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Integration by substitution
By Carl Evans (P2080) on Friday, February 18, 2000 - 04:10 pm:

Hi, I've recently started working on some integration questions that require substitution. I can follow the method generally, but do not understand how one can change du to dx by using the chain rule i.e. dx/du ×du. I don't understand because I was under the impression that dx was only 'there' to remind you of what you are intergrating with respect to. how can a reminder be manipulated algebraically by division, multiplication etc?

I hope you understand why I'm having difficulty.

Thanks in advance

Carl

By Neil Morrison (P1462) on Friday, February 18, 2000 - 07:44 pm:

dx is short for very small change in x. In mathematical integrals, dx is tending to zero anyway, but if you use integrals in practical senses (areas, surface volumes, etc) you have an actual value of dx. If you do Physics or Mechanics, you may recall that the moment of inertia of a shape (especially a disc or bar) can be found by breaking up the shape into little tiny pieces, and considering each little piece, and then adding them all together (ie making an integration) How little the pieces are is usually defined by a very small change in x (or another letter instead of x, eg r for radius) and if you're doing the integral mathematically, then the change is almost zero, and so dx is used.

By Sean Hartnoll (Sah40) on Saturday, February 19, 2000 - 12:15 am:

Hi. You're quite right to be confused. Similar conceptual difficulties happen in differentiation when one is told first that dy/dx is not a fraction and then someone goes and justifies the chain rule by writing dy/dx = dy/dz dz/dx and "cancelling" the dz's.

What is happening is that there are two ways of looking at the problem, both of which are useful under different circumstances.

The first is the "formal" approach. Under this approach the dx is just a reminder as you say and dy/dx is not a fraction. Both integrals and derivatives are defined as limits. Proving rigurously the chain rule of differentiation and the substitution rule for integration (which, as it goes, are effectively the same thing) are then exercises in analysis that are usually done in the first or second year of a maths degree.

The second approach is more "informal" but retains the intuitive idea of what a derivate and an integral actually are. dy/dx is considered a fraction of very small numbers and integrals are summations with a small stepsize dx. Considered like this we can write things like dy = dy/dx × dx and make this change inside an integral. So the substitution rule is just saying that if you want to make the substituion inside the integral of, say, y = x2, then you also have to let dy=2×x×dx. You could see this as saying that a small step in x results in a bigger step in y (for x>1). As you change variable, the size of the steps also changes, as this is reflected in how you change dx for dy.

Hope this helps,

Sean

By Carl Evans (P2080) on Monday, February 21, 2000 - 09:45 pm:

Sean and Neil, thanks for your help. I didn't know that the proof for substitution (and the chain rule) is studied at degree level. I was under the impression that I should understand both concepts now. So I'm a bit relieved. I guess I should just accept that whenever I change a variable e.g. from x to u, I multiply by dx/du?

By Sean Hartnoll (Sah40) on Tuesday, February 22, 2000 - 12:41 am:

What you should try to understand is what is happening geometrically, the "informal" understanding I talked about above. There is a clear reason why dx goes to dx/du×du. It is because when x increases by dx (a small amount), then u(x) increases by du/dx×dx = du, implying that dx = dx/du×du. Draw a picture to see what is happening.

Sean

By Carl Evans (P2080) on Wednesday, February 23, 2000 - 06:30 pm:

Sean, I'm getting closer to understanding this, but I haven't quite got it. Sorry to come across as being a bit slow, but once I get it, I won't forget it. I'm going to paste what I have written to another help column as this, I'm sure, will give you a better understanding of why I'm struggling with this particular topic. That's not to say that I don't understand what you have said above; I found that helped a lot. Cheers.


Hi, I'm struggling to grasp a fundamental concept behind integration via substitution. I can do most of the set questions without difficulty, but do not fully understand why it works. My teacher has tried to explain it to me, and I've received help from your service, but I'm still not satisfied. The following is an easy question to illustrate what it is I don't understand:

Integrate: x3 (x4 + 7)5 dx

Now, from the integration I have studied thus far, the 'dx' after the equation is there to remind me that I am to integrate this function with respect to x. Unless I'm mistaken, it isn't 'multiplied' in anyway to x3 (x4 + 7)5.

So, with this in mind, I'm going to solve this equation. The following method is the one I prefer; I'm aware of the other.

Let u = x4 + 7

du/dx = 4x3

du = 4x3 × dx (note the multiplication)

So, ¼ × du = x3×dx

Now, if x3 (x4 + 7)5 was 'multiplied' to dx, I could substitute the x3 dx for ¼ du:

I would therefore have to integrate:

¼ u5 du

= u6/24 + c

= (x4 + 7)6/24 + c

So, what I really don't understand is how , x3 dx can be substituted for x3×dx unless the former was actually multiplied to dx (or using the other method why dx/du×du = dx)

In trying to understand this, I've gone back to elementary differentiation and integration 'to see what's going on'. However. I still haven't quite got it. Thus:

If y = x2
dy/dx = 2x

Is my problem linked in to the fact that 2x × dx = dy?

I hope I have made clear what it is I do not understand, and I would be VERY grateful for any help you may be able to give.

Thanks

Carl
By Anthony Curtis (P684) on Monday, February 28, 2000 - 10:41 am:

Hi Carl,
The way I would explain it is as follows:

You said if y=x2
then dy/dx=2x

Now, I don't know if you have done differential equations but if you have you should be able to see that this is a very simple example of one. (Don't worry if you haven't, just read below and tell me if you understand)

if you had been using u=x2 as a substitution for a certain integral, then used du/dx=2x,
dx=du/2x (i.e. treating the du/dx as a fraction and multiplying the dx) and then subbing in, you would be able to integrate.

going back to the differential equation for a sec,
if dy/dx=2x
then the integral of 1 with respect to y = the integral of 2x with respect to x, and so y=x^2

Now, if you accept the chain rule that dy/dx=(dy/du)(du/dx), then we should be able to sort out your problem.

lets choose an integral y=integral of(2x(x2 + 9))dx
if we differentiate both side w.r.t x we get dy=2x(x2 + 9)dx
i.e. dy/dx = 2x(x2 + 9)
but we know that if we make a substitution of u=x2 + 9 we can say
(dy/du)(du/dx)=2x(x2 + 9)
we know du/dx=2x
so (dy/du)(2x)=(2x)(x2 + 9)
so the 2x's can cancel and we can put in our u=x2 + 9 to give

dy/du=u
another differential equation.

so we can say that the integral of 1 w.r.t y= integral of u w.r.t u
which gives, y=(1/2)(u2)
therefore y= ((x2 + 9)2)/2

which is what you would get if you did the initial integral by substitution. You can also check it with normal integration as a substitution isn't actually needed because 2x(x2 + 9) is just 2x3 + 18x.

(w.r.t.= with respect to)

I hope some of this helps. write back if you need anything explained.

By Carl Evans (P2080) on Tuesday, February 29, 2000 - 05:51 pm:

Hi Anthony,

Thanks for your help. I can follow what you are saying, at least in principle, but unfortunately I haven't started differential equations yet. However, it is the very next topic on my scheme of work. Once I've started working through that section I'll have a better understanding of the above. I'll write back if I have any further problems.
Cheers.

P.S. Sean: by studying maths at Cambridge I imagine you are extremely competent at your subject, but did you 'accept' certain things at A level until you could understand 'why' at university?

Carl