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## Applying properties of definite integrals

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# Finding derivative with fundamental theorem of calculus: x is on both bounds

AP Calc: FUN‑6 (EU), FUN‑6.A (LO), FUN‑6.A.1 (EK), FUN‑6.A.2 (EK)

## Video transcript

So let's see if we can take the
derivative of this expression right over here, if we can
find capital F prime of x. And once again,
it looks like you might be able to use
the fundamental theorem of calculus. A big giveaway is
that you're taking the derivative of a
definite integral that gives you a function of x. But here I have x on both the
upper and the lower boundary, and the fundamental
theorem of calculus, is at least from
what we've seen, is when we have x's only
on the upper boundary. And then, of course,
it's an x squared, but we've seen examples
of that already when we used the
chain rule to do it. But how can we break
this up and put this in a form that's a little
bit closer to what we're familiar with when we apply
the fundamental theorem of calculus? And to realize
that, we really just have to attempt to graph
what this is representing. So let's say that
this is our lowercase f of x, or I should say f of t. So let's call this
lowercase f of t. And let's graph it
over the interval between x and x squared. So let's say this is my y-axis. This is my t-axis. And let's say that
this right over here is y is equal to f of t. I'm drawing it generally. I don't know what this
exactly looks like. And we're going to
talk about the interval between x and x squared. So if we're going to talk about
the interval between x, which is right over here,
it's the lower bound, so x and x squared. It's the lower bound, at least
for this definite integral. We don't know for sure. It depends on what x you
choose on which one is actually smaller. But let's just say that for
the sake of visualizing, we'll draw x right
over here, and we will draw x squared
right over here. So this whole expression,
this entire definite integral, is essentially asking for,
is essentially representing this entire area, the
entire area under the curve. But what we could do
is introduce a constant that's someplace in
between x and x squared. Let's say that constant
is c, and break this area into two different areas
with c as the divider. So that same exact
whole area we can now write it as two
separate integrals. So one integral that
represents this area right over here, and then another integral
that represents this area right over there,
and where we just say c is some constant
between x and x squared. Well, how can we denote
this area in purple? Well, that's going to be--
So this thing is going to be equal to the sum
of these two areas. The purple area we can show
is the definite integral from x to c of our function
of t, cosine t over t dt. And then to that we're
going to add the green area. And then we'll get
the original area. So for the green area, our
lower bound of integration is now our constant c, and
our upper bound of integration is x squared, and it's going
to be of cosine t over t dt. And this is a form where, if
we know how to apply the chain rule, we can apply the
fundamental theorem of calculus. And this is almost in a form. We're used to seeing it where
the x is the upper bound. And, well, we already
know what happens. We can swap these two
bounds, but it'll just be the negative
of that integral. So this is going to be equal
to-- let me rewrite it-- the negative of the
definite integral from c to x of cosine t over t dt. And then we have plus
the definite integral that goes from c to x squared
of cosine t over t dt. So all we've done is we've
rewritten this thing in a way that we're used to applying
the fundamental theorem of calculus. So if we want to find
F prime of x, well, applying the derivative
operator over here, we're going to have
a negative out front. It's going to be equal to
negative cosine x over x. Once again, just the
fundamental theorem of calculus. And then plus--
we're first going to take the derivative of
this thing with respect to x squared, and
that's going to give you cosine of x squared
over x squared. Wherever you saw t, you
replace it with an x squared. And then you're going
to multiply that times the derivative of x
squared with respect to x. So that's just going
to be-- derivative of x squared with
respect to x is just 2x. And we're done. We just need to
simplify this thing. So all of this is going to
be equal to negative cosine x over x plus-- well, this
is going to cancel out with just one of those-- plus
2 cosine x squared over x. And I guess we could simplify
it even more as being equal to-- and we can swap these--
everything over x 2 cosine of x squared minus cosine of x. And we are done.

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