Intuitive explanation of (why/how) integration gives us the average value of a continuous function? I have come across this formula -
$$\langle f(t) \rangle = \frac{1}{T}\int_0^T f(t) \, \mathrm{d}t$$ 
For finding averages in many places...which is proved using definite integral as limit of a sum etc...However I am not able to get its intuitive explanation , which can make this result look obvious(or at least justify it logically)...
Any Help is appreciated.
 A: I think you should look at the geometric representation of integrals of continuous functions in $\mathbb{R}$. It's easy to see that while integrating, we are actually adding the values of a certain function at all the points in a certain interval (multiplied by a weight $\text{dt}$). Now, when we divide it by the length of the interval, we get the average value of the function (as $\sum{\text{dt}}$ is the length of the interval) It's not formal, but I think it's easy to visualize it this way. 
A: 
Hint: You might find pages $8$ to $24$ of this presentation helpful. They provide some nice graphics about the relationship of
  \begin{align*}
\frac{1}{b-a}\sum_{i=1}^{n}f(x_i)\Delta x\qquad\text{and}\qquad \frac{1}{b-a}\int_a^b f(x)\,dx
\end{align*}
  with respect to the average value of a function.

A: Here's an intuitive one. Consider the discrete version of your function $f(n)$, with $n\in \mathbb{N} $. The average of this function over $0,1,...n$ is $$\langle f(n) \rangle=\dfrac{1}{n+1} \sum_{i=0}^{n}f(i)$$ When $f$ is continous, you replace the summation with integration. 
