Question Video: Finding the Derivative of a Function Defined by an Integral Where Its Limits Contain a Variable

Definite Integral
Definite Integral

Video Transcript

Find the derivative of the function
𝑔 of 𝑥 is equal to the definite integral from three 𝑥 to four 𝑥 of 𝑢 squared
minus three all divided by 𝑢 squared plus five with respect to 𝑢.

We’re given a function 𝑔 of
𝑥. And we can see this is a definite
integral where both of the limits of integration are functions in 𝑥. And we’re asked to find the
derivative of this function. Since we’re trying to differentiate
a function where the limits of integration are functions in 𝑥, we’ll try and do
this by using the fundamental theorem of calculus.

So we’ll start by recalling the
following part of the fundamental theorem of calculus. If lowercase 𝑓 is a continuous
function on a closed interval from 𝑎 to 𝑏 and we have capital 𝐹 of 𝑥 is the
definite integral from 𝑎 to 𝑥 of lowercase 𝑓 of 𝑢 with respect to 𝑢, then
capital 𝐹 prime of 𝑥 will be equal to lowercase 𝑓 of 𝑥 for all values of 𝑥 in
the open interval from 𝑎 to 𝑏. In other words, this part of the
fundamental theorem of calculus gives us a way of differentiating a function defined
by an integral.

to differentiate in the question, 𝑔 of 𝑥, is very different to our definition of
capital 𝐹 of 𝑥. First, the lower limit of
integration is supposed to be a constant. However, we’re given this as a
function in 𝑥. There’s a similar story for our
upper limit of integration. It’s supposed to be 𝑥, but we’re
given a function in 𝑥. However, we can get around both of
these problems by using what we know about definite integrals and the chain
rule.

The first thing we’re going to need
to recall is one of our rules for definite integrals. To get around the problem of not
having a constant limit of integration, we’re going to want to split our definite
integral into two definite integrals. We know the following rule for
definite integrals. The integral from 𝑎 to 𝑏 of 𝑓 of
𝑢 with respect to 𝑢 is equal to the integral from 𝑐 to 𝑏 of 𝑓 of 𝑢 with
respect to 𝑢 plus the integral from 𝑎 to 𝑐 of 𝑓 of 𝑢 with respect to 𝑢. But there’s some important

The most important of these is this
will only be true if our function 𝑓 is integrable on all of these domains of
integration. In particular, our function 𝑓 must
be integrable on the closed interval from 𝑏 to 𝑐 and the closed interval from 𝑐
to 𝑎. We want to use this to rewrite our
function 𝑔 of 𝑥. So by using the upper limit of
integration as four 𝑥, the lower limit of integration as three 𝑥, and our function
lowercase 𝑓 of 𝑢 as 𝑢 squared minus three divided by 𝑢 squared plus five, we
would get the integral from 𝑐 to four 𝑥 of 𝑢 squared minus three divided by 𝑢
squared plus five with respect to 𝑢 plus the integral from three 𝑥 to 𝑐 of 𝑢
squared minus three all divided by 𝑢 squared plus five with respect to 𝑢.

But remember, we also need our
function to be integrable on both of these domains of integration. In particular, we need to choose a
value of the constant 𝑐. In this case, we could actually
choose any value of the constant 𝑐. To explain why, let’s take a look
at our integrand. Our integrand is the quotient of
two polynomials. This means it’s a rational
function. And we know rational functions will
be continuous across their entire domain. And if a function is continuous on
an interval, then it’s also integrable on that interval. So all we need to do is check the
domain of our function lowercase 𝑓 of 𝑢. It’s a rational function, so it
will be defined for all values of 𝑢 except where its denominator is equal to
zero.

And if we solve the denominator 𝑢
squared plus five is equal to zero, we’ll see that we get no real solutions. So our integrand is defined for all
real values of 𝑢. Meaning it’s continuous for all
real values of 𝑢, meaning it’s integrable on any interval. So in this case, we can actually
pick any value of 𝑐. However, this won’t always be the
case. In this case, we’ll choose 𝑐 is
equal to zero. We’ve now split this into two
definite integrals, each of which is closer to being able to use the fundamental
theorem of calculus.

There’s a little bit more
manipulation we need to do before we can use the fundamental theorem of
calculus. In our second integral, we want to
swap the order of our limits of integration. Remember, we can do this by
multiplying the entire integral by negative one. Doing this, we’ve rewritten our
second integral as negative one times the definite integral from zero to three 𝑥 of
𝑢 squared minus three all divided by 𝑢 squared plus five with respect to 𝑢.

