Fourier Transform involving frequency differentiation property

I am not sure about intuition in general, but regarding the step-function FT being a sync function:

Note that the shape will remain the same, but the frequencies over which the FT of a particular step-function resides is a function of the pulse-width of the original signal. Namely, expanding a function in the time-domain actually shrinks the corresponding frequency-domain function (think slowing down voice recordings, the sound gets very low i.e. lower frequency).

That being said, as you decrease the pulse-width of a particular step-function the frequency components of that signal increase because now there is more change happening (to use a loose descriptor) in a shorter amount of time.

In contrast, if we expand the step-function in the time-domain to have a longer pulse-width then there is less change and the corresponding frequency components must be much lower.

In general, I look at a function and try and get a feel for how quickly it might be changing to get a rough idea. But as I said, I don't know of any general rule of thumb here.

You are absolutely right. As your function has a value of zero everywhere except from the [0, T[ interval, you don't need to set the limits any wider. Those 'outer' intervals will add nothing to the result. The integral below will give you the answer.

$$X_3(j\omega)=\int_0^Tx_3(t) \cdot e^{-j\omega t} dt=\int_0^T(1-{{t} \over {T}})\cdot e^{-j\omega t} dt$$