Firstly, I apologise for the poorly formed title. The concept was quite hard to put into words.
Take the following Fourier Transform as a starter, where $f(t)=t$ in the domain $0<t<1$:
$\widetilde{F}(\omega)=\int_{-\infty}^{\infty}{(f(t)\cdot e^{-i\omega t})\mathrm dt}=\int_{0}^{1}{(t\cdot e^{-i\omega t})\mathrm dt}.$
We know that the integral of $f(t)$ over its domain is the area of a triangle, so:
$\int_{0}^{1}t\;\mathrm dt=\frac{1}{2}$.
My question is, is there any way that we can use the fact that the integral of $f(t)$ is the area of a triangle to simplify and evaluate the Fourier Transform described above?
The integral of $f$ only determines uniquely the value of the transform at $\omega=0$. All other values are pretty much free. For example, $f(t)=1/2$ has the same area as $f(t)=t$ on $[0,1]$ yet their transforms are very different.