[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.