= Solution
The profile is an even <function>, so its <Fourier transform> is real and
$$
\widehat T(\omega)=2\int_0^2(1-x/2)\cos(\omega x)\,dx.
$$
For $\omega\ne0$, <integration by parts> gives
$$
\int_0^2(1-x/2)\cos(\omega x)\,dx
=\frac1{2\omega}\int_0^2\sin(\omega x)\,dx
=\frac{1-\cos(2\omega)}{2\omega^2}.
$$
Therefore
$$
\boxed{\widehat T(\omega)=\begin{cases}\displaystyle\frac{2\sin^2\omega}{\omega^2},&\omega\ne0,\\2,&\omega=0.\end{cases}}
$$
At zero, direct integration gives the area of the triangle, namely two; the displayed quotient has the same limit. In particular \b[$\widehat T(\omega)\geq0$ for every real frequency]. This is the <Fourier transform of a triangular function>.
One can also see the nonnegative square structurally: $T=\tfrac12\mathbf1_{[-1,1]}*\mathbf1_{[-1,1]}$, so the <convolution theorem> gives $\widehat T=\tfrac12(2\sin\omega/\omega)^2$.
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