And now, we’re almost ready to use
the fundamental theorem of calculus to evaluate these definite integrals. The only problem is our upper limit
of integration is a function in 𝑥 instead of 𝑥. And there are two ways of getting
around this problem. First, we could use the
substitution 𝑣 of 𝑥 is equal to four 𝑥. Then we can differentiate our first
integral by using the chain rule. We could do the same with our
second integral, setting 𝑣 of 𝑥 to be three 𝑥 and then evaluating its derivative
by using the chain rule. And this would work. However, we can also do this in the
general case with the fundamental theorem of calculus.

We have if 𝑣 of 𝑥 is a
differentiable function, then by using the chain rule, we get if lowercase 𝑓 is a
continuous function on the closed interval from 𝑎 to 𝑏 and capital 𝐹 of 𝑥 is the
definite integral from 𝑎 to 𝑣 of 𝑥 of lowercase 𝑓 of 𝑢 with respect to 𝑢, then
capital 𝐹 prime of 𝑥 will be equal to lowercase 𝑓 evaluated at 𝑣 of 𝑥 times d𝑣
by d𝑥 as long as 𝑣 of 𝑥 is in the open interval from 𝑎 to 𝑏. So we were able to rewrite our
version of the fundamental theorem of calculus so long as 𝑣 of 𝑥 is a
differentiable function just by using the chain rule.

And now, we can see both of our
definite integrals are in this form. We’ll set 𝑣 one of 𝑥 to be four
𝑥 and 𝑣 two of 𝑥 to be three 𝑥. Usually, we would set our value of
𝑎 equal to zero in both cases and then check the intervals on which our integrand
is continuous. This is because to use the
fundamental theorem of calculus, we do need to check where our integrand is
continuous. However, we’ve already shown that
our integrand is continuous for all real values. Therefore, it will be continuous on
any closed interval.

We’re now ready to find an
expression for 𝑔 prime of 𝑥. We’ll differentiate each of our
integrals separately. To differentiate our first integral
by using the fundamental theorem of calculus, we do need to check that 𝑣 one of 𝑥
is differentiable. And it’s a linear function, so we
know it is differentiable. This means we can use this, and we
get 𝑓 evaluated at 𝑣 one of 𝑥 times d𝑣 one by d𝑥. And we get the same story in our
second integral. This time, 𝑣 two of 𝑥 will be
three 𝑥, which is also a linear function, so it’s differentiable. This means we need to subtract 𝑓
evaluated at 𝑣 two of 𝑥 times d𝑣 two by d𝑥.

So now, we found an expression for
𝑔 prime of 𝑥. And we can actually evaluate all of
these expressions. Let’s start by substituting in our
expressions for 𝑣 one of 𝑥 and 𝑣 two of 𝑥. This gives us 𝑔 prime of 𝑥 is
equal to lowercase 𝑓 evaluated at four 𝑥 times the derivative of four 𝑥 with
respect to 𝑥 minus lowercase 𝑓 evaluated at three 𝑥 multiplied by the derivative
of three 𝑥 with respect to 𝑥. And remember, from the fundamental
theorem of calculus, lowercase 𝑓 will be our integrand. So we’re now ready to find an
expression for 𝑔 prime of 𝑥.

First, to find 𝑓 evaluated at four
𝑥, we substitute 𝑢 is equal to four 𝑥 into our integrand, giving us four 𝑥 all
squared minus three all divided by four 𝑥 all squared plus five. Next, we need to multiply this by
the derivative of four 𝑥 with respect to 𝑥. Of course, this is a linear
function, so we know the derivative of this with respect to 𝑥 will be the
coefficient of 𝑥, which is four. And we can do exactly the same to
find an expression for our second term.

To find 𝑓 evaluated at three 𝑥,
we substitute 𝑢 is equal to three 𝑥 into our integrand, giving us three 𝑥 all
squared minus three all divided by three 𝑥 all squared plus five. And of course, we need to multiply
this by the derivative of three 𝑥 with respect to 𝑥, which we know is three. Finally, we’ll simplify this
expression by evaluating all of our exponents and multiplying through by our
coefficients. And by doing this, we get our final
answer. 𝑔 prime of 𝑥 is equal to four
times 16𝑥 squared minus three all divided by 16𝑥 squared plus five minus three
times nine 𝑥 squared minus three all divided by nine 𝑥 squared plus five.

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