[MUSIC]. 
Let's find the area under graph of a 
polynomial. 
So let's integrate from 0 to 1, the 
function x to the 4th dx. 
To solve this problem, I can use the 
fundamental theorem of calculus. 
So the first step is to find an 
anti-derivative of x to the 4th. 
Well, what's an antiderivative just for x 
to the n? 
Right, as long as n isn't minus 1, an 
antiderivative for this is x to the n 
plus 1 over n plus 1 plus c. 
So for this specific problem, right, 
what's an antiderivative for x to the 
4th? 
Well, that'll be x to the fifth over 5 
plus some constant c. 
And it doesn't matter which anti 
derivative I pick. 
So, just to emphasize that point, let's 
be a little bit ridiculous and, pick my 
anti derivative that I'm calling big F, 
to be x to the fifth over 5 plus 17, 
right? 
And this really is an anti derivative for 
x to the 4th. 
Because if I differentiate this function, 
I get the derivative next to the flipped 
over five, which is 5x to the 4th over 5 
plus the derivative 17, which is 0. 
And 5x to the fourth over 5 is just x to 
the 4th. 
So I really have found an anti-derivative 
for x to the fourth right here. 
It's x to the 4th over 5 plus 17. 
Now I apply the fundamental theorem. 
So the fundamental theorem tells me that, 
to integrate from 0 to 1. 
The function, x to the 4th is the same as 
evaluating an anti derivative for x to 
the fourth at 1. 
And subtracting that anti derivative 
evaluated at 0. 
In this case, my chosen anti derivative 
is this function. 
So I plug in 1. 
I get 1 to the 5th over 5, plus 17, minus 
what I get when I plug in 0, which is 0 
to the 5th over 5, plus 17. 
And one fifth plus 17, minus 17 Well this 
is just 1 5th. 
I can really see this fact. 
So here's a graph of the function y 
equals x to the 4th, and we just saw is 
that the integral from 0 to 1 of x to the 
4th dx is 1 5th, so there must be 1 5th 
of a square unit of area underneath this 
graph. 
Let me get out my scissors. 
Well here's a rectangle, and the width of 
this rectangle is one and the height of 
this rectangle is a 5th, so this 
rectangle contains 1 5th of a square unit 
of area, and it's the same as the area 
under this graph. 
So let me cut this rectangle a bit. 
I'll shove that piece Under there and 
I've got this leftover piece which I 
guess I have to do something with. 
Huh. 
Maybe I'll turn it over. 
Maybe I'll cut off this piece here. 
I'll put that there. 
I'll rip this there, and sort of shove 
that in there, and there'll be a little 
bit up there... 
I mean it's not great, the job that I did 
here, but it's at least believable that 
there's a 5th of a square unit of area 
underneath this graph. 
Admittedly this particular case, the area 
under the graph of X to a power, this 
particular case was known to Fermont 
before Newton. 
And yet what we're really selling here 
isn't the answer to a particular problem. 
It's a general technique for approaching 
these area problems. 
If you want to find the area under a 
graph of a function instead of trying to 
do it by hand You can apply the 
fundamental theorem of calculus and 
reduce it to an antidifferentiation 
problem, which hopefully is easier. 
This is an example of the way in which 
mathematics is a democratizing force. 
Problems that at one time would have been 
only accessible to the geniuses on earth. 
are now accessible to everybody, right? 
At one time in history, you would've had 
to been the smartest person on Earth in 
order to calculate the area of some 
curved object. 
And yet now, armed with the fundamental 
theorem of calculus, we can all take part 
in these area calculations.