Average slope of a curve (right side of the equation) intuition Okay, I can somehow understand why average slope of a curve is defined in such a fashion: 
Let $f$ be a continuous function on an interval $[a,b]$. Then, we define its average over $[a,b]$ to be the number
$$
f_{\operatorname{avg}} := {1 \over b-a} \int_a^b f(t) dt
$$
Let $y = F(x)$ be a curve. Fix the starting point $x_0$ and the ending point $x_1=x_0+\Delta x$. Then, the slope at a point $z$ is the derivative $F'(z)$. Thus, the "average slope'' should mean
\begin{equation*}
(F')_{\operatorname{avg}} = {1 \over x_1 - x_0}\int_{x_0}^{x_1} F'(t) dt = {F(x_1) - F(x_0) \over x_1-x_0}.
\end{equation*}
To summarize: "Avg. slope = Sum all slopes of all tangent lines over an interval $[a,b]$ and divide it by the number of slopes over that same interval."
But right side of the equation somehow bothers me. I don't get intuitevely how average slope (sum of all slopes/number of slopes) equals slope of a function $F(x)$ between two endpoints $x_0$ and $x_1$. 
 A: Consider a partition of $[a,b]$ into subintervals $I_k:=[x_{k-1},x_k]$ $\ (1\leq k\leq N)$. You  would then define the measured slope of $f$ on the subinterval $I_k$ by
$$m_k:={f(x_k)-f(x_{k-1})\over x_k-x_{k-1}}\qquad(1\leq k\leq N)\ .$$
Note that
$$f(x_k)-f(x_{k-1})=m_k(x_k-x_{k-1})\qquad(1\leq k\leq N)
\ ,$$
and therefore
$$\sum_{k=1}^N m_k(x_k-x_{k-1})=\sum_{k=1}^N\bigl(f(x_k)-f(x_{k-1})\bigr)=f(b)-f(a)$$
(a telescopic sum), or
$$\sum_{k=1}^N {x_k-x_{k-1}\over b-a}m_k={f(b)-f(a)\over b-a}\ .\tag{1}$$
This shows that the weighted mean of the measured slopes over the small subintervals $I_k$ is equal to the measured slope over the total interval $[a,b]$, before any limit is taken.
How does $f'$ come in here? By the MVT for each $k\in[N]$ there is a $\xi_k\in I_k$ such that $m_k=f'(\xi_k)$. The LHS of $(1)$ can therefore be rewritten as
$$\sum_{k=1}^N f'(\xi_k)\>{x_k-x_{k-1}\over b-a}\ ,$$
which is a Riemann sum for
$${1\over b-a}\int_a^b f'(t)\>dt\ .$$
Since $(1)$ is true also "in the limit" we arrive at your formula in a "natural way", i.e., without making use of the FTC.
A: The right side is the slope of the line drawn between $(x_0,F(x_0))$ and $(x_1,F(x_1))$.  I would see that as the definition of average slope.  Averaging an infinite number of local slopes is more problematic.  The right side only depends on two values of the function.
