Change of variable theorem.

quasar987

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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.

TD

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

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

quasar987

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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?

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

quasar987

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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".

Hurkyl

Staff Emeritus
Gold Member
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.

quasar987

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Wow that clears some things up!

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

Hurkyl

Staff Emeritus
Gold Member
P.S. Is there a calculus book you've not read Hurkyl?
I've read all but &epsilon; of them for some &epsilon; > 0...

quasar987

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Gold Member
Yep, that's definitely too much of them... HallsofIvy

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

Brown Arrow

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

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