[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.