[music] Let's integrate x cubed from 0 to
2.
In this case, the function x cubed is a
continuous function, therefore it's an
integrable function.
So, all I need is just to start with a
fine enough partition and it won't matter
how I pick those sample points inside the
partition.
My Riemann sum will be close to the true
value of the integral.
So, I'll cut the interval from 0 to 2 into
n pieces.
Pictorially, I want to start with the
interval between 0 and 2, and I want to
cut it up into n pieces.
Now, the length of this whole interval is
two units long, so each is these pieces
should have a went of 2 over n.
And that, let me write down the formula,
the cut point x sub i will be 2 over n
times i, and that works out great .
If i is 0, then I'm at the left hand
endpoint.
And if i is n, I'm at the right hand
endpoint.
And in between, I'm just marching from the
left hand side to the right hand side
moving 2 over n each time.
Now, how should I choose my sample points?
I'm just going to choose my sample points,
x sub i star, to be the right hand end
points.
So, they'll be, in fact the same as x sub
i.
Now, I just have to write down the Riemann
sum for this particular equal size
partition and our chosen sample points.
In general, the formula for Riemann sum is
the sum, I think it's the sum, i goes from
1 to n of the function evaluated at the
i-th sample point times the width of the
i-th subinterval in my partition, right?
This is the area of the rectangle.
And I'm adding up all those rectangles to
get an approximation of the area under the
graph.
Well, in this specific case, what do I
have?
X sub i star is the same thing.
If I go back to this, x sub i star is the
same thing as x sub i.
So, I can just plug in 2 over n times i,
and x sub i minus x sub i minus 1, that's
the width of the i-th subinterval in my
partition.
But, I rigged it so that all of the
subintervals have the same equal width.
They've all got width 2 over n, okay?
So, that's the sum that I want to
calculate.
Now, the function in this case is the
cubing function.
So, this is the same as the sum i goes
from 1 to n of 2 over n times i cubed
times 2 over n, which I could simplify
somewhat.
This is the sum i goes from 1 to n of 16,
2 cubed times 2, over n to the 4th, it's n
cubed over n times i cubed.
So that's the particular Riemann sum in
this case.
Quizzes are just a specific Riemann sum
for a specific value of n.
To get the value of the intregal, I'm
going to take the limit an n approaches
infinity.
Yeah, the integral from 0 to 2 of this
function x cubed is the limit as n goes to
infinity of this Riemann sum.
Now, strictly speaking, right?
This integral would be a limit over all
the possible partitions.
But since the function's integrable, it
doesn't matter what partition I choose as
long as it's fine enough.
And n going to infinity makes finer and
finer partitions.
So, what I really want to calculate, in
order to calculate this integral, is to
calculate this limit.
Well, let's see if we can do that.
The good news here is that I've got a
formula for the sum of cubes.
So, we will pull out that formula in a
minute.
And I've also got a constant here.
16 over n to the fourth is a constant.
It doesn't depend on i at all, so I can
factor that out of this sum by
distributivity.
So, this is the limit as n approaches
infinity of 16 over n to the 4th times the
sum of i cubed, i goes from 1 to n.
And now, I happen to know what the sum of
the first n cubed is.
I can use that formula that we got from
before.
That's the same as the sum of the first n
whole numbers squared.
And in this case, that's the limit as n
goes to infinity of 16 over n to the 4th
times, that's n times n plus 1 over 2.
That's the sum of the first n whole
numbers and the sum of the first n perfect
cubes is that squared.
Now, I've got to figure out what this is.
Well, I can factor out a over 2 squared
and that'll cancel part of the 16.
So, this is the limit of n squared times n
plus 1 squared times 4 over n to the 4th.
And now this is the limit as n goes to
infinity.
This ends up being n squares time n
squares plus lower order terms.
So, this is, the biggest term here is 4n
to the 4th over n to the 4th, so this
limit ends up being equal to 4.
And that is in fact the integral from 0 to
2 of x cubed dx.
That integral, therefore, is equal to 4.
Now, what if I wanted to do this integral
over some other interval?
Well, I could repeat this same kind of
calculation to deduce that the integral
from 0 to 1 of x cubed dx is equal to a
quarter, right?
But that would be another whole
calculation that I'd have to work out.
Let's say, I wanted to integrate over the
interval 1 to 2.
Well, there's a trick that you could use.
So, the integral from 0 to 2 of x cubed
dx, we already calculated.
But forget the exact answer.
What I want to note here is that this is
related to two other integrals.
Integrating from 0 to 2x cubed dx is the
same as integrating from 0 to 1 of x cubed
dx, and adding to that the integral from 1
to 2.
Now geometrically, something like that
hopefully makes some sense, right?
If I draw a graph, here's 0, here's 1,
here's 2.
Here's the function y equals x cubed, say
this first interval from 0 to 1 is
calculating the area here, right?
That's what this integral's calculating.
The second integral from 1 to 2 is
calculating this area here, right?
And if I add together these two areas, I
end up getting the area from 0 to 2, which
is what this integral's calculating.
Well, the nice thing here is that, since I
know this integral and I know this
integral, that's enough information to
recover this integral.
The integral from 0 to 2, we saw a moment
ago, was 4.
The integral from 0 to 1, I'm claiming
here is a quarter, and that's enough
information to tell me that the integral
from 1 to 2 of x cubed dx, well, that must
be exactly what I need to add onto a
quarter to get 4, that must be 15 fourths,
right?
So, this red area in here must be 15 4th
square units.
What we're seeing here is that you can
evaluate some integrals just by going back
to the definition and using our knowledge
of sums, but the sums are ugly, right?
These are really terrible calculations.
This is the exact same situation that
we've been in before.
A long, long time ago, we were faced with
the definition of derivative and we
started computing some derivatives just by
going back to the definition of
derivative.
But then, over time, we learned a bunch of
rules which made differentiation a whole
lot easier.
We're going to see the same thing
happening here.
Right now, we're just using these bare
hand arguments to calculate integrals.
But over time, we're going to develop more
tools for evaluating integrals.