a continuous real function f satisfies the identity f of two x is equal to three times f
of x for all x if the integral from zero to one of f of x is one what is the value of the
integral from one to two f of x and at first i thought this was a really hard question because
i was really amazed by the solution that they (Harvard-MIT Math Tournament) provide
and i couldn't come with that solution on my own
but i would still like to show you guys what that amazing solution looks like so have a look right
here they started by calling this to be S but i don't want to use s because my s looks like a 5 so
let's use I for integral and then we are going to make some observations now the integral going from
one to two and not the integral going from zero to one though but rather the integral going from
one half to one of the function f of x and we are going to see why right now because we have f of x
right here and we can use this relationship after we divide the 3 on both sides so we know this is
the integral going from one half to 1 and we will have f of 2x over 3 and then we are still in the
x world and now you see the inputs to x so it's a good idea to do some use up and that's why we
started with one half to one and then you will see it works out pretty well all right so i'm going to
let u equal to the inside which is 2x and then we can just divide the 2 on both sides and then
differentiate both sides we can see that dx is equal to du over 2. so the integral is going to be
here we have the 1 3 that's putting the front and then this is x going from one half to one
after this substitution we will just say u goes from one to two right multiply this and that by 2.
and then we have f of 2x which is now the f of u and then the dx is just du over 2. and now check
this out one half can be brought up to the front as well and we multiply that with one over three
so we get one over six and we have the integral going from one to two f of u d u this is exactly
that the dummy variable doesn't matter and one way to see that is and one way to see that is
you can just interpret this as the area under the curve f of x where f of u they have the same area
regardless which variable that you use so we can say the rest is just 1 over 6 times i cool
now we continue with the same fashion look at the integral going from 1 over 4 to 1 over 2.
again we divide this and not by 2 because similar things will happen
so here let's go ahead put on f of x dx and this is the integral and we will just have 1 over 4 to
one half and again this will be f of 2x divided by 3 and then we have the dx and we are going to
use the same use substitution this time though we have 1 over 6 just like earlier but this integral
is the integral going from one half to one of the function f so it's not just i but rather we would
have to utilize what we did earlier so the rest of the integral here is actually 1 over 6 i like so
and now we can see that this right here is just going to be 1 over 6 squared times i cool
and you can imagine we can just keep on going with the same thing right so the next one that we
will do is just the integral going from one over eight to one over four of the function f of x dx
and we will get one over six to the third power here and then multiply that by i yeah and then
so on so on so on and in general we can write down the following the integral going from 1 over 2 to
the n to 1 over 2 to the n minus 1 because this right here is just like 2 to the 2 this is 2 to
the third so this right here has one less factor of 2. and then the function f of x dx this right
here is just going to be 1 over 6 to the nth power and then multiply that by i cool yeah this is it
no what exactly is happening what exactly have we done right we haven't solved the question yet
well let's just go back to the original gift information you always have to use the gift
information if we don't use the default information something's wrong so the only
integral that we know is this so we want to go to from zero to one all right we want to go there
um okay we can take the limit of this but in that case the button will be zero the top right here
will be zero as well we don't get one don't worry you see we do have the one right here
right this is one half to one if we combine this integral with this integral then these two
together it will be the integral going from one over four to one because of the definite integral
property and then you want to go from one over h to one just combine this this and that together
so if you want to go from zero to one then we just have to add up all this together and then
be sure to take the limit as n goes to infinity and let me write that down right here for you guys
and then be sure we take the limit as n goes to infinity
and now check this out this right here will be the limit as n goes to infinity
and then the first integral is just 1 over 6 i good and then the next one is 1 over
6 squared times i yeah and then the third one is 1 over 6 to the third power times i and so
on so on and we have the general form which is 1 over 6 to the nth power times i and as we can see
everybody has the i we can factor that out and then take the limit as n goes to infinity
we will actually get a convergent geometry series because the common ratio is just one over six so
this right here becomes we have the i and we multiply by the first term which is one over six
and then divided by one minus the common ratio which is also one over six
and then do whatever you need to do i will tell you this right here is equal to one over five and
then we will have the i right here cool and then don't forget that what's this it's just equal to
one so ladies and gentlemen now we know one is equal to one over five times i so multiply the
five on both sides i is just equal to 5 and that's the answer to this integral right here
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guys so much and of course thanks to brian for sponsoring this video and i think this is super
super cool yeah i was really really amazed by this approach however i guess ready for the easier way
yes okay that's good here we go we are going to first start with this integral and let's put it
down right here the integral from 0 to 1 f of x and we are going to do probability earlier
that is to write f of x as f of 2x over 3. and we still have to integrate this from 0 to 1
in the x world and right here we can just do a u-sub and now take this integral to the u-world
well put zero in here we get u is equal to zero and put one here we get u is going to be at
two and we have the 1 over 3 that's putting the front and then we have f of 2x that is f of u now
and dx is that which is d u over 2. we can put a 1 over 2 in the front so we have
and now remember this right here is equal to 1 so we have 1 is equal to that and 1 is equal to this
by looking at the last equation we can multiply both sides by 6 and this right here will tell
us the integral going from 0 to 2 f and right here we use u but we can use t we can use w
it doesn't matter it's just a dummy variable let's use the best letter ever that's x
so let's put on f of x dx all right and that will just be equal to six this is so cool do we see it
guess what we're gonna do yeah just go ahead and subtract and ladies and gentlemen on the left-hand
side when we go from zero to two and then subtract from zero to one this record will give us the
integral going from one to two and then we have f of x and then on the right hand side it's just
six minus one we get five and that's it that that that's the answer five yeah i come over to this
yeah myself i did it myself yeah but i was really impressed with the students that they
had all i had no idea how to come with that solution i thought was really beautiful
i think i was i'm just i'm still thinking maybe there's actually an integral like uh with about
maybe the phone like two to three or so that you do have to do that kind of approach in order
to get the answer or maybe from zero to three i don't know but based on this question this is it