While calculating field and potential due to a continuous distribution of mass and charge,we use the standard methods of integration.Does it imply the validity of principle of superposition?If it does,then why we do not see it in the case of calculation of energy of the same distribution?

I don't understand your point about not seeing it in energy calculations. After all, potentials are just energy per unit charge or mass.

Oh!Thank you! But one cannot superpose two energies like potential and field.Because the latter two are point function.It mathematically follows from the fact that a^2+b^2 may not be equal to (a+b)^2

I'm still not sure if I understand your point. Can you give an example?

Oh!Thank you! But one cannot superpose two energies like potential and field.
in electric circuit analysis dealing with two or more voltage sources the resulting currents are calculated by superposition.

example:
lets suppose you have a circuit with 3 sources all different, v1, v2, and v3.
v2 and v3 are replaced with their internal resistance and the circuit is solved for v1.
next v1 is replaced with its internal resistance and the circuit is solved for v2.
last, v1 and v2 are replaced and the circuit is solved for v3.

the results are them algebracly added for the final results.

Continuing on with Leopold's electric circuit example, say you have 2 voltage sources in a circuit and you want to calculate the output (let's say the output is the voltage across a capacitor in the circuit). For definiteness, consider a single loop circuit containing sources v_1, v_2, and capacitor C. You know already that you can use linear superposition to get the voltage v_C=v_1+v_2. So why can't you use linear superposition to get the energy delivered to the capacitor?

The energy w_C delivered to the capacitor is the time integral of the power.

w_C(t)=\int_0^t v_C(\tau)i_C(\tau)d\tau

w_C(t)=\int_0^t v_C(\tau)\cdot C\frac{di_C(\tau)}{d\tau}d\tau

w_C(t)=C \int_{v_C(0)}^{v_C(t)}v_C dv_C

w_C(t)=\frac{1}{2}C[v_C^2(t)-v_C^2(0)]

Let's say for simplicity that the capacitor was uncharged at t=0, so that v_C(0)=0. Then we have for w_C(t):

w_C(t)=\frac{1}{2}C[v_C(t)]^2.

And now it's obvious why superposition doesn't hold when calculating the energy delivered to the capacitor. The energy is quadratic in a variable for which superposition does hold.

Well,I meant the energy of a charge distribution or something like that.What Tom2 has posted may be correct.But I did not mean this.

But it's the same basic idea. Look at the integral for the energy of a uniform charge distribution.

U=\frac{1}{2}\epsilon_0\int_V|\vec{E}|^2dV

What do you have in the integrand? A quantity that is quadratic in a variable for which superposition holds, namely the electric field.

Oh! ofcourse,I needed to go through your post 2nd time.Right.Now we may agree:
And now it's obvious why superposition doesn't hold when calculating the energy delivered to the capacitor. The energy is quadratic in a variable for which superposition does hold.

As I recall, for superposition to hold the system must be linear. That is it does not apply to non linear systems.
Superposition and Linearity go hand in hand.

As I recall, for superposition to hold the system must be linear. That is it does not apply to non linear systems.
Superposition and Linearity go hand in hand.

Indeed. A linear function f(x) is precisely a function that satisfies the principle of superposition, which can be written as two requirements:

1.) f(x_1+x_2)=f(x_1)+f(x_2) (additivity)
2.) f(kx)=kf(x) (homogeneity),

or equivalently as a single requirement:

f(k_1x_1+k_2x_2)=k_1f(x_1)+k_2f(x_2).

Note that under this definition, first degree polynomial functions of the form f(x)=mx+b qualify as linear if and only if b=0. That trips up most students when they first learn about linearity.