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[MUSIC].

We dive into thinking more about volume and length and things, let's just focus a

bit more on area. So let's find the area inside here.

This orange curve is the graph y equals x squared, and this red line is just the

horizontal line y equals 1. So I want to figure out the area of this

region which is below the red line and above the orange curve.

We already know how to attack a problem like this.

I take this region and cut it up into a vertical strips.

Alright, I'm going to imagine cutting it up into a whole bunch of thin rectangles.

And then I just want to add up the areas of those rectangles integrate.

Well to set up the intergral, and I should say pick a value of x.

And think now, how tall is this thin rectangle?

Well the top edge of the rectangle is at, 1, and the bottom edge is at x squared.

So, the height of this rectangle, here say, is one minus x squared.

That's the length here, say is 1 minus x squared.

That's the length here. What's the width?

Well, I'm just going to call that dx. Right, I'm going to imagine that these

rectangles are real thin. So, now the area of just this one

rectangle is 1 minus x squared, that's its height times its width dx.

But I don't just want the area of this one rectangle.

I want to add up the areas of all these rectangles.

So, I'm going to integrate this from x goes minus 1 to 1.

Alright, and I get those end points by thinkin about how, small and how large x

can be in this region. So I put a minus 1 and a 1 there, and

this definite interval will calculate the area of this region.

And I can calculate that definitne integral.

Let me just copy it down here. The integral that I want to calculate is

the integral x goes from minus 1 to 1 of 1 minus x squared dx.

And to calculate that integral, it's enough to use the fundamental theorem of

calculus. So I'll write down an anti-derivitive.

x minus x cubed over 3 is an anti-derivitive and then the fundemental

theorum of calculus says evaluate this at 1 and at minus 1 and I take the

difference. So, I will plug in 1 and I'll get one

minus 1 cubed over 3 and I am going to subtract what I get when I plug in minus

1 which minus 1 minus, minus 1 cubed over 3.

But what's this, this is 1 minus a third that's 2 3rd minus, minus 1 plus a 3rd

which is negative 2 3rd, and 2 3rd minus negative 2 3rd, is 4 3rd.

So the area of this region is 4 3rd square units, but I could also do this by

cutting the region up into horizontal strips.

So, here I am cutting this region up into some horizontal rectangles, which instead

of being thin in terms of their width, they're now thin in terms of their

height. I just want to add up all of these not

very tall rectangles to compute the area of this region.

So to do that I'm going to pick some value of y and I'm going to think about,

for that value of y, how wide is is that rectangle?

Well let's think about this point over here, right?

This is the curve y equals x squared. So for this value of y what's the

corresponding value of x? Well it's the square root of y, and that

tells me how wide this whole rectangle is right.

And this point over here is negative the square root of y.

So, from here to here, is 2 square roots of y, that's the width of, of this

rectangle, let me write that down. So, I got the width of the rectangle is 2

square roots of y. Now, how tall is that rectangle?

Well, let's call that dy, right, I'm imagining that the rectangle isn't very

tall, really thin. So the product 2 squared of y its width,

and it's not very tall height d y. That gives me the area of one of these

thin rectangles and I'm going to integrate that.

But from where to where? Well, y could be as small as 0 and as big

as 1. So this integral will calculate the area

of this region, decomposed into horizontal strips.

I was going to do that integral, but I can definitely do that, right?

I can do this integral by using the fundamental theory of calculus.

I just copy that integral down over here. It would be integrating from 0 to 1, 2

square roots of y dy, l only rewrite that as the interval from 0 to 1 of 2y to the

1 half power dy. And now I can write down an

anti-derrivitave of this by using the power rule.

So an anti-derrivitave is 2y to the 3 halves power, divided by 3 halves.

Right, an anti-derivative y to a power, is y to one more than that power, divided

by one more than that power. I want to evaluate that at 0, and at 1,

and take the difference. Well, when I plug in 1, I get 2 over 3

over 2. When I plug in 0 I get 0, so the answer's

2 divided by 3 over 2, which I can rewrite as 4 3rd, which is the area of

this region. The neat thing here is that either way,

we're getting the same answer, and I think this is more amazing than it seems

at first. I mean look, when I cut it into

horizontal strips I ended up wanting to calculate this interval, the interval y

goes from 0 to 1 2 square roots of y dy. And when I cut the region up into these

vertical strips I wanted to do the interval x goes from minus 1 to 1 of 1

minus x squared dx. But these two intervals end up being

equal because they're both calculating the area of the same region, they're both

calculating 4 over 3 square units.