Average slope of a curve (right side of the equation) intuition

Question

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 $x_1$.

Answer 1

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.

Answer 2

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.