[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.