here we are going to do the Berkeley math tournament the 2022 calculus tie breaker and
we have three questions and we're supposed to be doing them in 15 minutes so here we go the
first one is the limit as n goes to Infinity of the integral from 1 to n of Ln X over n * Ln in
the X world because we're in the X world this denominator can be put on outside so actually
let's write it down like this the limit as n goes to Infinity on the top let's just look at that in
blue the integral going from 1 to n of lnx DX and then that just go down to the bottom so this is
what the question is really asking and if you put Infinity to the end you get infinity over
infinity um the official solution use litos rule but I'm not a big fan of using litos rule when
we have n so I'm just going to integrate this and then plug in the one and n it actually works out
pretty well too so if we integrate Ln X DX in fact we get X Ln X x - x yeah okay and I will
tell you let's just do a d method real quick on the side in case you want to see it we don't
integrate Ln X we differentiate that instead and integrate one right integration by parts
differentiating L and X we get 1/x integrating one we get X this times this first part of the
answer and then we are going to multiply this and that and put that integral so we get X Ln
X minus the integral and then we have 1 /x * X as you can see this right here is really just equal
to one minus the integral of 1 in The X world we get minus X and we are going from 1 to n so let's
just go here pluging one to n so n goes first so we have n * Ln n minus n that's the first part
and then minus pluging one and then minus one all right so I'm just going to write this down
for you guys right here now we have the limit as n goes to Infinity so all the blue part is
just n lnn n lnn - n and then Min - - one this is just equal to zero so it's just plus one and
then the bottom is just that so we have n * Ln n all right and now you see this is really really
nice because if you split the fraction we'll just get the limit as n goes to infinity and then we
have this over there is one this over there the N cancel so we have Min - 1 over n sorry 1 over
lnn and lastly this over that so we add 1 / n * lnn all right as Ang go to Infinity this goes to
zero because the bottom goes to Infinity likewise the bottom goes to Infinity here so this goes to
zero so yeah all that for the answer one question number two here's the function e to the x * sin x
and we're just going to differentiate this 222 times and then plug in zero just product R 2022
times right now big deal let's do it of course not don't do it like that have a look e to the x * X
it's a good idea to comp exify it what do I mean let me show you let's utilize the most powerful
formula in complex analysis the OAS formula which says e to the IX equal cine x + I * sin x so you
see here we have the s x now we just need to have another E to X so let's look at this equation and
multiply Everybody by E2 the X on the left hand side we can just add the exponents and
then factor out the X yeah so we have e to the 1 + I and then X right here and then the right
hand side I'm just going to distribute e to the X cos x plus i e to the X sin x so as you can see
our function is really just the imaginary part of that and this is so much better because when
we differentiate this 2022 times even though you want me to do like two million times I can do it
for you too because this is just e to a number * X take the derivative you keep that so e to
the 1 + IX power and then use the chain rule 2022 times which is just multiply by this number 2022
times so we will have 1 + I raised to the 2022 power over there and on the right hand side I'm
not going to bother with the derivative I'm just going to write that down right here for you guys
like this e to x cosine x + i e to X sin x so when we differentiate this 2022 times in fact it's just
the imaginary part of that and now we just have to plug in zero so the question F and then right here
2022 of Zer is just the imaginary part of that so we have 1 + I raised to the 2022 and then e 1 + I
and X but X is zero right so now we just have to figure this out and in fact this is not bad
because this is just one so figure this out how do we do it though well I'm going to look at this
like this 1 + I to the 2022 as 1 + I to the second power first and then raised to the 10 11 power why
because if you work this out you just get one square which is 1 plus 2 * this and that which
is 2 I and then plus this thing Square I sare I square is just netive 1 so just 1 + 2 I minus 1
and this is not cancel so just 2 I yeah so this is just 2 I raised to the 1,11 power split this
we get 2 to the 101's power time I to the 1,1 power and to figure this out I'm just going to
divide this by four so one11 divided by four and then we get 28 and then Min - 2 1 is 5 and then
2 0 - 11 and then this is 2 and 8 uh we just care about the remainder which is three so this right
here it's really the same as I to the thir power which is really the same as i s * I which is just
I so all in all this right here is netive 2 to the of course you don't have to figure this out just
