Q:

Average value of current or voltage

I had read that average value of current must be calculated for half time period.
For example  if we apply it for a sine value full time period then its value becomes zero. But for any arbritrary graph which doesn't have shape similar above and below x axis. I mean that they have different kind of shape for half timperiod and for another half, 
(Periodic function/graph)
Then by calculating avg. Value for current (calculation for half timeperiod) don't we lose the real meaning of average.
Should we include that negative part of waveform to our average I , by taking modulus?

A:

Generally, for AC currents we want to know the effect of the current. For example, how much power or heating does it provide. The standard method for doing this is to calculate or measure the true RMS (root mean square) of the current waveform. If you like, the resultant value gives you the equivalent DC current that would have the same effect.

simulate this circuit – Schematic created using CircuitLab
Figure 1. A strange AC waveform.
A simple example may help. In Figure 1 we have an AC waveform whose geometric average or integral will be zero. Clearly the power delivered is non-zero so let's calculate the effective current.

Power is proportional to \$ I^2 \$.
For the first second power is proportional to \$ 10^2 = 100 \$.
For the second second(!) power is proportional to \$ 5^2  = 25  \$.
For the third second power is proportional to \$ 0^2 = 0 \$.

That's the squared part of RMS done. Now get the mean.

\$ Mean \; of \; squares = \frac {sum\;of\;squares}{periods} = \frac {100 + 25 + 0}{3} = \frac {125}{3} = 42 \$.

Now get the root.

\$ RMS =  \sqrt {Mean \; of \; squares} = \sqrt {42} = 6.5 \$. So the effective current is 6.5 A.

For the negative half-cycle the result will be the same due to the squaring.
Note that if we just calculated the "average" current for one half-cycle (the the positive one, for example) we would have got \$ I_{AVG} = \frac {10 + 5 + 0}{3} = 5 \; A \$. The RMS value is much higher because the \$ 10^2 \$ term has a large effect.

