Many applications of physics theory involve computations of integrals. Examples are voltage, force due to liquid pressure, surfaces...

In some cases, when there is linear dependence between two terms - I guess -, the integral is exactly equal to the mean value of the extrema of integration. However, I still feel uncomfortable with such procedure. How can I recognize when it is possible to compute an average instead of an integral? Is there an example of reasoning I should go through?

Thanks very much!

Edit: I've added here an example, to be clearer:

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When computing the capacitance of a spherical capacitor, at the end we get $$\frac{4\pi\epsilon_0r_ar_b}{r_b-r_a},$$ where $4\pi\epsilon_0r_ar_b$ is the mean value between the two surfaces $4\pi{r_a^2}$ and $4\pi{r_b^2}.$

The result, therefore, is $$C=\frac{\epsilon_0A_{av}}{d},$$ where $A_{av}$ is the average surface or area.

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    $\begingroup$ The integral is the general solution. In some specific cases, you will find that the integral is solvable in closed form and alternative, simpler expressions can be used (which may or may not include an average). You see to know that. So what is the question? $\endgroup$ – Brick Jan 13 at 14:31
  • $\begingroup$ @Brick I'm aware it is possible, but usually I see that later, a posteriori. So, I'm asking for a simple clarification of why that is true (i.e. which are the circumstamces where we can use an alternative to an integral). $\endgroup$ – Shootforthemoon Jan 13 at 14:38

Average of a given function, say $f(x)$ from $x_1$ to $x_2$ is found via the following integral:

$$\langle f(x) \rangle =\frac {\int_{x_1}^{x_2} f(x) dx}{x_2-x_1}$$

Now when the function you are considering is a linear function i.e., $f(x)=x$ (say) then you get:

$$\langle f(x) \rangle =\frac {\int_{x_1}^{x_2} x dx}{x_2-x_1}$$

$$ \langle f(x) \rangle =\frac { x_2+x_1}{2}$$

For cases otherwise the integral isn't the arithmetic mean of these and hence an integral is necessary to find the average.


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