Paying close attention to the documentation for [`Fourier`](http://reference.wolfram.com/mathematica/ref/Fourier.html) and [`FourierTransform`](http://reference.wolfram.com/mathematica/ref/FourierTransform.html) one notes that the coefficients of the Sum/Integral terms are different; therefore, to obtain a discrete transform with amplitudes equal to those from the continuous transform, one must multiply the former by Sqrt[2 Pi / n] where n is the length of the dataset:

The continuous waveform:

    DiscretePlot[
     Evaluate[Abs@FourierTransform[2 Sin[x], x, w] /. 
       DiracDelta -> DiscreteDelta], {w, 0, 2}]

![Mathematica graphics](https://i.sstatic.net/PxEMG.png)

and the discrete waveform:

    With[{datalength = 100}, 
     ListPlot[(Sqrt[2 Pi]/Sqrt[datalength]) Abs[
        Fourier[Table[2*Sin[x], {x, 0, datalength}]]], Joined -> True, 
      PlotRange -> {{0, 2}, All}, DataRange -> {0, 2Pi}]]

![Mathematica graphics](https://i.sstatic.net/Z0f3V.png)