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[MUSIC].

Let's think a bit, about circles. Let's think about the function little f

of x equals the square root of 1 minus x squared.

What do we get if we graph that function? We draw my coordinate axes, right.

And I'm trying to graph the function. y equals the square root of 1 minus x

squared and yeah, when I graph that, I get a semi-circle.

Now, we can compute this integral geometrically.

Well, let's think about the integral from 0 to 1 of the square root of 1 minus x

squared dx, right? That just calculates the area Inside this

quarter circle, and what's that area? Well, the whole unit circle has area pi.

So that quarter circle has area pi over 4.

This is a geometric calculation of an integral.

We can also look at the accumulation function.

Well here's a zoomed in copy of the graph of that semi circle and here's 0 and

here's t. So this area in here is exactly what the

accumulation function calculates. It's the integral from 0 to t.

Of the square root of 1 minus x squared right, that's this graph, d x.

Now my claim is that, that accumulation function can be written like this.

That, that integral is equal to t times the square root of 1 minus t squared all

over 2 plus the arc sin of t over 2. We can work this out in a couple

different ways. One way just involves differenciating.

So I'm going to differentiate this side and get, I hope, the square root of 1

minus t squared, alright, that will be a proof that an anti-derivative of the

square root of 1 minus x squared is this. So that's with t's replaced with x's.

Okay so let's try it. Let's see if we can, if we can really

pull that off. So so let me just write down what I'm

trying to show. I'm trying to differentiate this whole

thing. So t times the square root of 1 minus t

squared over 2, plus the arc sin of t over 2.

And that's differentiating a sum. So I just had to differentiate this and

add it to the derivative of this. And this first term is a product.

It's the t times the square root of 1 minus t squared over 2.

So I'm going to differentiate that product using the product rule.

So it's the derivative of t, which is just 1 times the second thing.

Or the square root of 1 minus t squared over 2.

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Plus the first thing t times the deriveter of the second thing alright d

dt the squared of 1 minus t squared over 2 alright so that's the deriveter of this

first term plus the deriveter of[UNKNOWN] t over 2 well the deriveter of

art[UNKNOWN] Is 1 over the square root of 1 minus t squared, but then it's over 2.

So I'll put a 2 there. Okay, so I'm making some progress.

I've still got to differentiate the squared of 1 minus t squared over 2 so,

let's just, let's just keep going. So this first term is looking really

great. This is the square-root of of one minus

t-squared, over two. And, of course, I'm trying to get the

square-root of one minus t-squared. So that's, that's looking pretty good.

Plus t times, now what's the derivative of this square-root?

Well, I'll put the over-two here. So I just need to differentiate the

square-root of minus t-squared. And that's one over.

2 times the square root of 1 minus T squared.

And then it's times the derivative of the inside function.

And the derivative of 1 minus T squared is minus 2T.

All right, so that's the derivative of the square root of 1 minus T squared over

2. The over 2 is there, and that's the

derivative of just the square root of 1 minus T squared.

And I'll just add the derivative arc sign t, one over two the square root of one

minus t squared. Okay, now I can simplify this a bit.

Right, what do I have here? Well, I've got a minus two here and a two

there. So those will cancel and what I'm going

to be left with here is , The square root of 1 minus t squared over

2. Minus t squared, over 2 times the square

root of 1 minus t squared. That's this whole second term, plus 1

over 2 times the square root of 1 minus t squared.

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Now I'm going to combine these last two terms, all right?

This is equal to what? Well, I'll put 1 minus t squared over

this common denominator. So I've got this first term, the square

root of 1 minus t squared over 2 Plus 1 minus t squared over that common

denominator of 2 times the square root of 1 minus t squared.

Now I've got something divided by the square root of the same thing.

So I could replace this. With, well, the first term, 1 minus t

squared over 2 plus, this is something over the square root of the same thing,

so the square root of 1 minus t squared over 2.

Now I've got something over 2 plus something over 2, that's just the square

root of 1 minus t squared, and indeed We have made it to our goal.

I started by wanting to differentiate this expression and I ended up with the

sqaure of 1 minus t squared. The second method is a bit more

geometric. So I want to find geometrically the area

in here. And I can break that region up by drawing

this line, so that I've got this circular sector and this triangle.

Now what's the area of the triangle piece?

Well, the height of that triangle is the square root of 1 minus t-squared, right.

The base is length t, and this hypotenuse is length 1.

So by the pythagorean theorem the height of that triangle is the square root of 1

minus t squared and that means the area of this triangle is its base t times its

height the square root of 1 minus t squared divided by 2.

So that's the area of this triangle, What's the area of the circular sector?

Well, I've got an angle here which is theta and by a bit of geometry that angle

is the same as that angle up there. This length here is t, right.

We're going to need that in a minute. So if I've got a circular sector with a

radius 1 and a angle here, theta, that area is theta over 2.

This length is t. So that means that I know a formula for

theta in terms of t. Theta is arcsine t, right, because, you

know, really, sine theta is t. And that means instead of writing theta

over two for that area, I could write the area of that circular sector to be

arcsine t over two. So the total area here is arc sin t over

2 plus t times the square root of 1 minus t squared over 2.

Alright? And that's exactly what I'm claiming in

the accumulation function. This term is calculating the area of that

triangle. And this term is calculating the area of

that circular sector. Now what happens if we evaluate the

accumilation function at 1. So what I would be calculating is the

interval from 0 to 1 of the square root of 1 minus x squared dx.

And that's what I get when I just plug in t equals 1 here.

So 1 times the square root of 1 minus 1 squared over 2, 0.

Plus arc sign of 1 over 2 well that term is 0 and arc sign of 1 is pie over 2, so

I've got pie over 2 over 2 that is just pie over 4 which is you know really the

area of a quarter circle these pieces are coming together We can compute the

accumulation function just by pure geometry or some trigonometry.

We can verify that its derivative really is the square root of 1 minus x squared.

And then we can recover the area of a semi circle this way as well, right.

No mathematical fact is an island.