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Okay, now as I promised we should think about sufficient conditions for
integrability.
And well we've actually recently come up with more integrable example,
and now we should move on to some much more simpler cases actually.
So we're not going to try to prove this one for
you, but we're going to just state it as a basic cases.
For example, if the function is monotonically common,
monotonically rises or falls whatsoever or has no discontinuations
on the whole segment, then it is integrable on this segment.
And it's nice thing, for example,
if you can see the x squared which is defined and
plotted in the figure like this, that's our x squared from 1 to 0.
It's quite easy to understand that first of all,
it is monotonically rises and second of all, it has no discontinuations.
Well Nash's functions of all, so it's totally integrable.
Thus just let us try to consider how one should calculate as natural integral here.
And in order to do so, let us just consider some well partial case.
It's not a general case, you should understand.
But we know that for any partition and any tagging,
we should get the same limit here since the function is integrable.
Thus for any special partition and special tagging,
the answer should still be the same.
So let us consider uniformal partition here.
So we have 1 to 0, and we have for example, n dots in partition.
Thus we have the same length of each segment and
dots actually 0, 1/n, 2/n, and
whatsoever, whatsoever, whatsoever,
up to n/n, which actually equals to 1.
And let us assume for
the sake of simplicity that we are considering some point here.
Well, for example, I just can't assume that we can see
the taken with the right end of which segment, right.
So for each segment of this partition,
the area under the curve should be written as follows.
The length of which partition is 1/n, right.
And we should multiply it by the value of function in our tact point.
Tact point in our case, for example, for
the first was and is (1 / n) squared, right?
Okay, that was nice.
So for the second one is (2 / n) squared
multiplied by the length, which is the same and
whatsoever, and goes and goes and goes and goes.
So let us just try to rewrite it in the more comprehensive way.
First of all, we should consider moving this common
multiplier and this common divisor here.
So we are getting something like a denominator n power 3.
And in the nominator, we have 1 squared + 2 squared + 3 squared,
and this goes up till n squared, which is actually
the value for the last segment, no right here.
Okay, so it's time to remember your discrete mathematics course,
which was the first one in this online degree program.
And remember the formula was this sum of squares for
first n natural numbers, and it's quite easy.
I'm going to write this down for you.
It is n multiplied by (n +1) multiplied
by (2n + 1) / 6 n power 3.
So that's it.
If you don't remember,
please revisit the video from the discrete mathematics course concerning induction.
So and if you don't want to remember induction, well that's sort of easy,
and just believe me that this formula's right.
So the last thing we need to do is we need to actually find the limit,
calculate the answer.
In order to do so,
we need to find the limit when the diameter of this partition approaches 0.
In other words in our case, if the partition
is uniformly as the length of each segment
coincides with the diameter and equals to 1/n.
Thus if it is approaches 0,
then n should approach infinity.
And we need to consider the limit of this sequence whilst n approaches infinity.
And it's kind of an easy one for us because it's polynomial
divided by polynomial, and we all do know how to handle this one.
We need to divide both numerator and
denominator by the highest power, in our case it's the third power.
And we get 1, multiply that
by (1 + 1 / n) multiplied
by (2 + 1 / n) / 6.
And if n approaches infinity, 1 / n approaches 0.
Thus, this is 0, this is 0, and we get 2 / 6,
or in other words one-third.
And one-third is actually our answer, that's nice.
And I'm going to congratulate you with the first definite
integral you've actually calculated.
So here we are going to move forward with understanding how
we avoid this extremely complex cases of taking limits and
using some discrete mathematics rules and induction and
for more general case of calculating definite integrals.
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