### Video Transcript

Use the fundamental theorem of
calculus to find the derivative of the function π
of π¦ is equal to the integral
between π¦ and five of three π‘ squared sin two π‘ with respect to π‘.

For this question, we first
recognize that weβve been given a function defined by an integral capital π
of
π¦. And weβve been asked to find the
derivative of this, capital π
prime of π¦. 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 π is a continuous function on the closed interval between π and
π and the function capital πΉ of π₯ is defined by the integral between π and π₯ of
lowercase π of π‘ with respect to π‘. Then πΉ prime of π₯ is equal to π
of π₯ for all values of π₯ on the open interval between π and π. Now, the equation given in our
question does not have a capital πΉ of π₯, but instead we have a capital π
of
π¦.

Before we start to think about π
prime of π¦, letβs dial things back to π
of π¦ 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 weβre
operating on, π¦, 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 π
of π¦. And of course, it doesnβt 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
π
prime of π¦ is equal to π of π¦. And remember the variable is π¦ not
π₯. Looking back at our integral, we
have to find lowercase π of π‘ to be negative three π‘ squared sin two π‘. This means that for lowercase π of
π¦ weβll replace all of our π‘s for π¦s. We, therefore, find that capital π
prime of π¦ is equal to negative three π¦ squared sin two π¦. And we have, therefore, answered
our question. We have found the derivative
capital π
prime of π¦ 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.