Q:

How do I fill the area from 0 to pi in my Sine wave and in the process get the area value?

I am new to Mathematica and I tried searching the docs but no specific examples there.
This is what I have achieved so far, returns an error, probably wrong syntax.
Plot[ Sin[x], {x, 0, 2 Pi}, Filling -> Range {x, 0, Pi}]

A:

No, there is no "special function that allows me to integrate and in the process, it does the shade", but you can shade the area and plot the points used in the integration. For example:
f[x_?NumericQ] := Sow[{x, Sin[x]}][[2]]
{#[[1]], Plot[f@x, {x, 0, Pi}, Epilog -> Point@#[[2]], 
              Filling -> Axis]} &@Reap@NIntegrate[f[x], {x, 0, Pi}]

A:

I don't really understand the question but you can try this:
Manipulate[Plot[{Sin[x], Sin[x] Boole[x < y]}, {x, 0, Pi}, 
    Filling -> 2 -> Axis, Frame -> True, 
    ImageSize -> 500, Epilog -> Inset[NIntegrate[Sin[x], 
      {x, 0, y}], {y, 1/2 Sin[y]}, {Right, Center}]], {y, 0, Pi}
]

A:

Taking the question at face value: you can calculate the area of the fill from the Graphics object that is generated by Plot, specifically the Polygon expression.
I will use this fill method and my old polyarea code.
The plot:
gr = Plot[{If[x < Pi, Sin[x]], Sin[x]}, {x, 0, 2 Pi}, Filling -> {1 -> Axis}]

The area:
polyarea = 
  Compile[{{v, _Real, 2}}, 
   Block[{x, y}, {x, y} = Transpose@v; 
    Abs[x.RotateLeft@y - RotateLeft@x.y]/2]];

Cases[Normal @ gr, Polygon[a_?ArrayQ] :> polyarea[a], {-4}]

{1.99894}

Normal is needed to convert the GraphicsComplex data into plain coordinate data.
Additional Plot precision will yield additional numeric precision:
gr = Plot[{If[x < Pi, Sin[x]], Sin[x]}, {x, 0, 2 Pi},
      Filling -> {1 -> Axis}, 
      Method -> {MaxBend -> 0.5}];

Cases[Normal @ gr, Polygon[a_?ArrayQ] :> polyarea[a], {-4}]

{1.99998}

For a function that crosses the axis you will need to account for the sign.  For example:
Plot[Re[Zeta[1/2 + I t]], {t, -7, 20}, Filling -> Axis, Method -> {MaxBend -> 0.2}]

Cases[Normal @ %, Polygon[a_?ArrayQ] :> Sign@Last@Mean@a * polyarea@a, {-4}]

Total @ %

{-0.00312878, -1.15988, 7.2798, 3.74814, 11.4189}

21.2838

Confirmation of the area calculated with NIntegrate:
NIntegrate[Re[Zeta[1/2 + I t]], {t, -7, 20}]

21.2838

