Challenging definite integration 2010 IIT JEE Paper 1 Problem 52 Periodic Definite Integral. The second term at about minute 14 should have a positive sign. Luckily, it doesn't effect the final answer!
Challenging definite integration
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- For any real number x, let [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 -10 to 10, including the boundaries,
- by f(x) = x - [x], if the greatest integer of x ([x]) is odd and
- 1 + [x] - x if the greatest integer of x is even.
- Then the value of pi squared over 10 times the definite integral from -10 to 10 of f(x) cos(pi*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(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 what this function will look like.
- So this is x = 0, this is x = 0 , this is x = 1, x = 2,
- x = 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 zero and one, what is the absolute, Oh! Sorry. What is the greatest integer of x?
- So between zero and one, so let me just write over here - so between zero and one,
- until you get to one, so may be I should do this,
- from including zero, until one, the greatest integer of x, the greatest integer of x, is equal to zero.
- If I am at point five, the greatest integer of 0.5 ([0.5]) is equal to zero.
- As I go from one to two, as I go from one to two, the greatest, this [x] is equal to 1, it's the greatest
- integer..., from like 1.5 the greatest integer is one. If I am at 1.9, the greatest integer is one!
- And then, if I go to above, from between two and three, so If I go between two and three,
- then the greatest integer is going to be two. If I am at 2.5, the greatest integer is going to be two.
- So with that let's at least try to draw this function over these intervals.
- So between 0 and 1, the greatest integer is 0. Zero - we can consider (it) to be even.
- Zero is even, especially for alternating - 1 is odd, 2 is even, 3 is odd.
- So 0 is even, so we will look at this circumstance right over here, if [x] is even.
- And then over this time frame, or over this part of the x- axis, the greatest integer of x is just zero,
- So the equation in, or the line, or the function is just going to be 1 - x over this interval, because [x] = 0.
- So 1 - x will look like this. If this is 1, 1 - x just goes down like that. That's what it looks like
- from 0 to 1. Now 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, [x], is odd.
- So we use this case. And over here, we're going to have x - [x].
- Over this interval, [x] = 1. So its going to be the graph of x - 1.
- So x - 1 at (x = ) 1 is going to be 0, and at (x = ) 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 - 1 and this over
- here essentially was 1 - x. And we could keep doing it. As we go from 2 to 3,
- [x] = 2, we would look at this case. So we are going to have 1 + 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 = 2, or a little bit above that,
- we're going to have 3 - 2, we're going to go to 1, it's going to be right at 1
- And then, as x = 3, 3 - 3 = 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.
- It's keep, it's going to keep going up and down like this - with a slope of negative 1and positive 1,
- then negative 1 then positive 1; It's just going to keep doing that over and over again,
- You could keep trying that with other intervals, but it's pretty clear that this is the pattern.
- Now, what is, what we want to do is to evaluate the integral from 10 to -10 of this function times cos(pi*x)
- So let's think about cos(pi*x) - 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. May be we can simplify it into an simpler integral.
- So cos(pi*x), cos(0) = 0. Cosine... sorry cos(0) = 1. No one could get that wrong!
- That's cos(pi*0) = cos(0), so that's 1. Cos(pi) = -1, so when x = 1, this becomes cos(pi), so then the
- value of the function is -1, it will be over here, and then cos(2*pi (two times pi)) is then 1 again,
- So it will look like this. I mean, this of course will be, this is at one-half, when you put it over
- here, it will become pi/2 , cos(pi/2) = 0, zero. So it will look like this. It will look, let me draw it as neatly as possible...
- So it will look like this. cosine, and then it will keep doing that! And then it will 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 -10 to 10, can we simplify that,
- and it looks like it would just be, cause we have this interval, Let's look at this interval over here
- Let's look at just from 0 to 1. So just from 0 to 1, we are going to take this function, and take the
- product of this cosine times, essentially, (1-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-1), it is actually going to be the
- same area. Because these two, going from 0 to 1, going from 0 to 1, and going from 1 to 2, 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 are 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 2 to 3, is going to be...
- First of all, it's clearly going, the integral from 2 to 3 is clearly the same thing as the integral
- from 0 to 1, both functions look identical over that integral, over that interval.
