Find a function $f(x)$ in an integral (Related question here).
Is there a way to calculate the function $f(x)$ in this integral in terms of $x$ without using $a,b,c$:
$$\int_{a}^{b} f(x)dx=c$$

Two examples $\rightarrow$ how do find these functions $f(x)$:
1) How do find that $f(x)$ can be $x^2$?:$$\int_{0}^{1}f(x)dx=\frac{1}{3}\Longleftrightarrow f(x)=x^2$$
2) How do find that $f(x)$ can be $\frac{x^4(1-x)^4}{1+x^2}$?:$$\int_{0}^{1}f(x)dx=\frac{22}{7}-\pi\Longleftrightarrow f(x)=\frac{x^4(1-x)^4}{1+x^2}$$
 A: Short answer: You can't, even if you limit the kind of function to polynomial or rational.
You can't tell what the limits $a$ and $b$ are, since you can choose any for your given value. For example, if you have $\int_a^b f(x)\,dx=c$ where $a\ne b$ then you also have
$$\int_0^1 (b-a)f(a+(b-a)u)\,du=c$$
The value of $c$ is also meaningless, since you also have, if $c\ne 0$,
$$\int_0^1 \frac{(b-a)}cf(a+(b-a)u)\,du=1$$
If the original $f$ was polynomial/rational, so is our new integrand.
A: Fix $a,b,c$. Take ANY integrable function $g(x)$, such that $G:=\int_a^b g(x)dx\neq 0$ . Then $f(x)=cg(x)/G$ gives you what you want. For example, take $g(x)=x^2$, with antiderivative $x^3/3$. Then $\int_0^1 g(x)dx=1/3$. So:
$$\int_0^1 \frac{22/7-\pi}{1/3}x^2dx=\frac{22}{7}-\pi.$$
You could also use $f(x)=g(x)-G/(b-a) + c/(b-a)$ whenever $(b-a)\neq 0$. 
A highbrow answer: An integral can be thought of as a function (more specifically, a functional). For example we can define $I(f):=\int_a^b f(x)dx$, for fixed $a,b,c$. So it's a function that maps functions to numbers. This shows that integrals are not one-to-one. Indeed if $I(f)=c$ then so does $I(f+g)=c$ for any $g$ such that $I(g)=0$. This is related to the fact that $I(f+g)=I(f)+I(g)$, is a linear functional.
