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)

# Appendix: Why isn't the amplitude "right" #

In the comments, the OP asks why the amplitude of the function `2 Sin[x]` appears to be ~2.5 in the transformed data.  Let's take a pedagogical approach here, in part to show off how one might use Mathematica to answer these types of questions on their own.

**Is there a relationship between the amplitude of the wave and the height of the peak in the transform?**

Hopefully, the answer is yes, but let's generate some data, plot it and see.

    testdata = 
     Table[{i, 
       Evaluate[
        Abs@FourierTransform[i Sin[x], x, w] /. {DiracDelta -> 
           DiscreteDelta, w -> 1}]}, {i, 1, 10, 1}]

![Mathematica graphics](https://i.sstatic.net/hso12.png)

**There is a linear relationship - yeah!  What is that relationship?**

Let's perform a linear least squares analysis on `testdata` to see what the slope of that line is.  I will assume that the plot should go through zero.

    LinearModelFit[testdata, x, x]["BestFitParameters"] // Chop
    (* {0, 1.25331} *)

So that means that the height of the peak in the transformed data will be 1.25331 times larger than the amplitude of the time-domain function.  Where does this 1.25331 come from?  Taking a closer look at `testdata`:

![Mathematica graphics](https://i.sstatic.net/sVfL2.png)

We see a Sqrt[2 Pi]/2 nestled in there.  Evaluating `N@Sqrt[2 Pi]/2` yields:   

    (* 1.25331 *)

Nice.