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[MUSIC]. 

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

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

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

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

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

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Now, we can compute this integral 
geometrically. 

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

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

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

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So that quarter circle has area pi over 
4. 

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This is a geometric calculation of an 
integral. 

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We can also look at the accumulation 
function. 

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

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

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

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

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

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

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

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different ways. 
One way just involves differenciating. 

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

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

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

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

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

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

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

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

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

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

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

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

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

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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 

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

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

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

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

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

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

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

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

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

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

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

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

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2 times the square root of 1 minus T 
squared. 

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

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

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

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

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

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

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

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

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

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

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2. 
Minus t squared, over 2 times the square 

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

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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? 

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

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

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

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

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

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

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

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

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

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

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

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

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geometric. 
So I want to find geometrically the area 

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

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

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Now what's the area of the triangle 
piece? 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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triangle. 
And this term is calculating the area of 

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that circular sector. 
Now what happens if we evaluate the 

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

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

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

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

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

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

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

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accumulation function just by pure 
geometry or some trigonometry. 

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

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

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No mathematical fact is an island.