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Here's a version of the theorem:

$$\int_a^b f(g(x))g'(x)dx=\int_{g(a)}^{g(b)} f(u)du$$ provided that:

  1. $f$ is continuous on an interval $I$,

  2. $g'$ is continuous on $[a,b]$,

  3. $g[a,b]=I$ (i.e. the image of $g$ on $[a,b]$ is $I$),

  4. $[a,b]\subseteq I$.

1) Is condition 4 strictly necessary?

2) Is condition 3 correct, or should it be changed to "$g[a,b] \subseteq I$"?

EDIT: I have restated the above theorem more succinctly. $$\int_a^b f(g(x))g'(x)dx=\int_{g(a)}^{g(b)} f(u)du$$ provided that:

  1. $f$ is continuous on $g[a,b]$,
  2. $g'$ is continuous on $[a,b]$,
  3. $[a,b]\subseteq g[a,b]$.
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1 Answer 1

up vote 1 down vote accepted

You're only assuming that $f$ and $g'$ is continuous on the interval $I$ which could mean that $f$ and $g'$ doesn't behave "nicely" outside of $I$. The extreme case would be that $f$ and $g'$ aren't even defined outside of $I$. So you need condition 4 (where I think you mean $[a,b]\subseteq I$) in order to make sure that the left hand side is well-defined.

Condition 3 could be relaxed to saying that $g([a,b])\subseteq I$, i.e. the image of $[a,b]$ under $g$ is contained in $I$. This is to ensure that the right hand side as well as $f(g(x))$ on the left hand side is well-defined. Again think of the extreme case where $f$ is not defined outside of $I$.

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Thanks. I've edited out the "$\subseteq$" typos. About condition 3: since the interval I is arbitrary, do you think that editing "$=$" to become "$\subseteq$" is really a relaxation though (noting that the order of the conditions isn't important)? –  Ryan Feb 26 '13 at 9:48
    
@Ryan: If the inclusion $g([a,b])\subset I$ is strict, then you could just pick $I'=g([a,b])$ and then $I'$ would clearly satisfy 1.-3., but I don't think you can be sure that 4. holds. But this doesn't really matter, because 4. is only there to ensure that the integrals and composition makes sense. –  Stefan Hansen Feb 26 '13 at 9:53
    
This is exactly the unsureness I'm having. I can't decide if $=$ or $\subseteq$ would be the "simpler" version or if the $=$ is even correct. Thanks for your help! –  Ryan Feb 26 '13 at 9:56
    
Hey Stefan, would you refer to the cleaned-up version above and tell me if you think it's correct? Thank you. –  Ryan Feb 26 '13 at 10:17
    
@Ryan: Looks fine to me :) –  Stefan Hansen Feb 26 '13 at 10:19

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