# Change of variable theorem.

1. Oct 5, 2005

### quasar987

There's this thing I've always wondered about but never really found the answer. However, I'm sure it's obvious to many of you. It goes as follow: How come the change of variable theorem (for simple integrals) is so complicated but the fact is that it never fails to say "let's set u =... Then, du = ...", which is not at all what the change of variable theorem is saying imho.

2. Oct 5, 2005

### TD

Perhaps I'm not understanding you, do you mean the substitution rule?

$$\int_{\varphi \left( \alpha \right)}^{\varphi \left( \beta \right)} {f\left( x \right)dx} = \int_\alpha ^\beta {f\left( {\varphi \left( t \right)} \right)\varphi '\left( t \right)dt}$$

3. Oct 5, 2005

### quasar987

Yeah, that's totally what I'm talking about. Why is it formulated in terms of function phi of a variable t? For exemple, compare the two following ways of changing variables for this integral:

$$\int_a^b e^{k(x-2)}dx$$

The "fast-substitution" way everyone use:

Set t = x - 2. then by definition of differential of a function of one variable, dt = (dt/dx)dx = (1)dx = dx. The bounds transform like so: x = a --> t = a-2, x = b --> t = b - 2. Hence, plugging all these substitutions in the original integral yields

$$\int_a^b e^{k(x-2)}dx = \int_{a-2}^{b-2} e^{kt}dt$$

The "rigourous-substitution" way goes:

Set $\varphi (t) = t + 2$. Then $\varphi '(t) = 1[/tex]. [itex]\varphi (t_1) = a \Leftrightarrow t_1 + 2 = a \Leftrightarrow t_1 = a - 2$. $\varphi (t_2) = b \Leftrightarrow t_2 + 2 = b \Leftrightarrow t_2 = b - 2$. So now according to the change of variable theorem,

$$\int_a^b e^{k(x-2)}dx = \int_{a-2}^{b-2} e^{k(\varphi(t)-2)}(1)dt = \int_{a-2}^{b-2} e^{k(t)}dt$$

It's so much complicated. SO why is it formulated in this complicated way rather than in the simplest way, which seem to always work anyway?

4. Oct 5, 2005

### amcavoy

I believe it is just a generalization. The second way in which you did it was practically the same as the first, you just showed all of your work.

Alex

5. Oct 5, 2005

### quasar987

What do you mean precisely in this case by "a generalisation", and what do you mean by "you just showed all your work"? It seems to me I did all the steps for both ways only the "rigourous way" has naturally more steps then the "usual way".

6. Oct 5, 2005

### Hurkyl

Staff Emeritus
The statement you're using of the change of variable theorem is the one that looks most like a direct application of the chain rule.

I.E. it's more natural to apply the transformation df(x) &rarr; f'(x) dx than it is to transform f'(x) dx &rarr; df(x)

But, I seem to remember at least one calculus text that stated the reverse formula as a corollary.

As to how people actually do the integral by hand, the forward direction would be written as the substitution x = t + 2, and worked from there.

Sometimes, the forward direction is most convenient. For example, when integrating &radic;(1-x²), the most natural thing to do is say x = sin u. Sometimes the reverse direction is more convenient, as you've observed.

Sometimes, it's most convenient to do it neither way -- I've had situations where using an implicit relation, like t² = x² + 2, gives the simplest integrand.

7. Oct 5, 2005

### quasar987

Wow that clears some things up!

P.S. Is there a calculus book you've not read Hurkyl? :tongue:

8. Oct 5, 2005

### Hurkyl

Staff Emeritus
I've read all but &epsilon; of them for some &epsilon; > 0...

9. Oct 5, 2005

### quasar987

Yep, that's definitely too much of them...

10. Oct 8, 2005

### HallsofIvy

Staff Emeritus
Fortunately calculus books have the property of all containing exactly the same thing!

11. Dec 9, 2010

### Brown Arrow

most of the calculus book contain the same example i have read a few.