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AP Calc: CHA‑5 (EU), CHA‑5.A (LO), CHA‑5.A.1 (EK)

What I've got here is the graph
of y is equal to cosine of x. What I want to do is
figure out the area under the curve y is equal to
f of x, and above the x-axis. I'm going to do it
over various intervals. So first, let's think about
the area under the curve, between x is equal to 0 and
x is equal to pi over 2. So we're talking about
this area right over here. Well, the way we denote it is
the definite integral from 0 to pi over 2 of cosine of x dx. And remember, all
this is, is kind of, it's reminiscent of
taking a sum of a bunch of super thin rectangles
with width dx and height f of x for each of
those rectangles. And then you take
an infinite number of those infinitely
thin rectangles. And that's kind of what this
notation is trying to depict. But we know how to
do this already. The second fundamental
theorem of calculus helps us. We just have to figure out what
the antiderivative of cosine of x is. Or what an antiderivative
of cosine of x is. Evaluate it at pi
over 2, and from that subtract it evaluated at 0. So what's the antiderivative
of cosine of x? Or what's an antiderivative? Well, we know if we take
the derivative-- let me write this up here. We know that if we take the
derivative of sine of x, we get cosine of x. So the antiderivative of
cosine of x is sine of x Now, why do I keep saying sine
of x is an antiderivative? It's not just the
antiderivative. Well, I could also
take the derivative of sine of x plus any
arbitrary constant, and still get cosine of x. Because the derivative
of a constant is 0. This could be pi, this could
be 5, this could be a million, this could be a googol, this
could be any crazy number. But the derivative of this is
still going to be cosine x. So when I say that we just have
to find an antiderivative, I'm just saying, look, we just have
to find one of the derivatives. Sine of x is probably the
simplest, because in this case the constant is 0. So let's evaluate. So one way that we
can denote this, the antiderivative cosine, or
an antiderivative of cosine of x is sine of x. And we're going to
evaluate at pi over 2. And from that, subtract
and evaluate it at 0. So this is going to
be equal to-- let me do it right over here. This is going to be equal
to sine of pi over 2 minus sine of 0, which is equal
to sine of pi over 2 is 1. Sine of 0 is 0. So it's 1 minus 0 is equal to 1. So the area of this region
right over here, this area is equal to 1. Now let's do
something interesting. Let's think about the area. Let's think about the area
under the curve between, let's say pi over
2 and 3 pi over 2. So between here and here. So we're talking
about this area. We're talking about that
area right over here. This is 3 pi over 2. So once again, the
way we denote the area is the definite integral from pi
over 2 to 3 pi over 2 of cosine of x dx. The antiderivative-- or an
antiderivative of cosine of x is sine of x. Evaluated at 3 pi
over 2 and pi over 2. So this is going
to be equal to sine of 3 pi over 2 minus
sine of pi over 2. What's sine of 3 pi over 2? If we visualize the unit
circle really fast, 3 pi over 2 is going all the way 3/4
around the unit circle. So it's right over there. So sine is the y-coordinate
on that unit circle. So it's negative 1. So this right over
here is negative 1. This right over here,
sine of pi over 2 is just going
straight up like that. So sine of pi over 2 is 1. So this is interesting. We get negative 1 minus 1,
which is equal to negative 2. We've got a negative area here. We got a negative area. Now how does that make sense? We know in the real world,
areas are always positive. But what is negative 2
really trying to depict? Well, it's trying to
sign it based on the idea that now our function
is below the x-axis. So we could kind of think
we have an area of 2, but it's all below the
x-axis in this case. And so it is signed
as negative 2. The actual area is 2, but
since it's below the x-axis, we get a negative
right over here. Now let's do one
more interesting one. Let's find the definite
integral from 0 to 3 pi over 2 of
cosine of x dx. Now this is just denoting
this entire area. Going from 0 all
the way 3 pi over 2. And what do you think
is going to happen? Well, let's evaluate it. It's going to be sine of 3
pi over 2 minus sine of 0. Which is equal to
negative 1 minus 0, which is equal to negative 1. So what just happened here? The area under, all this orange
area that I just outlined, is clearly not negative, or any
area isn't a negative number. And the area isn't even 1. But what just happened here? Well, we saw in the first case
that this first area was 1. The area of this first
region right over here is 1. And then the area of the second
region, we got negative 2. And so one way to
interpret it is that your net area above
the x-axis is negative 1, or another way to say it,
your net area is negative 1. So it's taking the 1
region above the x-axis and subtracting the 2 below it. So what the definite
integral is doing when we evaluate it using the
second fundamental theorem of calculus, it's
essentially finding the net area above the x-axis. And if we get a
negative number, that means that the net area is act--
that actually most of the area is below the x-axis. If we get 0, then that
means it all nets out. And if you want to
see a case that's 0, take the integral from 0 all
the way all the way 2 pi. And this will evaluate to 0
because you have an area of 1 and another area of
1, but it nets out with this area of negative 2. Let's try that out. So if we go from 0 to
2 pi of cosine of x dx, this is going to be equal to
sine of 2 pi minus sine of 0, which is equal to 0 minus 0,
which is indeed equal to 0. Now clearly there was
area here, but all of the area that
was above the x-axis netted out with all the area
that was below the x-axis.

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