If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:15:43

for any real number X let brackets around X denote the largest integer less than or equal to X often known as the greatest integer function let F be a real valued function defined on the interval negative 10 to 10 including the boundaries by f of X is equal to X - the greatest integer of X if the greatest integer of X is odd and 1 plus the greatest integer of X - X if the greatest integer of X is even then the value of pi squared over 10 times the definite integral from negative 10 to 10 of f of X cosine PI of X DX is so before we even try to attempt to evaluate this integral let's see if we can at least visualize this function f of X right over here so let's do our best to visualize it so let me draw my x-axis and let me draw my y-axis so let me draw my y-axis and then let's think about this what this function will look like so this is X is equal to 0 this is X is equal to 0 this is X is equal to 1 X is equal to 2 X is equal to 3 we could go down to negative 1 negative 2 we could just keep going if we like hopefully we'll see some type of pattern because it seems to change from odd to even so between 0 & 1 what is the absolute sorry what is the greatest integer of X so between 0 & 1 so let me just write it over here so between 0 & 1 until you get to 1 so maybe I should do this from including 0 until 1 the greatest integer of X the greatest integer of X is equal to 0 if I'm at point 5 the greatest integer below point 5 is 0 as I go from 1 to 2 as I go from 1 to 2 the greatest it this brackets around X is equal to 1 it's the greatest inter from it 1.5 the greatest integer is 1 if I'm at 1.9 the greatest integer is 1 and then if I go to above from between 2 & 3 so if I go between 2 & 3 see then the greatest integer is going to be 2 if I'm at 2.5 grades ginger is going to be 2 so with that let's try to at least draw this function over these intervals so between 0 and 1 the greatest integer is 0 0 we can consider to be even 0 is even especially for alternating 1 is odd 2 is even 3 is odd so 0 is even so we would look at this circumstance right over here if X is even and then over this prime frame or over this part of the x axis the greatest integer of X is just 0 so the equation in or the line or the function is just going to be 1 minus x over this interval because the greatest integer is 0 so 1 minus X will look like this if this is 1 1 minus X just goes down like that that's what it looks like from 0 to 1 now let's think about what happens as we go from 1 to 2 as we go from 1 to 2 or not including 2 but including 1 all the way up to 2 not including it the greatest integer is 1 or the greatest integer the greatest integer is odd so we use this case and over here we're going to have X minus the greatest integer of X over this interval the greatest integer of X is 1 so it's going to be the graph of x minus 1 so X minus 1 at 1 is going to be 0 and at 2 it's going to be 1 again so it's going to be this it's going to look just like that so this right here is X minus 1 and then this over here essentially was 1 minus X and we could keep doing it as we go from 2 to 3 the greatest integer of X is 2 we would look at this case so we're going to have 1 plus 2 so we're going to have 3 we're going to have 3 here minus X so when we start over here when X is 2 or a little bit above that we're gonna have 3 minus 2 we're going to go to 1 it's going to be right at 1 and then as X is equal to 3 3 minus 3 is 0 it's going to oscillate back down like this so I think we have an appreciation for what this graph is going to look like if keep it's going to keep going up and down like this with the slope of negative 1 then pause one the negative one and positive one it's just going to keep doing that over and over again you can keep trying it out with other intervals but it's pretty clear that this is the pattern now what is what we want to do is evaluate the integral from 10 to negative 10 of this function times cosine of PI of X so let's think about cosine of PI of X and things and think about whether that also is periodic and of course it is and then if we can simplify this integral so we don't have to evaluate it over over this entire period over here maybe we can simplify it into a simpler integral so cosine of PI X cosine of 0 is 0 cosine sorry cosine of 0 is 1 I want to get that wrong because that's just cosine PI 0 is cosine of 0 so that's 1 cosine of PI is negative 1 so when X is equal to 1 this becomes cosine of pi so then the value of the function is negative 1 it'll be over here and then cosine of 2 pi 2 times pi is then 1 again so it'll look like this let me this of course will be this is at 1/2 when you put it over here become PI over 2 cosine of PI over 2 is 0 0 so it'll look like this it will look let me draw it as neatly as possible so it will look like this cosine and then it'll keep doing that and then we'll go like this so it is also periodic so if we wanted to figure out if we wanted to figure out the the integral of the product of these two periodic functions from all the way from negative 10 to 10 can we simplify that and it looks like it would just be because we have this interval let's look at this interval over here let's look let's look at just from 0 to 1 so just from 0 to 1 we're going to take this function and take the product of this cosine x essentially 1 minus X and then find the area under that curve whatever it might be then when we go from 1 to 2 when we take the product of this and X minus 1 it's actually going to be the same area because these two going from zero to one going from zero to one and going from one to two it's completely symmetric you could flip it over you could flip it over this line of symmetry and both functions are completely are completely symmetric so you're going to have the same area when you take their product so what we see is every interval over every interval when you go from two to three it's going to be first of all it's clearly going the integral from two to three is clearly the same thing as the integral from zero to one both functions look identical over that integral over that interval but it'll also be the same as going from one to two because it's completely symmetric when you when you take the products of the function that function will be completely symmetric around this axis so the integral from here to here will be the same as the integral from there to there so with that said we can rewrite this thing over here so what we want to evaluate PI squared over ten times the integral