### Video Transcript

The fundamental theorem of
calculus: evaluating definite integrals. In this video, we will learn how to
evaluate definite integrals using the fundamental theorem of calculus. The theorem is usually stated in
two parts. For the purpose of this video,
weβll be focusing on the second part. This tells us that if lowercase π
is a continuous function on the closed interval between π and π and capital πΉ is
any antiderivative of lowercase π. Here expressed as capital πΉ prime
of π₯ is equal to lowercase π of π₯. Then the integral between π and π
of lowercase π of π₯ with respect to π₯ is equal to uppercase πΉ of π minus
uppercase πΉ of π.

Now here, an interesting point to
note is that weβve said capital πΉ is any antiderivative of lowercase π. This means that if there are many
antiderivatives, it does not matter which one we pick to use our theorem. To make sense of this, let us think
back to what we mean by an antiderivative. By now, we should be familiar with
the fact that the first part of the fundamental theorem of calculus essentially
tells us that differentiation and integration are inverse processes. This means that we can find the
general form for the antiderivative of a function, lowercase π of π₯, by evaluating
the indefinite integral of that function. Letβs consider an example
function. Lowercase π one of π₯ is equal to
two π₯. Its antiderivative, uppercase πΉ
one of π₯, would be equal to the indefinite integral of two π₯ with respect to
π₯.

To solve this, we use the power
rule of integration, raising the power of π₯ by one and dividing by the new
power. Of course, here, we mustnβt forget
to add on our constant of integration, π. More simply, our answer is written
as π₯ squared plus π. Now, this constant of integration
π can take any value we like. And our expression would still be
one of an infinite number of antiderivatives of our original function πΉ one of
π₯. If we took the case of π equals
zero, πΉ one of π₯ would simply be π₯ squared. π equals five would give us an
antiderivative of π₯ squared plus five. π could even be negative π, which
would mean our antiderivative would be π₯ squared minus π. All of these are antiderivatives of
two π₯.

Given that the fundamental theorem
of calculus allows us to use any of these antiderivatives to evaluate a definite
integral, what do we do. Letβs continue to use our example
function π one of π₯ equals two π₯. That now, weβre looking to the
definite integral between π and π of this function. The theorem tells us that this is
equal to the antiderivative capital πΉ one of π minus capital πΉ one of π. We can arbitrarily pick one of our
antiderivatives, for example, π₯ squared plus five. If πΉ one of π₯ equals π₯ squared
plus five, then πΉ one of π equals π squared plus five. Similar reasoning follows for πΉ
one of π. Simplifying this, we see that there
is a plus five and a minus five term, which cancel each other out. And weβre therefore left with π
squared minus π squared.

In fact, if we had used the general
form which would involve any constant π, the same thing wouldβve happened. So regardless of the constant we
use, we arrive at the same result. Given this fact, we can simply
choose to ignore the constant of integration altogether when evaluating a definite
integral. And this is essentially equivalent
to taking the case where π is equal to zero. If we had considered this case to
begin with, we wouldβve reached the same answer, but with fewer steps of
working.

Let us now see an example of our
theorem in practice.

Let π of π₯ equal six π₯ squared
plus one. Evaluate the definite integral of
π from π₯ equals two to π₯ equals three.

For this question, weβve been asked
to evaluate a definite integral, which using standard notation would look like
this. We see that the integrand is our
function π of π₯. And the limits of integration are
two and three, as specified by the question. To evaluate this definite integral,
weβre gonna be using the second part of the fundamental theorem of calculus. This tells us that if lowercase π
is a continuous function on the closed interval between π and π and capital πΉ
prime of π₯ is equal to lowercase π of π₯. In other words, capital πΉ is an
antiderivative of lowercase π. Then the integral between π and π
of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π minus capital
πΉ of π.

Letβs now apply this theorem to
solve our problem. Our function, lowercase π of π₯,
is equal to six π₯ squared plus one. To find this antiderivative,
capital πΉ of π₯, we can integrate lowercase π of π₯. If we apply the power rule of
integration, raising the power of π₯ in each of our terms and then dividing by the
new power, we get an answer of six π₯ cubed over three plus π₯. And we also add our constant of
integration π. At this point, we remember that the
fundamental theorem of calculus allows us to use any antiderivative, which means π
can take any value. It makes sense for us to choose the
simplest case possible, which is when π is equal to zero. Essentially, this means we can
ignore this constant. Simplifying, the antiderivative
that weβll then use is two π₯ cubed plus π₯. Great. Letβs set this antiderivative to
one side and go back to our original calculation.

