Modulus Function Integration Problem

Added: 2026-05-15

[00:00:00]Okay, so we're going to solve this problem where we need to find the value of A so that this integral is equal to 12. And instead of actually integrating this function, which would be really hard to do cuz we've got lots of iterations of the modulus function in there, instead we'll think of the integral as the area underneath the graph of this function and we'll try and sketch this function to help work with the area to find the value of A.

[00:00:20]So we'll do this starting from the inside and working outwards. So we could just start with y = 1 - x. Start really simple. Work our way outwards and build up to what does the graph of this complicated looking function actually look like? So y = 1 - x is nice and easy to draw. We have a y intercept of 1 and the gradient is -1. So we have an x intercept there at 1 as well. And if we want to just go up to the modulus of 1 - x, we'll do this one small step at a time. All we do here is change all of the negative values of y would become positive and then the rest of the graph over on the left here where y is positive would stay the same. So we'd still have this y intercept of 1 and an x intercept of 1. And then if we want to go to just the negative of modulus of 1 - x. Let's draw this on a separate graph now so it doesn't get too messy. So we've got the negative of this. We're just going to take the red graph here and reflect it in the x axis. So we've now got a y intercept instead of 1 it's down at -1 and our x intercept is still at 1. And if we want to go up to 2 - the modulus of 1 - x. This is effectively going to be the same graph but we've just added two to all of our y values so it moves up two units. So this point here that's got coordinates 1 0 this would go up to around here. This point that's got coordinates 1 2. And our y intercept would move up to to the point with coordinates 1 there. So 0 1.

[00:01:48]And if we continue this over to the left and to the right it's the same shape. So going from 1 we just go along 1 unit and then we're down to the x axis. We've got -1 as our x intercept there. Then we can think of this just one little step at a time. 1 2 goes along to 2 1. We go along one, down one, then along another one, down another one. We have our x intercept is at 3 here. So our x is 3.

[00:02:11]I'll just draw this in a little bit neater. And then if we want to do the modulus of this red function, so y is the modulus of 2 minus the modulus of 1 minus x, then we're just going to reflect these two vertices on the left and on the right. So it'll mostly look the same as what we had before. So now we're going to get We've still got this point at -1 here. We start to get a kind of w shape here for the graph of modulus of 2 minus modulus of 1 minus x. We've still got the same x intercepts, everything there.

[00:02:44]We've still got this point with coordinates 1 2. Then the next thing we do is just take the negative of this. So to take the negative of this, we again just reflect this in the x axis. We get now an m kind of shape. And all the coordinates have changed over here. So we've got 1 2 becomes 1 -2. All the y values become negative. We've still got the same x intercepts. We'll clear a bit of space now, and then we're almost ready to see what this overall function looks like when we include the 3 in that final modulus function.

[00:03:15]And of course you can see the y intercept is now at -1 there. So if you want to go to the next step, we're effectively just adding 3 to all of this. So the whole graph is going to move up 3 units. So this point here 1 -2 is going to move up to the point with coordinates 1 1. And we could take this point here at -1 0. This is going to go up to -1 3. So this will go over around here. And then we've also got the point with coordinates 3 0 is going to move up to 3 3. So let's just get all of these labeled nicely. So we've got 3 3.

[00:03:51]We've got -1 3 here.

[00:03:54]We've got 1 1, so we can connect in between all of these.

[00:03:58]And then everything else is just going to go off to the left like this of gradient 1 and this is going to go off to the right. Let me make this a bit wider to fit it in. This is going to go off to the right with gradient -1. So, you can think of this effectively if we're moving this three units up, this is like moving the x-intercept here three units to the right. So, we're going to get an x-intercept at six here, which you could check by substituting in x is six if you like. And here we're moving three to the left from -1, so we get an x-intercept at -4. So, then all that remains now is actually just to draw the graph of y equals the modulus of this and then we'll have the function we're trying to integrate. So, we have y equals the modulus of all of this. We just need to take the negative parts where y is negative and reflect them in the x-axis. So, we just reflect this bit up over here. We keep everything else, so I'm just going to go over this in red so it's really clear what the red function is. And then the only real difference that's going to affect our problem cuz we're working with values of x greater than zero anyway is that this bit over on the right, instead of going off down underneath the x-axis, it's going off above the x-axis from this point six here.

[00:05:09]So, now we can try and actually find some of these areas to work out the value of A so that as we move along, the area is actually going to be 12. So, we can actually just start with this weird shape we've got here. So, instead of breaking this up into maybe a trapezium, a trapezium, a triangle, we can redraw this as we can think of it as one really big triangle going from the origin and this is the point six on the x-axis.

[00:05:35]So, this has got a width of six and this point at the top is 3 3, so it's got a height of three units. Then we've also got one smaller triangle to add in, so this smaller triangle is going to look like this, where this is crossing the x-axis, so the y-axis at two now, cuz our original y-intercept was -1. We've moved up three units. We've got the intercept at two there. So, the height of this, we can think of this as the base of this triangle as being two. Then we go along to this point with coordinates 1 1.

[00:06:07]So, we're going to get effectively the perpendicular height is going to be one. So, you can do 2 * 1 / 2 gives us an area of one for this little triangle. Then this big triangle we do 6 * 3 / 2.

[00:06:21]That's going to give us an area of nine.

[00:06:23]So, you see we've only got an area of 10 so far. So, we're going to have to move a little bit further along into this region over on the right. So, this is a line it's got gradient one. So, we can think about how far do we actually need to go along here so that we get we only need an area of two. So, cuz it's got gradient of one, this is a 45° angle in here. We're going to have an isosceles triangle. So, we need effectively the width and the height they're going to be the same. We'd need these to be two and two. So, we get 2 * 2 / 2 gives us an area of two to make the total area 12.

[00:06:58]So, then we can see from this picture it becomes a really simple geometry problem after dealing with this nasty-looking nested modulus function. This tells us then that we go along six and then another two to get an area under the curve of 12. So, then we can conclude that we need A to be equal to eight to solve this problem so that the integral's actually 12.