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[MUSIC]. 

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Let's find the area under graph of a 
polynomial. 

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So let's integrate from 0 to 1, the 
function x to the 4th dx. 

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To solve this problem, I can use the 
fundamental theorem of calculus. 

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So the first step is to find an 
anti-derivative of x to the 4th. 

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Well, what's an antiderivative just for x 
to the n? 

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Right, as long as n isn't minus 1, an 
antiderivative for this is x to the n 

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plus 1 over n plus 1 plus c. 
So for this specific problem, right, 

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what's an antiderivative for x to the 
4th? 

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Well, that'll be x to the fifth over 5 
plus some constant c. 

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And it doesn't matter which anti 
derivative I pick. 

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So, just to emphasize that point, let's 
be a little bit ridiculous and, pick my 

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anti derivative that I'm calling big F, 
to be x to the fifth over 5 plus 17, 

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right? 
And this really is an anti derivative for 

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x to the 4th. 
Because if I differentiate this function, 

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I get the derivative next to the flipped 
over five, which is 5x to the 4th over 5 

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plus the derivative 17, which is 0. 
And 5x to the fourth over 5 is just x to 

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the 4th. 
So I really have found an anti-derivative 

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for x to the fourth right here. 
It's x to the 4th over 5 plus 17. 

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Now I apply the fundamental theorem. 
So the fundamental theorem tells me that, 

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to integrate from 0 to 1. 
The function, x to the 4th is the same as 

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evaluating an anti derivative for x to 
the fourth at 1. 

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And subtracting that anti derivative 
evaluated at 0. 

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In this case, my chosen anti derivative 
is this function. 

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So I plug in 1. 
I get 1 to the 5th over 5, plus 17, minus 

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what I get when I plug in 0, which is 0 
to the 5th over 5, plus 17. 

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And one fifth plus 17, minus 17 Well this 
is just 1 5th. 

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I can really see this fact. 
So here's a graph of the function y 

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equals x to the 4th, and we just saw is 
that the integral from 0 to 1 of x to the 

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4th dx is 1 5th, so there must be 1 5th 
of a square unit of area underneath this 

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graph. 
Let me get out my scissors. 

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Well here's a rectangle, and the width of 
this rectangle is one and the height of 

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this rectangle is a 5th, so this 
rectangle contains 1 5th of a square unit 

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of area, and it's the same as the area 
under this graph. 

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So let me cut this rectangle a bit. 
I'll shove that piece Under there and 

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I've got this leftover piece which I 
guess I have to do something with. 

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Huh. 
Maybe I'll turn it over. 

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Maybe I'll cut off this piece here. 
I'll put that there. 

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I'll rip this there, and sort of shove 
that in there, and there'll be a little 

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bit up there... 
I mean it's not great, the job that I did 

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here, but it's at least believable that 
there's a 5th of a square unit of area 

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underneath this graph. 
Admittedly this particular case, the area 

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under the graph of X to a power, this 
particular case was known to Fermont 

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before Newton. 
And yet what we're really selling here 

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isn't the answer to a particular problem. 
It's a general technique for approaching 

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these area problems. 
If you want to find the area under a 

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graph of a function instead of trying to 
do it by hand You can apply the 

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fundamental theorem of calculus and 
reduce it to an antidifferentiation 

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problem, which hopefully is easier. 
This is an example of the way in which 

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mathematics is a democratizing force. 
Problems that at one time would have been 

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only accessible to the geniuses on earth. 
are now accessible to everybody, right? 

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At one time in history, you would've had 
to been the smartest person on Earth in 

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order to calculate the area of some 
curved object. 

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And yet now, armed with the fundamental 
theorem of calculus, we can all take part 

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in these area calculations.