#### quasar987

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[tex]\int_a^b e^{k(x-2)}dx[/tex]

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

[tex]\int_a^b e^{k(x-2)}dx = \int_{a-2}^{b-2} e^{kt}dt[/tex]

The "rigourous-substitution" way goes:

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

[tex]\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 [/tex]

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?

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

Sometimes, the forward direction is most convenient. For example, when integrating √(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

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Wow that clears some things up!

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

P.S. Is there a calculus book you've

I've read all but ε of them for some ε > 0...P.S. Is there a calculus book you've not read Hurkyl?

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

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

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

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