Modus operandi for proving Evaluation Fundamental Theorem of Calculus (Abbott p 200, Spivak p 272 T14.2)

Question

1. How can we presage to use Mean Value Theorem to start the proof?

2. Mean Value Theorem engenders a point in an open interval. Shouldn't this be $x_i \in (t_{i - 1}, t_i) $?

After reading LittleO's post I discovered intuition at artofproblemsolving.com.

Answer 1

1. You have a function $g$ and you know nothing about it except that it is differentiable on an interval, and you want to show something about its behaviour in that interval. The mean value theorem is a good bet.
2. If $x_i \in (t_{i-i},t_i)$ then $x_i \in [t_{i-i},t_i]$ so the statement is not wrong.
3. If you start driving your car at time $a$ and stop at time $b$, and your speedometer is printing out a graph of your speed over time given by $f(t)$, and your odometer is printing out a graph of the distance that you've travelled given by $g(t)$, then the total distance that you travel $g(b)-g(a)$ is equal to the area under the graph of $f$ from $a$ to $b$.

Answer 2

The intuition for the (second) fundamental theorem of calculus is that \begin{equation} \text{"the total change is the sum of all the little changes".} \end{equation}

The total change is $g(b) - g(a)$. The little changes are \begin{equation*} g(t_i) - g(t_{i-1}) \approx g'(\xi_i)(t_i - t_{i-1}) \end{equation*} (where $\xi_i$ is any point in $[t_{i-1},t_i]$.)

By adding up all the little changes, you get the total change. \begin{align*} g(b) - g(a) &= \sum g(t_i) - g(t_{i-1}) \\ &\approx \sum g'(\xi_i)(t_i - t_{i-1}). \end{align*} Notice that this last expression is a Riemann sum for the integral $\int_a^b g'(t) \, dt$.

It's wonderful that the mean value theorem allows us to replace the approximate equalities with exact equalities, which helps us turn this intuitive argument into a rigorous proof.