from negative 10 to 10 of f of X cosine cosine of PI X using the logic we just talked about this is going to be the same thing as being equal to PI squared over 10 PI squared over 10 times the integral well times the integral of from 0 to 1 but 20 times that because we have 20 integers between negative 10 and 10 we have 20 intervals of length 1 so times 20 times the integral times the integral from 0 to 1 of f of X of f of X cosine of PI X DX forgot to write forgot to write the DX over there and I want to make sure you understand it because this is really the hard part of the problem just realizing that the integral over this interval is just one twentieth of the whole thing because over every interval from zero to one it's going to be this the integral is going to evaluate to the same thing as going from one to two which will be the same thing as going from two to three we're going from negative 2 to negative one so instead of doing the whole interval from negative 10 to 10 we're just doing 20 times the interval from zero to 1 from negative 10 to 10 you actually have 20 you're there's a difference of 20 here so we're multiplying by 20 and this simplifies it a good bit first of all this part over here simplifies to 20 divided by 10 is 2 so it's 2 pi squared so becomes 2 pi squared that's just this part over here times the integral from 0 to 1 now from 0 to 1 what is f of X well we just figure it out from 0 to 1 f of X is just 1 minus X f of X is just 1 minus X from 0 to 1 times cosine of PI X cosine of PI X DX and now we just have to evaluate this integral right over here so let's do that so 1 minus x times cosine of PI X is the same thing as cosine of PI X minus X cosine of PI X now this right here taking the antiderivative take the we're going to take the well let's just focus on taking the antiderivative this is pretty easy but let's try to do this one because it seems a little bit more complicated so let's take the antiderivative of X cosine of PI X DX and what should jump in your mind is well you know this isn't that simple but if I were able to take the derivative of X that would simplify it's very easy to take the antiderivative cosine of PI X without making it more complicated so maybe integration by parts and remember integration by parts tells us that the integral I'll write it up here the integral of U DV is equal to UV minus the integral of V D U and we'll apply that here but there's I've done many many videos where I prove this and show examples of exactly what that means but let's apply it right over here and in general we're going to take the derivative of whatever the u thing is so we want you to be something that's simpler when I take the derivative and then we're going to take the antiderivative of DV so we want something that does not become more complicated when I take the antiderivative so the thing that becomes simpler when I take its derivative is X so if I set u is equal to X then clearly D U is equal to just DX or you say D u DX is equal to 1 or so D U is equal to DX and then DV then DV is going to be the rest of this this whole thing over here is going to be is going to be DV DV is equal to is equal to cosine PI X DX and so V would just be the antiderivative of this with respect to X V is going to be equal to 1 over PI 1 over PI sine of PI X right if I took the derivative here derivative of the inside you get 2 pi times 1 over PI cancels out derivative of sine of PI X becomes cosine of PI X so that's our U that's our V so this is going to be equal to this is going to be equal to u times V so it's equal to X this x times this so x over PI sine of pi X - minus the integral minus the integral of V which is 1 over PI sine of PI X D u D U is just DX D X and this is pretty straightforward the antiderivative of sine of PI X is 1 over PI or negative 1 over PI cosine of PI X and you could take the derivative if you don't believe me you could do u substitution but hopefully you can start to do these in your head especially if it going to take the ìit joint entrance exam so this whole expression is going to be this part over here is going to be x x over pi sine of PI X and then this over here is going to be well you have the antiderivative sine of PI X is 1 over negative 1 over PI cosine of PI X the negatives cancel out so you have plus and then you the 1 over pi times 1 over PI 1 over PI squared cosine of PI X that's the antiderivative right then you can verify derivative of cosine of PI X is going to be negative pi sine of X 1 pi will cancel out here you get a negative sign then you have a sine of PI X so this is the antiderivative of that and so if we want the antiderivative of this whole thing right over here this is this is what we care about from 0 to 1 DX the antiderivative of cosine PI X pretty straightforward we've actually already done it right over here it is 1 over PI 1 over PI sine of PI X that's this first term and the antiderivative of X cosine PI X is this thing over here but we're subtracting it so we'll put a negative sign out front so minus X over PI sine of PI X minus 1 over PI squared cosine of cosine of PI X and of course we took the antiderivative but it's a definite integral we need to evaluate it from 0 to 1 0 to 1 and we don't want to figure it forget that 2 pi squared out front that 2 pi squared out front so let's evaluate this so the first thing we're going to have 1 over PI sine of 1 pi sine of 1 pi is 0 so you've 0 minus 1 over pi times sine of 1 pi again that's again 0 minus 1 over PI squared cosine of PI cosine or cosine of 1 PI cosine of PI is negative 1 negative 1 times negative 1 over PI squared is plus 1 over PI squared so we've evaluated it at 1 and from that we want to subtract it evaluated at 0 sine of 0 is 0 minus this is clearly zero because you have a 0 out front minus 0 and then you have minus cosine of 0 cosine of 0 is 1 so then you have a minus 1 over PI squared 1 over PI squared and so this term we could just say this is a negative these don't matter negative positive positive and we're just left with a 1 over PI squared plus 1 over PI squared which is equal to which is equal to 2 over PI squared that's what this part evaluates to it's 2 over PI squared but we can't forget we can't forget that we're going to multiply this whole thing times 2 pi squared so we're going to multiply the whole thing times 2 pi squared that's this thing out front here and so the PI squared cancels out the PI squared we're left with two times two which is equal to four and we're done this thing that looked pretty complicated just evaluates out to four