Fourier Transform of a half triangle
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.

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