[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. 
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. 
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.