[MUSIC]. Let's find the area between two
parabolas. I want to find the area between the graph
y equals x squared and the graph of y equals 1 minus x squared.
As usual let's start by graphing. Well here's my coordinate plane.
I'll draw a graph of y equals x squared. And the other curve is y equals 1 minus x
squared, so I'll draw a downward-facing parabola.
there's the one side of it and there's the other side of it.
I'll draw some rectangles to get an idea about what Riemann Sum I really want to
take the limit of. So it's really the area inside here that
I want to calculate. Draw some some rectangles then for the
for the Riemann Sum. That would approximate that area.
Right, so there's some rectangles for the the Riemann sum that would approximate
the area in there. And then I can write down the heights and
the widths of one of those rectangles. Let's take a look and say this rectangle
here. Well, what's the width of that rectangle,
will be real thin, so I'll call the width dx.
And what's the height of that rectangle. Well the top of that rectangle is on the
curve 1 minus x squared and the bottom of that rectangle was on the curve x
squared. So if we take the top coordinate minus
the bottom coordinate, that will give me the height of this rectangle.
The top coordinate is 1 minus x squared. The bottom y coordinate is just x
squared. So this quantity is giving you the height
of this rectangle. Oh, now I can write down the integral.
So I know the height and the width of these rectangles so the thing I want to
add up are the areas of those little rectangles.
So I want to integrate the height. 1 minus, this is 2x squared, times the
width, dx. But I'm trying to integrate from where to
where, right? What are the possible x values in this
region? Well, if we think about exactly where
these two curves cross, right? What's the x coordinate where these
curves cross? And this coordinate here is minus 1 over
the square root of 2 and this coordinate is 1 over the square root of 2.
So I'm going to be integrating to compute the area from x equals negative 1 over
the square root of 2 up to 1 over the square root of 2.
And it's this integral that'll calculate the area in between those two curves.
I can use the fundamental theorem of calculus to evaluate this integral.
Let's find an antiderivative. So what's an antiderivative of 1, well x
is an antiderivative of 1. What's an antiderivative of 2 x squared,
well 2 x to the 3rd over 3 differentiates to 2 x squared.
And I'm looking for an antiderivative of a difference, so it'll be the difference
of antiderivatives. Now I'll evaluate my antiderivative at
the right and left endpoints. Right, so I have to evaluate my
antiderivative at the end points, which are 1 over the square root 2 and minus 1
over the square root of 2 and then take the difference.
So what do I get when I plug in 1 over the square root of 2.
I get 1 over the square root of 2 minus 2 times 1 over the square root of 2 to the
third divided by 3. So this is what I get when I plug in x
equals 1 over the square root of 2 into my antiderivative.
And I subtract what I get when I plug in negative that, which is negative 1 over
the square root of 2, minus 2 times negative 1 over the square root of 2
cubed, all over 3. I can simplify this a bit.
Well, first off, I've got a quantity minus negative that same quantity.
That's just two copies of this same quantity.
Now I can keep calculating this and try to simplify this even a little bit
further. Now let's keep going.
So what do I have here? Well it's 2 times 1 over the square root
of 2, minus 2. And here I've got 1 over the square root
of 2 to the 3rd power. So I could write that as a half.
Times 1 over the square root of 2. That's really 3 copies of the square root
of 2 divided by 3. Okay.
But, 2 times 1/2, alright, cancels. And then I've got a common factor of 1
over the square root of 2. So I could, pull out that common factor
of 1 over the square root of 2. And when I pull that out I've got a one
there minus a third is left. now 2 times 1 over the square root of 2,
that's just the square root of 2 and 1 minus a third well that's really 2 3rds,
so what's left all over here is the square root of 2.
Times 2 3rds, this is the value of the integral and also then, the area between
those two curves. Before committing to this as my final
answer, I should check to see that the answer's somewhat reasonable.
Well if we numerically approximate this, this is about 0.94.
So is a bit less than one square unit a reasonable answer?
Well the region that we're interested in fits inside a rectangle of height 1, and
width the square root of 2 units. The area of this rectangle is about 1.41
square units. And the area of the region we just
calculated to be about 0.94 square units, right?
This region inside here. And it's pretty reasonable.
I mean, you know? It looks like.
If this thing is about 1.41 square units it's not unreasonable to think that this
is a little bit less than 1 square unit in this curved region, here.
Yea, we're really finding answers that accord with our geometric intuition.