- But it will also be the same as going from 1 to 2. Because it's completely symmetric, when you, when you take
- the product of the functions, 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 we want to evaluate pi squared over 10 times the integral from -10 to 10 of
- f(x) cos(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 -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(x), of f(x) cos(pi*x) dx. Forgot
- to write, forgot to write the dx over there. I want to make sure you understand, cause 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 0 to 1, it's going to be the, the integral is going
- to evaluate to the same thing as going from 1 to 2, which will be the same thing as going from 2 to 3,
- or going from -2 to -1, So in stead of doing the whole interval from negative to the 10, we are just
- doing 20 times the interval from 0 to 1. From -10 to 10, you actually have 20, you, you, there's a difference
- of 20 here, so we are multiplying by 20. And this simplifies it a good bit. First of all, this part over
- here simplifies to 20/10 = 2. So it's 2 pi squared, so it 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(x)? We just figured it out. From
- 0 to 1, f(x) is just (1-x). f(x) is just (1-x) from 0 to 1, times cos(pi*x), cos(pi*x), dx. And now we
- just have to evaluate this integral right over here. So let's do that. So (1-x) times cos(pi*x) is the
- same thing as cos(pi*x) - x cos(pi*x). Now, this right here, taking the anti-deri... Let's take the,
- we're going to take the, well let's just focus on taking the anti-deri, This is pretty easy, but let's
- try to do this one, cause it seems a little bit more complicated. So let's take the anti-derivative of
- x cos(pi*x) dx and what should jump in your mind is that well yeah, this isn't that simple, but if I
- were already able to take the derivative of x, that would simplify, it's very easy to take the anti derivative
- of cos(pi*x) without making it more complicated. So may be integration by parts. And remember, integration
- by parts tells us that the integral, I'll write it up here, the integral of u dv = uv minus the integral of v du
- And we'll apply that here. But there's, I have 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 are going to take the
- derivative of whatever the u thing is. We want u to be something that's simpler when I take the derivative.
- And then we are going to take the anti derivative of dv - so we want something that does not become more
- complicated when I take the anti derivative. So the thing that becomes simpler when I take its derivative
- is x. So if I said u = x, then clearly, du = just dx, or you say du/dx = 1, so du = dx, and then dv,
- then dv is going to the the rest of this, this whole thing over here is going to be dv. dv = (is equal
- to) cos(pi*x) dx and so v, which is be the anti derivative of this, with respect to x, v is going to
- be equal to 1/pi, 1/pi sin(pi*x). Right? If I took the derivative here, derivative of the inside, you
- get a pi, times 1/pi, cancels out, derivative of sin(pi*x) becomes cos(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/pi sin(pi*x) minus, minus, the integral, minus the integral of v, which is 1/pi
- sin(pi*x) du. du is just dx. dx. And this is pretty straightforward. The anti derivative of sin(pi*x),
- is 1/pi, or negative 1/pi cos(pi*x). And you could take the derivative if you don't believe me. You could
- do, use substitution, but hopefully, you can start to do these things in your head. Especially if you
- are going to take IIT joint entrance exam. So this whole expression is going to be, this part over here is
- going to be x over pi sin(pi*x), and this over here is going to be, well you have the anti derivative
- of sin(pi*x), 1/pi, negative 1/pi cos(pi*x), the negatives cancel out. So you have plus, and then you the 1/pi times 1/pi,
- 1/(pi squared) cos(pi*x). That's the anti derivative right there. You can verify. Derivative of cos(pi*x)
- is going to be negative pi sin(pi*x), one pi will cancel out - here you get the negative sign, and you
- have the sin(pi*x). So this is the anti derivative of that. And so if we want the anti derivative of
- this whole thing right over here, this is, this is what we care about, from 0 to 1 dx. the anti derivative
- of cos(pi*x), pretty straightforward. We have actually already done it right over here. It is 1/pi, 1/pi
- sin(pi*x), that's this first term, and the anti derivative of x cos(pi*x) is this thing over here, but
- we are subtracting it, so we will put a negative sign out front. So minus, x/pi sin(pi*x) minus 1/(pi squared)
- cosine, of cosine, of pi*x, and of course we took the anti derivative, 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 out, 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/pi sin(1*pi), sin(1*pi) = 0. So it's 0 minus 1/pi times sin(1*pi) again, that's again 0! Minus 1/(pi squared)
- cos(pi), cosine of, or cos(1*pi), cos(pi) = negative 1, negative 1 times negative 1/(pi squared) is plus
- 1/(pi squared). So we have evaluated at 1. And from that we want to subtract it evaluated at 0. sin(0) = 0
- Minus, this is clearly zero because you have a zero out front, minus zero, and then you have minus
- cos(0), cos(0) is 1, so then you have a minus 1/(pi squared), 1/(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/(pi squared) plus 1/(pi squared) which is equal to, which is equal to, 2/(pi squared). That's what
- this part evaluates to. It's 2/(pi squared), 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 two pi squared.
- That's this thing out front here. And so the pi squared cancels out the pi squared, we're left with two
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