The question has asked us to
evaluate this definite integral. Our upper limit of integration is
three. And our lower limit is two. The fundamental theorem of calculus
then tells us that this integral is equal to capital πΉ of three minus capital πΉ of
two. Now, since we have just found
capital πΉ of π₯, our antiderivative, we can substitute in the values of three and
two into this function. After we have input these values,
we can perform a few simplifications on our new expression. After working through these
simplifications, we reach an answer of 39. With this step, weβve completed our
question. The definite integral given in the
question evaluates to 39. We evaluated our integral using the
antiderivative of the given function πΉ of π₯. And the tool that we used to help
us was the second part of the fundamental theorem of calculus.

Okay, before moving on, a quick
word on notation. Given the definite integral between
π and π of a function π of π₯ with respect to π₯, very often, youβll see the next
step written in the following form. With the antiderivative of the
given function written in brackets like so and the limits of integration carried
over to the right-hand bracket. And this is simply a shorthand. Itβs a way of expressing the
antiderivative as a function before we substitute in our limits of integration and
evaluate. It is equivalent to what we should
now be familiar with, which is capital πΉ of π minus capital πΉ of π. Youβll almost always see this step,
since itβs a very useful shorthand, which helps to tidy up our working.

Looking back at our previous
example function, π one of π₯ is equal to two π₯. If we were to consider the definite
integral between one and three of this function with respect to π₯, our next step
would look something like this, with the antiderivative of π one of π₯ in brackets,
as shown. We would then proceed to continue
our evaluation by substituting in three and one, the limits of integration. And weβd eventually reach an answer
of eight.

Let us look at an example question
using this notation.

Evaluate the integral between zero
and two of two sin π₯ minus three π to the π₯ with respect to π₯.

To answer this question, weβre
going to be using the second part of the fundamental theorem of calculus. This tells us that if lowercase π
is a continuous function on the closed interval between π and π and capital πΉ is
any antiderivative of lowercase π. Then the integral between π and π
of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π minus capital
πΉ of π. Looking back at our question, the
first thing we might notice is that the function lowercase π, which is our
integrand, consists of two different terms. The first term involves the
trigonometric function sine. And the second involves the
exponential π. Now, we should be familiar with the
fact that both trigonometric and exponential functions of this form are continuous
over the entire set of real numbers. We have therefore fulfilled the
criteria that our function π must be continuous over the closed interval between π
and π, which in our case is the closed interval between zero and two.

Now, given that we do have two
terms, we might find that our working is clearer if we split these up into separate
integrals. We would do this like so,
remembering to keep the limits of integration the same across both terms. We can now evaluate each of these
integrals separately. The antiderivative of two sin π₯ is
negative two cos π₯. And the antiderivative of negative
three π to the π₯ is negative three π to the π₯. Of course, remember, we can ignore
the constant of integration in both cases since weβre working with definite
integrals.

Here, we note that weβve expressed
our antiderivative in brackets, with the limits of integration being carried over to
the right-hand bracket in both cases. Given that both of these brackets
have the same limits of integration, we can simply combine them. Now, you might notice that we
couldβve moved directly from our original integral to this step, by treating each of
the terms individually. Instead of splitting our integral
into two and then recombining, we would simply have found the antiderivative of each
of our terms. If youβre not sure, however,
thereβs no harm in writing out the method in full.

To move forward with our question,
we now substitute in the limits of our integration, which are zero and two. We then reach the following
expression. With our first set of parentheses,
thereβre no simplifications necessary. So we can just leave this. For the second set of parentheses,
we might recall that cos of zero is equal to one. So negative two cos zero is equal
to negative two. Alongside this, π to the power of
zero is also one. So negative three π to the power
of zero is negative three. Our second set of parentheses
therefore becomes negative two minus three, which is negative five. But for our final answer, weβre
subtracting this. So weβre left with a positive
five. And weβve now reached our final
answer. The definite integral given in the
question evaluates to negative two cos of two minus three π squared plus five.

