[MUSIC] 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. [MUSIC]