okay let's do this integral right here from negative 2 to 1 of 1/x^2 dx this right here
doesn't seem that bad right so that's C this is the integral from negative 2 to 1 this is X
to the negative 2 and you got the DX so now let's add 1 to the power and becomes negative 1 divided
by the new power we get 1 over negative 1 and and enter the property for that is this negative x to
the negative 1 power which is 1 over X and then we can plug in numbers now 1 in first so we have
negative 1 over 1 and then minus plug in negative 2 so 1 over negative 1 over negative 2 like this
and now you see this is pretty much parsley 1/2 and you are saying negative 1 minus 1/2
so just worked out your fractions you end up with negative 3 1/2 and as always that's it [Applause]
wait wait wait wait somebody told me this is wrong wait wait somebody told me that I didn't
get any answers there at the end any points on the test on this no this is wrong you serious why why
yeah subtracting strong I don't think so but why did you show to me is wrong okay here this right
here it was an innocent-looking integral huh but sometimes when you have an innocent-looking thing
it's actually pretty evil so be careful so this depth were pretty much fundamental theorem of
calculus part two you find an inside the reverti then you plug in numbers and then you just work
it out right however that only works if you have the function being continuous on this interval and
as we all know when we have x squared in the denominator X cannot be 0 and you do go from
negative 2 to 1 so you cannot broke the rule right we did unfortunately so this right here
is not correct at all in fact let me cross our obvious okay because this right here we didn't
have a continuous function so this didn't work and another thing that you can do to see that
if I this is false I will write a note for you guys when you have the integral from negative
2 to 1 1 over x squared DX we were pretty much trying to find the area under this curve from
negative 2 to 1 so you guys can see if you draw a picture real quick and you see 1 over x squared is
always positive it's always above the x-axis so this area should have been a positive but
earlier we end up with a negative value for the integral so there is that this true there's a
disagreement that's another hint that why this is actually incorrect now let me make it right
for you guys right so this is pretty much of things that you say to your girlfriend when you
messed up how can I make it right I will make it red for you guys so here is the deal first
of all you have to recognize when you have the function you're integrating is this continuous on
this interval it's not and more specifically you have to recognize there is a first class at all
at X equal to 0 it's right here huh and whenever you have this kind of situations in fact this is
what we call the type to improper integral when we have a particular acetal but this is how you
are going to make it right if the integral right here actually converges ah let's see here is 0
we have to avoid that so one of the ways to do it is of course take some limits so instead of going
to hit exactly 0 we can just kind of pretend we're almost over there from left hand side so
this right here is what we call this 0 minus we approach 0 from the left hand side so this is the
idea of taking the limit and you're going to break this integral into two integrals the first one it
goes from negative 2 to 0 minus and the proper way to write this is that we have to take the
limit and let me write on integral first and you see for this part right here we go from negative
2 to 0 - so we start the ad negative 2 and want to end up with no - but this right here is where
we require the limit so we will have to take up a river right here that's a B you can use any other
the letters that you want but of course usually only have the integral it's the integral from A to
B that's why I put a pea on the top and we put B approaches to zero - so this is the proper way to
deal with that and then the rest 1 over x squared DX like that and of course we have to pick up the
second part right here so we are going to add and this right here also requests the limit so I will
say it's a limit and this is going from its remind surplus to one so I will say that say a approaches
to zero plus and then we have the integral a to one right because a is approaching zero plus and
then the rest is the same of course just like this and now you pretty much have this integral and you
have to limit questions right here and here is the deal if both of them converges then of course in
the end just and then you add up your finite value that's it that would be the answer for
that improper integral but if one of the limits right here diverges Tandi or you can choose a
divergent that's it right because the idea is that you see this is pretty much edema question
you have B approaches to zero - and this is also another limit question but you have a right here
a is approaching zero plus a and B technically you don't have any connections even though sometimes
uma and I was infinity minus infinity and what depends you can check out my other video I cut
out to be the debate but for Cal two situations just say it diverges and move on that's it if
you see one of the limits right here diverges you see infinity is whatsoever as long as it's
not finite just don't diverge you don't even have to worry about the other one you can move
on okay so keep that in mind now let's work this out two steps no your integration and believing
the limits that's it so let's do the integration which is pretty much we need 1 negative 1 of X
and of course we still have to take the limit as B approach 0 minus and then we have to plug in some
numbers so we have negative 2 to be like this and let's just focus on this I'm not going to worry
about that the only worry about that if this is financial anyway by plugging B so we are taking
the limit as B approaches 0 minus and we have negative 1 over B and this is the first part we
have to - plug in negative 2 right here alright so here is the fun part you screen for this why
not I will plug in cereal - right here for the B because this is just - positive 1 out that's
finite let's see what we'll get from here we have negative and a lot times 1 and then when you put
0 - in here it gets through - like this can we do any conclusion right here yes and perhaps let
me write a note for you kiss on the side whenever you have any nonzero number over a 0 in the limit
situation this is the limit form alright this is the limit form so this right here is that because
you're - whatsoever usually you will just deal with this kind of forms for you have this this is
either approach to positive infinity or negative infinity depends on the sign so now let's just
walk outside here you have negative 1 on the top so it's negative 0 - means a number that's just a
little bit less than 0 such as negative 0.0001 you can just imagine like that anyway negative
divided by negative you end up with positive infinity so this right here approaches past
infinity of course infinite - one half it's just a saying if you take away 50 cents from Jeff besos
besos Jeff Bezos doesn't matter if you take a woman 1 million from him it's still rich right
so here's the conclusion this right here diverges I'm not just saying this right here diverges I'm
also saying the whole thing right here diverges okay but again you have the infinitive right here
this is finite so infinitely minus 1/2 is showing infinity so this right here for sure is diverges
if you work this out yes you also get past infinity but you didn't have to suppose if you
work this out somehow again negative infinity you just say that's divergent seriously just trust me
SoCal - later on if you study what they call the Cauchy principle value yes we continue but at the
moment the conclusion is that the integral from negative 2 to 1 1 over x squared DX this right
here you don't put an equal sign you don't say equals to diverge you just say this right here
diverges that's a verb that's it differ come down below and let me know if you have any questions