Okay, if we look back to the
fundamental theorem of calculus, another important condition to consider is the
continuity of the function lowercase π for which weβre integrating. Remember that the theorem states
that our function must be continuous on the closed interval between π and π. This π and π form the limits of
our integral. If lowercase π is not continuous
over this interval, then we cannot confidently say that this relationship is
true. To illustrate this, let us consider
the function one over the square root of π₯. If we were to draw the graph of
this function, it might look something like this. Now letβs consider the definite
integral of our function between one and two with respect to π₯. This could be interpreted as the
area under the curve between one and two, as shown on our graph.

To evaluate this, we might want to
reexpress one over the square root of π₯, so that the power of π₯ is in a more
manageable form. We then use the familiar power rule
of integration. And we would simplify. If we continued, weβd find no
problems. And weβd reach numerical
answer. Okay, what would happen instead if
we were asked to evaluate the definite integral between negative one and one? Here, we might start to run into
some problems. It should be clear from our graph
that π of π₯ is undefined when π₯ is less than or equal to zero. Trying to imagine the area under
our curve between these bounds would be meaningless. Since our function is undefined
over part of the closed interval between negative one and one, it cannot be said to
be continuous. Given this fact, thereβs no sense
in continuing, since we cannot use the fundamental theorem of calculus to evaluate
this integral.

Letβs take a look at an example to
illustrate this.

Evaluate the integral between four
and nine of negative two times the square root of π₯ with respect to π₯.

For this question, weβve been asked
to evaluate a definite integral. With questions of this type, it can
sometimes be useful to move constant factors, such as the negative two, from the
inside of the integrand to the outside. Next, we might also find it useful
to reexpress our square root of π₯ as π₯ to the power of a half or π₯ to the power
of 0.5. And weβll see why in a moment. To move forward with this question,
weβre gonna be using the second part of the fundamental theorem of calculus. This gives us a way to evaluate
definite integrals using the antiderivative of the function which forms the
integrand.

At this point, we note that the
theorem states the function, lowercase π, must be continuous on the closed interval
between π and π. π and π are the limits of
integration, which in our case are four and nine. Now, the function that weβre now
working with, lowercase π, is the square root of π₯, which weβve just expressed as
π₯ to the power of a half. This function is not continuous
over the entire set of real numbers, but rather is only continuous when π₯ is
greater than or equal to zero. Luckily, both of the limits of our
definite integral, four and nine, are greater than or equal to zero. And so we can therefore say that
the square root of π₯ is continuous on the closed interval between four and
nine. This means that itβs perfectly fine
to use our theorem.

To proceed with our evaluation, we
use the power rule of integration, raising the power of π₯ by one and dividing by
the new power. The antiderivative of π₯ to the
power of half is therefore two over three times π₯ to the power of three over
two. Again, weβre gonna shift this
constant factor outside of our brackets to make our calculations easier. Next ,we substitute in our limits
of integration. Now, at this stage, it might be
more useful to see our power of three over two expressed as the cube of a square
root. Conveniently, nine and four are
both square numbers. And we can simplify our parentheses
to be three cubed minus two cubed. We now move forward with a few more
simplifications. And eventually, we reach an answer
of negative 76 over three. This is the final answer to our
question.

We evaluated the given definite
integral using the second part of the fundamental theorem of calculus to help
us. Along the way, we were sure to
confirm that our function lowercase π was continuous over the closed interval
between the limits of integration. One final point that we didnβt
really go into earlier. Weβre able to say that the square
root of π₯ is not continuous when π₯ is less than zero, because, in fact, the square
root of π₯ is undefined over the real numbers when π₯ is less than zero. And of course, a function cannot be
continuous at points where it is not defined.

Letβs now move on. Another thing worth considering are
the cases of piecewise functions or cases involving the absolute value of a
function. The reason we might need to think
about these functions carefully is that they can be thought of as having different
behaviors over different regions of their domain.

Let us see how to deal with this in
the following example.

Evaluate the definite integral
between negative four and five of the absolute value of π₯ minus two with respect to
π₯.

