𝑥, so d by d𝑥 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 𝑥, the derivative of our integral, is the rate of change of
this area with respect to 𝑥. In other words, what’s the rate at
which the area of this geometric shape changes due to a change in 𝑥? 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 𝑥 is equal to lowercase 𝑓 of 𝑥. 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 𝑥 one. And the rate of change of this area
is the function lowercase 𝑓 evaluated at 𝑥 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 𝑥 changes. In our example, we see this
represented in the fact that 𝑓 of 𝑥 two is greater than 𝑓 of 𝑥 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 𝑡 squared sin two 𝑡 with respect to 𝑡.
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 𝑎 and
𝑏 and the function capital 𝐹 of 𝑥 is defined by the integral between 𝑎 and 𝑥 of
lowercase 𝑓 of 𝑡 with respect to 𝑡. Then 𝐹 prime of 𝑥 is equal to 𝑓
of 𝑥 for all values of 𝑥 on the open interval between 𝑎 and 𝑏. Now, the equation given in our
question does not have a capital 𝐹 of 𝑥, 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
𝑥. Looking back at our integral, we
have to find lowercase 𝑓 of 𝑡 to be negative three 𝑡 squared sin two 𝑡. This means that for lowercase 𝑓 of
𝑦 we’ll replace all of our 𝑡s 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 𝑎 and the variable
on which our function capital 𝐹 is operating on which is, of course, 𝑥. What if, instead, this limit was no
𝑥 but 𝑥 squared. Or indeed, the limit was some other
function of 𝑥. Let’s say 𝑢 of 𝑥. 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 𝑥 equals the integral between
two and 𝑥 to the power of four of five cos squared five 𝜃 with respect to 𝜃.
Here, we have a function 𝑦 of 𝑥,
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 𝑎
and 𝑏 and we have another function capital 𝐹 of 𝑥, which is defined by the
integral between 𝑎 and 𝑥 of lowercase 𝑓 of 𝑡 with respect to 𝑡. Then 𝐹 prime of 𝑥 is equal to
lowercase 𝑓 of 𝑥 for all 𝑥 on the open interval between 𝑎 and 𝑏. Let’s now check that we have the
correct form to use.
The function defined by the
integral is 𝑦 of 𝑥 instead of capital 𝐹 of 𝑥. Our integrand is a function which
is in fact continuous over the entire set of real numbers. And this is lowercase 𝑓 of 𝜃
instead of lowercase 𝑓 of 𝑡. 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 𝑥 but rather is 𝑥 to the power of
four. And this is a function of 𝑥. To use the fundamental theorem of
calculus, we’ll therefore need to come up with a modification. First, we express 𝑦 of 𝑥 in a
slightly tidier form. Next, we’ll define our troublesome
upper limit as something else, for example, the variable 𝑢. We then have that 𝑦 of 𝑥 is equal
to the integral between two and 𝑢 of 𝑓 of 𝜃 d𝜃.
Now, we need to find 𝑦 prime of
𝑥, which is 𝑦 of 𝑥 differentiated with respect to 𝑥. We can write this as d by d𝑥 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 𝑥. One tool that we can use to move
forward is the chain rule which tells us that d𝑦 by d𝑥 is equal to d𝑦 by d𝑢
times d𝑢 by d𝑥. Of course, here we have a d𝑦 by
d𝑥. At this crucial step, the chain
rule allows us to re express this as d𝑦 by d𝑢 times d𝑢 by d𝑥. 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 𝑢 and our upper limit, is
indeed 𝑢.
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 𝑢. And hence, 𝑦 Prime of 𝑥 is equal
to 𝑓 of 𝑢 d𝑢 by d𝑥. In fact, this is an extremely
useful generalization that we can use when the limits of our integral involve a
function of 𝑥 rather than 𝑥 itself. To be forward with our question, we
can now substitute back into the equation using our definition, which is that 𝑢 is
equal to 𝑥 to the power of four. We now recall that 𝑓 of 𝜃 is
equal to five cos squared five 𝜃. Replacing our 𝜃 with 𝑥 to the
power of four, we get that 𝑓 of 𝑥 to the power of four is equal to five cos
squared five 𝑥 to the power of four.
Next, we need to differentiate 𝑥
the power of four with respect to 𝑥. And of course, this is four 𝑥
cubed. Multiplying these two things
together, we are left with 20𝑥 cubed times cos squared five 𝑥 to the power of
four. And finally, we have reached the
answer to our question, since this is 𝑦 prime of 𝑥.
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 𝑥. Let’s now take a look at one final
example.
Find the derivative of the function
𝑔 of 𝑥 is equal to the integral between one minus two 𝑥 and one plus 𝑥 of five
𝑡 sin 𝑡 with respect to 𝑡.
For this question, we’ve been given
a function defined by an integral, 𝑔 of 𝑥. And we need to find its derivative,
𝑔 prime of 𝑥. 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 𝑡 equal to five 𝑡
sin 𝑡. Now, the first thing we will notice
when looking at our integral is that not only other top and bottom limits functions
of 𝑥 instead of 𝑥 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 𝑎 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 𝑥 is a function of 𝑥. 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 𝑎 and 𝑏, and uppercase 𝐹 of 𝑥 is
defined by the integral between 𝑎 and some function of 𝑥, which we’ll call 𝑢 of
𝑥 of 𝑓of 𝑡 with respect to 𝑡. Then capital 𝐹 prime of 𝑥 is
equal to lower case 𝑓 of 𝑢 of 𝑥 multiplied by d by d𝑥 of 𝑢 of 𝑥, where 𝑢 of
𝑥 is in the open interval between 𝑎 and 𝑏. Okay, so looking back at 𝑔 of 𝑥,
it is now expressed is the sum of two separate integrals.
Given the rules of differentiation,
we confined 𝑔 prime of 𝑥 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 𝑥. Our modified theorem tells us that
this is, therefore, equal to negative 𝑓 of one minus two 𝑥 times d by d𝑥 of one
minus two 𝑥. Our second term takes the same
form, but using the upper limit of one plus 𝑥. We now remember that 𝑓 of 𝑡 is
equal to five 𝑡 sin 𝑡. For 𝑓 of one minus two 𝑥, we’ll
replace all of the 𝑡 terms in this equation with one minus two 𝑥. We must then multiply this by the
derivative of one minus two 𝑥.
For our second term, we follow
exactly the same pattern, first substituting in to lowercase 𝑓 and then finding the
derivative of one plus 𝑥 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 𝑥. 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
𝑥 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 𝑎 and 𝑏, and the function capital 𝐹 of 𝑥 is defined by
the integral between 𝑎 and 𝑥 of lowercase 𝑓 of 𝑡 with respect to 𝑡. Then capital 𝐹 is continuous on
the closed interval between 𝑎 and 𝑏, is differentiable on the open interval
between 𝑎 and 𝑏, and capital 𝐹 prime of 𝑥 is equal to lowercase 𝑓 of 𝑥 for all
values of 𝑥 on the open interval between 𝑎 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 𝑥
instead of 𝑥 itself. And here this function we’ve called
𝑢 of 𝑥. When a function is defined by an
integral in which both of the limits have an 𝑥 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.
