Video Transcript
Let 𝑓 of 𝑥 be equal to four if 𝑥
is less than one and four 𝑥 if 𝑥 is greater than or equal to one. Determine the integral from
negative one to three of 𝑓 of 𝑥 with respect to 𝑥.
We’re given a piecewise-defined
function 𝑓 of 𝑥, and we’re asked to determine the indefinite integral of this
function. There’s a few different ways we
could do this. For example, we could sketch a
graph of the function 𝑓 of 𝑥. We could then remember that the
indefinite integral of 𝑓 of 𝑥 between negative one and three will be the signed
area under the curve of 𝑓 of 𝑥. And of course, we would also need
to check that our function was continuous on this interval. This would work. However, we’re going to do this by
using the fundamental theorem of calculus.
So we’ll start by recalling the
fundamental theorem of calculus. In fact, we’ll only recall the part
which relates to evaluating definite integrals. This says if lowercase 𝑓 is
continuous on a closed interval from 𝑎 to 𝑏 and capital 𝐹 prime of 𝑥 is equal to
lowercase 𝑓 of 𝑥, then the integral from 𝑎 to 𝑏 of lowercase 𝑓 of 𝑥 with
respect to 𝑥 is equal to capital 𝐹 evaluated at 𝑏 minus capital 𝐹 evaluated at
𝑎. In other words, if our integrand is
continuous on the interval of integration and we have an antiderivative of our
integrand, then we can evaluate our definite integral by just evaluating our
antiderivative.
Normally, we would jump right into
using the fundamental theorem of calculus. However, in this case, we’re given
a piecewise-defined function 𝑓 of 𝑥. And since 𝑓 of 𝑥 is defined
piecewise, we’ll want to split our integral into the parts where our function
changes definition, in this case when 𝑥 is equal to one. So we want to split our definite
integral into the integral from one to three of 𝑓 of 𝑥 with respect to 𝑥 add the
integral from negative one to one of 𝑓 of 𝑥 with respect to 𝑥. However, we have to be very careful
at this stage. For example, we don’t yet know if
our function 𝑓 of 𝑥 is continuous when 𝑥 is equal to one.
This is a reason why whenever we’re
asked to evaluate a definite integral, it makes sense to first check that our
function is continuous across the entire domain of integration. In this case, we can actually just
check this directly. For example, we can see that our
function 𝑓 of 𝑥 is not only a piecewise function; in fact, it’s a piecewise
continuous function. And we know that piecewise
continuous functions are guaranteed to be continuous everywhere except possibly at
the endpoints. In fact, they’ll definitely be
continuous if their endpoints meet up.
So to check the continuity of 𝑓 of
𝑥, we’ll want to check that our endpoints meet up. So we want to substitute 𝑥 is
equal to one into both of our pieces. And if we do this, we can see in
both instances we get four. So our endpoints meet up. So, in fact, 𝑓 of 𝑥 is not only
continuous when 𝑥 is equal to one. 𝑓 of 𝑥 is continuous for all real
values of 𝑥. This tells us that we can split our
integral up in this way. Now we’re ready to use our
piecewise definition of 𝑓 of 𝑥 to change our integrands.
Let’s start with our first
integrand. We can see that our interval of
integration is the closed interval from one to three. So our values of 𝑥 will be between
one and three. And we can see that 𝑓 of 𝑥 is
just equal to the function four 𝑥 on this interval. So because 𝑓 of 𝑥 is exactly
equal to the function four 𝑥 on this interval, we can just change 𝑓 of 𝑥 with
four 𝑥. We want to do the same with our
second integrand. We can see, in this case, our
interval of integration is the closed interval from negative one to one. However, this time, we need to be
slightly more careful. We want to substitute 𝑓 of 𝑥 with
four.
From our piecewise definition of 𝑓
of 𝑥, we can see this is true for all values of 𝑥 less than one. However, our interval of
integration also includes when 𝑥 is equal to one. This is another reason it’s very
important we check our function 𝑓 of 𝑥 is continuous. We’ve already checked the endpoints
of our two pieces meet up. So when 𝑥 is equal to one, we
already know 𝑓 of 𝑥 is equal to four. So we know that 𝑓 of 𝑥 is exactly
equal to the function four on the closed interval from negative one to one. So we can replace our second
integral with the integral from negative one to one of four with respect to 𝑥.
And now we can just apply the
fundamental theorem of calculus to evaluate each of our integrals separately. Let’s start with our first
integral. First, we can see the lower limit
of integration is one and the upper limit of integration is three. So we’ll set 𝑎 equal to one and 𝑏
equal to three. Next, we’ll set our function
lowercase 𝑓 to be our integrand. Remember, on our interval of
integration, lowercase 𝑓 is actually equal to this. We’re now ready to try and use the
fundamental theorem of calculus. First, we need to show that our
integrand is continuous on the interval of integration. And we’ve already done this.
Next, we need to find an
antiderivative of our integrand. And there’s a few different ways of
doing this. For example, we could try and use
our rules of differentiation in reverse. However, the easiest way is to use
what we know about indefinite integrals. We want to calculate the integral
of four 𝑥 with respect to 𝑥. And we can evaluate this integral
by using the power rule for integration. We want to add one to our exponent
and then divide by this new exponent. This gives us two 𝑥 squared plus
our constant of integration 𝑐.
In fact, this means we have the
general antiderivative of four 𝑥. This is an antiderivative for any
value of 𝑐. So we’ll just use 𝑐 is equal to
zero. Therefore, by using the fundamental
theorem of calculus, we were able to show the integral from one to three of four 𝑥
with respect to 𝑥 is equal to our antiderivative evaluated at three minus our
antiderivative evaluated at one, which we represent by using the following
notation.
We now want to do exactly the same
for our second definite integral. First, our value of 𝑎 is negative
one, 𝑏 is equal to one, and 𝑓 of 𝑥 is equal to four. We’ve already shown that our
function 𝑓 is continuous on this interval. This means all we need to do is
find the antiderivative of our integrand. In fact, we can do this once again
by using the power rule for integration. We get four 𝑥 plus 𝑐, and again
we can use any value of 𝑐. So we’ll choose 𝑐 equal to
zero. Therefore, by using the fundamental
theorem of calculus, we’ve shown, to evaluate our definite integral, we just need to
evaluate our antiderivative at the limits of integration.
Now, all we need to do is evaluate
this expression. We’ll start by evaluating our first
antiderivative at the limits of integration. Doing this, we get two times three
squared minus two times one squared. And we can just calculate this;
it’s equal to 16. We then want to do the same with
our second antiderivative. Doing this, we get four times one
minus four times negative one. And we can also calculate this;
it’s equal to eight. So, evaluating both of our
antiderivatives at the limits of integration, we get 16 plus eight, which we know is
equal to 24.
Therefore, by splitting our
piecewise continuous function up into its individual pieces, we were able to
determine the definite integral from negative one to three of 𝑓 of 𝑥 with respect
to 𝑥 is equal to 24.