For this question, weβve been asked
to evaluate the definite integral of a function, which weβll call lowercase π. This function is the absolute value
or the modulus of π₯ minus two. Now, for any real number, we can
express an absolute value function as a piecewise function. We can do this by recalling that if
π₯ minus two evaluates to a negative number or absolute value, weβll multiply this
by negative one to turn it into a positive number. Okay, so when π₯ minus two is
greater than or equal to zero, our function is simply π₯ minus two. But when π₯ minus two is less than
zero, our function is multiplied by negative one. So it is negative π₯ minus two. Of course, itβs probably more
useful to us to isolate π₯ on one side of these inequalities. We do so by adding two to both
sides. Now, it might also be useful for us
to simplify this as negative π₯ plus two.

Okay. Now that we have reexpressed our
function piecewise, we can think about how it might look graphically. Here we see the graph. Although the scale is not exact,
both the graph and the piecewise definition should show us the difference in
behavior of our function either side of π₯ equals two. We see a sharp corner at the point
two, zero on our graph. In fact, we would say that our
function is not differentiable when π₯ equals two. But it is continuous when π₯ equals
two. This is important, because in order
to evaluate our definite integral, weβll be using the second part of the fundamental
theorem of calculus. This allows us to evaluate a
definite integral using the antiderivative, uppercase πΉ, of the function which
forms our integrand, lowercase π. The condition for doing so is that
lowercase π must be continuous on the closed interval between π and π, which are
the limits of the integration. Given that our function lowercase
π is continuous when π₯ is equal to two, we are able to conclude that it is
continuous over the entire set of real numbers. And hence, the continuity condition
is satisfied.

Okay, onto evaluating the definite
integral. Now weβve already said that our
function behaves differently either side of the line π₯ equals two. A useful first step for us then is
to split up our integral into two parts. The first going from the lowest
bound, negative four to two, and the second going from two to five. Since the upper limit of our first
integral is the same as the lower limit of our second integral, the sum of these two
will be the same as our original integral. Now that weβve split our integral
into two parts, weβre able to substitute in the two different subfunctions that we
defined using the piecewise definition of the absolute value of π₯ minus two. We can understand this by
considering our integrals as the area under these lines. From negative four to two, our
function behaves like minus π₯ plus two. And from two to five, our function
behaves like π₯ minus two. We can interpret the sum of these
two areas as being the same as our original integral.

From this point, we can now move
forward using the familiar power rule of integration. We raise the power of π₯ for each
of our terms and divide by the new power. Letβs tidy up to make some room for
the next steps. Here, weβve input the limits of
both integrals. And some color has been added to
help follow the calculation. Weβll need to go through a few more
simplification steps. Again, weβll clear some space. And weβll continue to simplify. Eventually, we reach a point where
weβll express everything in terms of halves. And we reach a final answer of
forty-five halves or 45 over two. With this, weβve completed our
question. We did this by first expressing the
absolute value of π₯ minus two as a piecewise function. Then, by splitting our original
integral into two parts and using the second part of the fundamental theorem of
calculus to help us evaluate each individual part.

Okay. To finish off, letβs go through
some key points. The second part of the fundamental
theorem of calculus tells us if lowercase π is a continuous function on the closed
interval between π and π and if capital πΉ is any antiderivative of lowercase π,
which we can express as capital πΉ prime of π₯ is equal to lowercase π of π₯. Then the integral between π and π
of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π minus capital
πΉ of π. Remember, we can use any
antiderivative of the function lowercase π. This means we can choose to take
the case, which makes our calculations as simple as possible. Which is when π, our constant of
integration, is equal to zero, essentially allowing us to ignore this.

Often when given a definite
integral, our next step will be to write the antiderivative in brackets with the
limits of integration carried over to the right-hand bracket. This is a way to express the
antiderivative as a function before inputting the limits of integration. But this is usually an intermediate
step on your way to capital πΉ of π minus capital πΉ of π. In order to use this theorem,
remember to check that lowercase π is continuous and indeed defined on the closed
interval between π and π. If it is not, then the integration
may fail. Finally, piecewise functions or
functions involving absolute values may require you to split the integral into
multiple parts, since these types of functions behave differently over different
regions of their domains.