How To Set The Area Between Two Functions Equal To A Constant? Please pardon the broad nature of this question. Suppose two functions encompass an area between them. What approach might be taken to adjust either function through adding constants to set the area between them equal to a constant? Part of the problem I have encountered is that every time one graph is moved the intercepts change, which makes the process rather time consuming. Is there a way to do this without writing a computer program? 
 A: It sounds like you have $f(x)$ and $g(x)$, each of which ranges from $x=a$ to $x=b$ and the area between them is what you are talking about.  If $f(x) \gt g(x)$, then the area is $\int_a^b (f(x)-g(x)) dx$.  If you add a constant $c$ to $g$, the area will decrease by $c(a-b)$.  Is this what you are thinking?
Added:  if you are integrating between the intersection points of $f$ and $g$ you are correct that the area between them will be influenced by changes in the intersection point.  If the intersection points are $a$ and $b$, adding a constant $c$ to $g$ will subtract $c(a-b)$ from the integral from $a$ to $b$.  Now you need to deal with the shift of the intersection points.  Without knowing more about $f$ and $g$ there isn't much to say.  If $f$ and $g$ are differentiable, adding a small constant $\epsilon$ to $f$ will shift the lower intersection point to the left by $\frac {\epsilon}{f'(a)-g'(a)}$ so the added area will be about $\frac 12 \frac {\epsilon}{f'(a)-g'(a)}\epsilon$.  Similarly at the right end, it will shift the intersection point left by $\frac {\epsilon}{f'(b)-g'(b)}$ but you expect the denominator to be negative here.  The added area is similar, $\frac {\epsilon^2}{2(g'(b)-f'(b))}$.  The total increase by raising $f$ is then $\epsilon (b-a)+\frac {\epsilon^2}{2(f'(a)-g'(a))}\frac {\epsilon^2}{2(g'(b)-f'(b))}$.  If the derivatives are (almost) constant at $a,b$ you can solve the quadratic to find (an approximate) solution.
Added again:  If you can integrate $f$ and $g$ analytically, you can turn the problem into an algebraic one.  Taking the cosine example, we have $\int_{-a}^a \cos x \ dx = 2 \sin a$.  To get an area $A$ under one hump of $\cos x - b$, we note that the intersection is $\pm \arccos b$, so we have $\int_{-\arccos b}^{\arccos b} (\cos x - b) dx=2 \sin (\arccos b)-2b\arccos b=2\sqrt {1-b^2}-2b\arccos b$  You need to solve $A=2\sqrt {1-b^2}-2b\arccos b$ for $b$.  You won't be able to do this algebraically, but a one dimensional root finder will make quick work of it.
A: Are you looking for the area between the two curves, as defined by their points of intersection? So you're not just looking for $\int_a^b f(x)-g(x) dx$ for fixed a and b, you want to be able to evaluate it for the points where $f(x)=g(x)$ (assuming a well-defined pair of intersection points).
Unfortunately, I don't think there's any well-defined theory to look at what happens to those points of intersection and the area between the graphs when you add constants, with one major barrier being that for two arbitrary curves it can be quite difficult to prove that there are two points of intersection in the first place, although you can look at the multiplicity of roots of $[f(x)+c]-g(x)$ as c varies to work out where $f(x)+c$ and $g(x)$ are tangential, and hence the area of intersection is 0.
