Video Transcript
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.
