this is the one that I got from the Berkeley math tournament where I was
a guest this year at about like six weeks ago in November on Berkeley campus
for the intercution B for with high school students so Berkeley afternoon people it gets
up watching this is for you guys and also all the participants you guys are awesome so here's
the one integral going from zero to two and we have x times the square root it's a big square
root it's an infinite square root it keeps on going on forever and then we have x to the Ln
X power and then inside here we have cube root and again this right here keeps on going forever
and we have X not to the lnx powered up but rather L and second power x so means Ln x squared
and then let's do again first root x to the Ln third power x and so on so on so on the next
one will be the fifth root but with the pattern already so this is it and we have the DX [Music]
it's the question shout out to the one who came with this question
and shout out to all the Berkeley map to number you guys also shout out to all the [Music]
and here is perhaps I will say the cleanest solution for this question anyway let's clean
up this expression if it's infinite you can know something about infinite series
so we are looking at the integral going from 0 to 2. this is X to the first yeah
next I'm going to look at this as X this is L and X inside the square root which is the one half
power so we can look at this as Ln X over 2. times the next one it's X this is Ln second power x
over 3 for the cube root but then over 2 for the square root and you multiply so it's over three
times two and you kind of see the pattern it's not so bad multiply X and then L and three x
still power technically and then fourth root right so like this and so on so on and then DX [Music]
all right so how do we deal with this let's put this time in summation one so this right here
is the integral going from 0 to 2 x is X and I'm going to write this as the summation let
me use n going from uh let me use n equals one to Infinity and we see that here we have L and X
but this right here does not have any Ln right so it's technically n minus one power over [Music]
this is one this is two this is three two three two one technically four two one AKA factorial
so this right here is the same as n factorial
yeah and then we have the DX here so now the question is how do we simplify this
hmm I will tell you we should know that when we have the series s n goes from zero to Infinity
of x to the nth power over n factorial this right here gives us nice D the nice e to the X function
and this works for any x value right this works for all x value so they're quite similar except
for here we have L and x and this is X Also let's start with one The Source your rule so let's fix
this a little bit if you want to start with one it's okay you can write it out the first term of
this is just one and then this is your plugin when n is equal to zero and then the remaining
is the summation and goes from 1 to Infinity of x to the nth power over n factorial this
is equal to e to the x put this to the other side we see the summation and goes from 1 to Infinity
let me still use the Red X here x to the nth power over n factorial is equal to e to the x minus 1.
that's okay okay and then how should we deal with the minus one oh let me show you
right here I can write this as the integral going from 0 to 2. X and then
this is the same as saying this 1 over L and x times because here it's in the in the world right
here you can put the X on the outside I know this has the N but that's Ln yeah and goes from 1 to
infinity and then this is L and x and we can put the N right here over n factorial and then we have
the DX so how does this help this is the integral going from 0 to 2 x here we have 1 over L and X
and this thing here is when X being equal to Ln X so we can just have
the following summation and goes from 1 to Infinity
Ln X raised to the N Over N factorial this right here is just what e to the Ln X power minus 1 AKA
x minus 1. so this thing here is just x minus 1. ah good but it doesn't look any better right
it does trust me right here for the X we can do the following we can write this as
integral going from zero to two look at the X as e to the L and X power very cool huh and then we
take this raised to the power of that namely we have the over L and X here and then the top
here is x minus 1. and then we have the DX so now what the L and x and the L and X cancel so in fact
this is actually just asking you to integrate from 0 to 2 and we have e to the x minus 1 DX
how do we integrate e to the x minus one it's just e to the x minus one and we go from zero to two
and now let's do this in our head put two in here we get e to the first 0 we get minus 1 for the
power and it's minus so it's equal to negative one and that is 1 over e and then we are done