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I'm wondering if this is possible for the general case. In other words, I'd like to take $\int_a^b{g(x)dx} + \int_c^d{h(x)dx} = \int_e^f{j(x)dx}$ and determine $e$, $f$, and $j(x)$ from the other (known) formulas and integrals. I'm wondering what restrictions, limitations, and problems arise.

If this is not possible in the general case, I'm wondering what specific cases this would be valid for, and also how it could be done. It's a curiosity of mine for now, but I can think of some possible problems and applications to apply it to.

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    Well in principle yes: let $j(x)=1$. Then let $e=0$, and let $f=\int_a^bg(x)\,dx + \int_c^d h(x)\,dx$. You get$a$nice equality, though not at all practical as a way to evaluate integrals.2012-07-06

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Here's a method that should allow one a large degree of freedom as well as allowing Rieman Integration (instead of Lesbegue Integration or some other method):

Let $\tilde{g}$ be such that: $\int_a^b{g(x)dx} = \int_e^f{\tilde{g}(x)dx}$ ...and $\tilde{h}$ follows similarly. Then they can be both added inside a single integral.

The first method that comes to mind is to let $\tilde{g}(x) = \dot{g}\cdot g(x)dx$, where $\dot{g}$, a constant, is the ratio of the old and new integrations. A similar method that comes to mind is to actually let $\dot{g}$ be a function.

Another method that I'm exploring, and is somwhat questionable, is to attempt to use $e$ and $f$ as functions, possibly even of $x$, although this may be undefined or just plain wrong.

I'll add ideas to this as I hopefully come up with better methods.

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    @Mercy: Sorry for the slow reply. I was trying to denote a multiplier by $\dot{g}$. I realize that that $\dot{g}$ sometimes means other things, but it is supposed to evaluate to a real number in this case. I believe that JL$_{344}$ did something like what I was thinking, especially with the $\frac{b-a}{f-e}$ in front of $g$.2012-07-06
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Assume $a< b$ and $c< d$. Let
$ e \le \min\{a,c\}, \ f \ge \max\{b,d\}, $ and set $j=\tilde{g}+\tilde{h}$ where $ \tilde{g}:=g.1_{[a,b]},\tilde{h}:=h.1_{[c,d]}: [e,f] \to \mathbb{R}. $ Then $ \int_a^b g+\int_c^d h=\int_e^f(\tilde{g}+\tilde{h})=\int_e^f j. $

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    @MattGroff If $X$ is a set, the function $1_X$ is the indicator function, and is designed to equal $1$ if an argument is a member of $X$, and $0$ otherwise. Mercy is using a period for multiplication, though I think it would be better as $\cdot$ (which is `\cdot` in $\LaTeX$).2012-07-06
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Certainly. In fact $e$ and $f$ can be anything you want, as long as they are not equal. An affine transformation is one way to do it. Namely if $j(x)=\frac{b-a}{f-e}g\left(\frac{b-a}{f-e}(x-e)+a\right) +\frac{d-c}{f-e}h\left(\frac{d-c}{f-e}(x-e)+c\right),$ then $\int_a^bg(u)du+\int_c^dh(v)dv=\int_e^fj(x)dx.$

This transformation follows from the change of variables $u=\frac{b-a}{f-e}(x-e)+a,\qquad v=\frac{d-c}{f-e}(x-e)+c.$

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    Why do we do all o$f$ that?2012-07-07