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[MUSIC]. Let's find the area between two

Â parabolas. I want to find the area between the graph

Â y equals x squared and the graph of y equals 1 minus x squared.

Â As usual let's start by graphing. Well here's my coordinate plane.

Â I'll draw a graph of y equals x squared. And the other curve is y equals 1 minus x

Â squared, so I'll draw a downward-facing parabola.

Â there's the one side of it and there's the other side of it.

Â I'll draw some rectangles to get an idea about what Riemann Sum I really want to

Â take the limit of. So it's really the area inside here that

Â I want to calculate. Draw some some rectangles then for the

Â for the Riemann Sum. That would approximate that area.

Â Right, so there's some rectangles for the the Riemann sum that would approximate

Â the area in there. And then I can write down the heights and

Â the widths of one of those rectangles. Let's take a look and say this rectangle

Â here. Well, what's the width of that rectangle,

Â will be real thin, so I'll call the width dx.

Â And what's the height of that rectangle. Well the top of that rectangle is on the

Â curve 1 minus x squared and the bottom of that rectangle was on the curve x

Â squared. So if we take the top coordinate minus

Â the bottom coordinate, that will give me the height of this rectangle.

Â The top coordinate is 1 minus x squared. The bottom y coordinate is just x

Â squared. So this quantity is giving you the height

Â of this rectangle. Oh, now I can write down the integral.

Â So I know the height and the width of these rectangles so the thing I want to

Â add up are the areas of those little rectangles.

Â So I want to integrate the height. 1 minus, this is 2x squared, times the

Â width, dx. But I'm trying to integrate from where to

Â where, right? What are the possible x values in this

Â region? Well, if we think about exactly where

Â these two curves cross, right? What's the x coordinate where these

Â curves cross? And this coordinate here is minus 1 over

Â the square root of 2 and this coordinate is 1 over the square root of 2.

Â So I'm going to be integrating to compute the area from x equals negative 1 over

Â the square root of 2 up to 1 over the square root of 2.

Â And it's this integral that'll calculate the area in between those two curves.

Â I can use the fundamental theorem of calculus to evaluate this integral.

Â Let's find an antiderivative. So what's an antiderivative of 1, well x

Â is an antiderivative of 1. What's an antiderivative of 2 x squared,

Â well 2 x to the 3rd over 3 differentiates to 2 x squared.

Â And I'm looking for an antiderivative of a difference, so it'll be the difference

Â of antiderivatives. Now I'll evaluate my antiderivative at

Â the right and left endpoints. Right, so I have to evaluate my

Â antiderivative at the end points, which are 1 over the square root 2 and minus 1

Â over the square root of 2 and then take the difference.

Â So what do I get when I plug in 1 over the square root of 2.

Â I get 1 over the square root of 2 minus 2 times 1 over the square root of 2 to the

Â third divided by 3. So this is what I get when I plug in x

Â equals 1 over the square root of 2 into my antiderivative.

Â And I subtract what I get when I plug in negative that, which is negative 1 over

Â the square root of 2, minus 2 times negative 1 over the square root of 2

Â cubed, all over 3. I can simplify this a bit.

Â Well, first off, I've got a quantity minus negative that same quantity.

Â That's just two copies of this same quantity.

Â Now I can keep calculating this and try to simplify this even a little bit

Â further. Now let's keep going.

Â So what do I have here? Well it's 2 times 1 over the square root

Â of 2, minus 2. And here I've got 1 over the square root

Â of 2 to the 3rd power. So I could write that as a half.

Â Times 1 over the square root of 2. That's really 3 copies of the square root

Â of 2 divided by 3. Okay.

Â But, 2 times 1/2, alright, cancels. And then I've got a common factor of 1

Â over the square root of 2. So I could, pull out that common factor

Â of 1 over the square root of 2. And when I pull that out I've got a one

Â there minus a third is left. now 2 times 1 over the square root of 2,

Â that's just the square root of 2 and 1 minus a third well that's really 2 3rds,

Â so what's left all over here is the square root of 2.

Â Times 2 3rds, this is the value of the integral and also then, the area between

Â those two curves. Before committing to this as my final

Â answer, I should check to see that the answer's somewhat reasonable.

Â Well if we numerically approximate this, this is about 0.94.

Â So is a bit less than one square unit a reasonable answer?

Â Well the region that we're interested in fits inside a rectangle of height 1, and

Â width the square root of 2 units. The area of this rectangle is about 1.41

Â square units. And the area of the region we just

Â calculated to be about 0.94 square units, right?

Â This region inside here. And it's pretty reasonable.

Â I mean, you know? It looks like.

Â If this thing is about 1.41 square units it's not unreasonable to think that this

Â is a little bit less than 1 square unit in this curved region, here.

Â Yea, we're really finding answers that accord with our geometric intuition.

Â