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[MUSIC].

Let's use calculus not just to calculate the area under the graph of a function,

let's use calculus to find the area between two curves.

Specifically, what's the area between the graph of y equals the square root of x.

And the graph of y equals x squared. As usual, let's start by drawing a graph

of the situation. Here's my plane.

the graph of y equals x squared maybe looks like that.

And the graph of the square root of x, maybe looks like this in red.

So what's this area between curves even mean?

Well I really mean the area inside, this little wing shaped region here.

this point is worth thinking about, that point has coordinates 1,1.

let's set it up as an integration problem.

I can imagine carving this region up into lots of thin rectangles, right?

Setting up a[UNKNOWN] sum. Let's draw some of those thin rectangles,

in this region here. And, you have to worry a little bit about

exactly how tall all these rectangles are.

And how wide all these rectangles are. Well, how tall are these rectangles?

Let's just pick one of the rectangles in particular to think about.

Let's pick this rectangle, and think about it.

How tall is it? Well, its top edge.

Is on y equals the square root of x, and its bottom edge is on this orange curve.

The y equals x squared term. So the height of that particular

rectangle is its top coordinate minus its bottom coordinate, so it will be square

root of x minus x squared tall And how wide is it?

Well, the width of this little tiny tri-, rectangle, is, not supposed to very big,

right? So let's call its width dx, right?

I'm imagining that its, width is, very narrow.

It's a very thin rectangle. So that amounts to evaluating a certain

integral. Yeah, but which integral?

Well, I want to add up the areas of all these little boxes.

And these little boxes have heights given by this, and widths given by some small

quantity, I'm[INAUDIBLE] dx. So the integral that I want to calculate

will be the integral of the area. So it'll be the height.

The square root of x minus x squared. Times the width which is DX.

And then what does my X go between? Well the smallest value of X here is zero

and the biggest value of X here is one so I'll have X going from zero to one.

So that's the integral that I want to calculate which will find the area in

between these two curves. I can evaluate that integral by using.

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The fundamental theorem of calculus. So, the fundamental theorem of calculus

tells me to find an anti-derivative for this.

So, let's first thing. What's an anti-derivative of the square

root of x? Well that's an anti-derivative of x to

the one half power. But remember my rule for

anti-differentiaing x to a power. It's x to that power plus one, so 3

halves is one more than a half, divided by that power plus 1, 3 halves so that

when I differentiate this I get back x to the one half power and I'll just add some

constant C on there to give myself the general anti-derivative.

What if I want to anti-differentiate x squared?

Well, that's maybe easier. And it's the same power rule.

It's x to the third power, which is one more than two, divided by two plus one.

So that when I differentiate this, I just get back x squared, and I'll add a

constant. Now what's an antiderivative of the

difference? Well, the difference of anti derivatives.

So, the aniti derivative of the square of x minus x squared is the anti derivative

for the squared of x which is x to the 3 halves over 3 halves, minus the anti

derivative for x squared. So, x to the 3rd over 3.

So, there's the general anti-derivative for this.

Armed with the anti-derivative, I can now evaluate the internal.

So, one way to write that is as follows. If I went to integrate from 0 to 1, the

squared of x, minus x squared dx. I'll just write down the anti-derivative

for this, or an anti-derivative for this. Which is x to the 3 halves over 3 halves,

minus x to the 3rd over 3. That's an anti-derivative for this, plus

0, if you like. And then, I'll evaluate this quantity at

0 and 1 and take the difference. So, this is a little bit of fancy

notation. Just for plugging in x equals 1, plugging

in x equals 0 and finding the difference. And if I want to emphasize what I'm

plugging in I might write even x equals 0 here just to emphasize that x is the

variable that I'm working with. Alright well let's do that.

let's plug in x equals 1 here. And when I do that I get 1 to the 3

halves power over 3 halves minus 1 to the third over 3.

And I'll subtract what I get when I plug in 0.

SO I plug in 0, 0 to the 3 halves over 3 halves minus 0 to the third.

Over three. Well this is easy to calculate now.

1 to the 3 halves is 1. So it's the reciprocal of three halves.

Which is 2 3rds, minus 1 to the 3rd, which is 1 over 3.

So that's just a third. And here I've got minus 0.

So I've 2 3rds minus 1 3rd. This integral ends up being equal to 1

3rd which is computing the area between the original curves.

I hope this problem wets your appetite for harder problems along these same

lines, right? You can imagine very complicated area

calculations.