Why does dividing the definite integral by the width give the average value? I understand how to apply the 'formula' but I'm having difficulties being convinced/understand that dividing the area by the width gives you the average value as the formula suggests.
I don't understand how the definite integral and the width are related in a way such that dividing them gives you the average function value?
Thanks
 A: That's easy to see: the definite (Riemann) integral is the limit of Riemann sums :
$$\int_a^b f(x)\,\mathrm d\mkern1mu x=\lim_{n\to\infty}\sum_{k=1}^nf(\xi_k)\frac{b-a}n,\quad\text{where}\quad\xi_k\in[x_{k-1},x_k],\quad x_k=a+k\,\frac{b-a}n.$$
Thus 
$$\frac1{b-a}\int_a^b f(x)\,\mathrm d\mkern1mu x=\lim_{n\to\infty}\frac{\sum_{k=1}^nf(\xi_k)}n$$ is the limit, as $n$ tends to $\infty$, of the averages of $n$ values of the function $f(\xi_k)$.
A: This is nothing more than a definition. But, to motivate the definition, we could think of an example where $f(t)$ is the speed at time $t$ of a car driving along a straight road. What is the average speed of the car during the time interval from $t=a$ to $t=b$? It should be a number $s$ such that $s\times(b-a)$ is equal to the total distance the car traveled during this time interval. In other words, 
$$
s(b-a)=\int_a^b f(t) dt.
$$
Divide by $b-a$ to get the average speed $s$.
A: I’m surprised that no one has yet mentioned areas explicitly.  
For simplicity, let’s say that $f(x)\ge0$ in $[a,b]$. Then the definite integral $\int_a^bf(x)\,dx$ gives you the area of the region bounded by the $x$-axis below, $f(x)$ above, and the lines $x=a$ and $x=b$ on the left and right. That is, it’s the area of a box with three straight sides and one wiggly side. If you think of finding the average value of $f$ as straightening out those wiggles, then it’s reasonable to define this average value as the height of a rectangle that has the same width and area as the box with the wiggly side. Since the area of a rectangle is its width times its height, this means that the average value of $f$ over the interval $[a,b]$ is the area of the wiggly-sided box divided by its width, namely $$\overline f=\frac1{b-a}\int_a^bf(x)\,dx.$$ A possibly useful physical model is the side view of a pan of water that’s sloshing around in the pan. The water level when completely still is the average depth of the sloshing water.
A: To gain some intuition, imagine a function that takes the value $1$ on the interval $[0,1]$ and value $3$ on the interval $[1,2]$, i.e.
$$ f(x)= \begin{cases}
    1, & \text{for } 0 \leq x \leq 1 \\
    3, & \text{for } 1 < x \leq 2 
  \end{cases}$$
When you think about it for a bit, it seems plausible to claim that the average value of that function should be $2$. And indeed, if you use your formula, you get that the area divided by the "length" is $$\frac{1+3}{2}=2.$$
If you would like to formally see why that formula works, you would need to see the proof of the "average value formula." You can find a very good explanation for example here.
A: The first thing to keep in mind is that the mean value is a definition.
For the sake of simplicity we will take the one dimensional problem. 
The only fixed quantity is the length of the interval, say $L$ in which a variable $\phi$ is averaged.
The goal is to find the most easiest way to compute the integral:
$$\int_{L}{\phi(x)\,dx}=\overline{\phi}L$$
This unknown value $\overline{\phi}$ coincides with the mean value of $\phi$ along $L$ and its explicit expression is:
$$\overline{\phi} = \frac{1}{L}\int_{L}{\phi(x)\,dx}$$
A: You can take it analogous to$$X =\frac{ \sum_{i=1}^{n}{f_ix_i}}{\sum_{i=1}^{n}{f_i}}$$
Where $X$ is the average of the distribution of data.
Here $dx$ is analogous to frequency, $f_i$, since assuming for that interval for x nalue of function $f$ is fixed and $f(x)$ is analogous to value,$x_i$. Since integral is equivalent to summation. We can say average of the function$$X= \frac{ \int_a^b f(x)dx}{\int_a^b dx}$$. Hope this will help understand.
