Can we possibly combine $\int_a^b{g(x)dx}$ plus $\int_c^d{h(x)dx}$ into $\int_e^f{j(x)dx}$? 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.
 A: 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
$$
\int_a^b g+\int_c^d h=\int_e^f(\tilde{g}+\tilde{h})=\int_e^f j.
$$
A: 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
$$u=\frac{b-a}{f-e}(x-e)+a,\qquad v=\frac{d-c}{f-e}(x-e)+c.$$
A: 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.