leave it and then we have uh the eye right here like this and then we are just going to put that
back so the imaginary part is just it has the ey so it's just this and we are just looking for the
coefficient so the answer in in fact it's just going to be -2 to the 101's power okay question
number three but by the way this is a t-shirt I got from BMT last year and let me show you guys
the back too I was a special guest last year and this year I will be going to BMT again November
4th on UC Berkeley campus hope to see you guys there for more information check out the link
in the description they're not sponsoring this video in fact I'm sponsoring them why because I
love them so much I gratuated found there and so much fun last year I would like you all to give
a nice warm welcome to black p thank you thank you thank you thank you thank you thank you thank
you oh get set
integr so we are going to do this again in fact we're are sponsoring them because of
you guys we can actually offer them to to have scholarships to the top three winning
teams we have a very special scholarship for you guys from one of our sponsors
um he generously donated $1,000 for the first place
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team for more information check out the links in the description all right we we have the
integral going from 1 over e to e inverse tangent of X overx and let me tell you this function does
not have an elementary anti-derivative so don't think about integrating this and then plugging
numbers so what do we do then thankfully this is a definite integral so we can maybe use some
integral properties but before that maybe let's make the limit of Integrations more symmetric here
this is e to the1 and this is e to the 1 maybe that's get R of the E first and to do so we can
say x is equal to e to some variable that say e to the U and then just differentiate both sides
DX will be e to the U du and take this integral to the U world this is when X is equal to e to the1
that means U will be1 so U starts at1 and when X is equal to e to the first which is telling us U
is equal to one ah so much better right and then we have inverse tangent of X which is just e to
the U over another X which is just e to the U and then DX is equal to e to the U du so nice because
this then that cancel however the answer right here is not equal to zero because even though we
have negative 1 to one even the inverse tangent is an odd function but e to the U is not OD function
so what do we do well hopefully you can end up with some symmetry with combine integrals that
kind of business right so as a usual business let's go ahead and call this to be I so I equals
the integral from -1 to 1 and we have inverse tangent of e to the U du and then perhaps let's
do another substitution especially here we have e to the U maybe that's make the power negative
so you get reciprocal you see why I'm going to say let V equal to U so that means U is equal
to V du is equal to negative DV so I is equal to the integral this is when U is equal to netive
1 put it here that means V will be one and then when U is equal to one put it here that means V
is going to be1 so it's kind of like opposite huh don't worry continue here we have inverse tangent
of e to the U which is V here and then du is negative DV good now what well here let's go
ahead and use this negative to change the water back so I is equal to the integral going from -1
to 1 thanks to the negative and then this right here we have inverse tangent of e to the V here's
the trick I'm going to look at that as 1 over e to the v d v and this is actually super super
nice because not only we know this is equal to that but here this input and that input if
you just forget about the Dain variable they're just the reciprocals better yet we have inverse
tangent when we have the reciprocal right here this right here is really just the same as the
inverse cotangent and the inverse cotangent is the same as Pi / 2 minus the in tangent of its
regular version let me write that down for you guys this is the same as integral going from 1
to 1 the expression right here can be written as Pi / 2 minus inverse tangent of e to the
V DV inverse cotangent all right is really just like that very nice now U is a d variable V is a
d variable so if you combine this integral and that integral together the function Parts can
be cancelled very nice so have a look adding them up we will get 2 I which is really just
I + I and that will just give us the integral going from 1 to 1 and we just really have to
integrate Pi / 2 and perhaps let's put down DV or du doesn't matter or DX doesn't matter this
is just a number when you integrate from negative - 1 to one you get two two times this number so
you'll just get Pi 2 I it's equal to Pi so I is equal to Pi / 2 and finally you may just write
that down I which is our original integral let me just write that down 1/ e to e and then inverse
tangent of x overx DX this right here ladies and gentlemen again it's another famous result p/ two
so this is the answer that I will present for you guys just like that so what do you guys [Music]
think