Anyone have a good proof for the second part of FTC? Does anyone have a good proof for the second part of the fundamental theorem of calculus? I haven't been able to find any good videos on it so far so I'd like someone to write it down and I can throw some questions on it if that's okay?
Prove: $$ \int_{a}^b f(x)\, dx = F(b) - F(a) $$
 A: My favorite proof doesn't use any Riemann sums, so I like it more. It is as follows:
We start by proving part one.
Let $G(x)=\int_a^{x}f(t)\,dt$. It then follows that $$G(x+h)-G(x)=\int_{a}^{x+h}f(t)\,dt-\int_{a}^{x}f(t)\,dt=\int_{x}^{x+h}f(t)\,dt.$$ Or, equivalently, $$\frac{1}{h}(G(x+h)-G(x))=\frac{1}{h}\int_{x}^{x+h}f(t)\,dt.$$ Now, consider the interval $I=[x,x+h]$. The function $f(t)$ has a maximum on this interval, so we will call it $M$. It also has a minimum on this interval, we'll call it $N$. It is a known fact that $Nh\leq\int_{x}^{x+h}f(t)\,dt\leq Mh$. Dividing through by $h$ gives $N\leq\frac{1}{h}\int_{x}^{x+h}f(t)\,dt\leq M$. Clearly, $\lim_{h\to 0}M=\lim_{h\to 0}N=f(x)$. Thus, by the squeeze theorem, $$G'(x)=\lim_{h\to 0}\frac{1}{h}(G(x+h)-G(x))=\lim_{h\to 0}\frac{1}{h}\int_{x}^{x+h}f(t)\,dt=f(x).$$
Now for part two.
Since $G(x)$ is an antiderivative of $f(x)$, any other antiderivative, $F(x)$, of $f(x)$ satisfies the relation $F(x)+C=G(x)$. Since $G(b)-G(a)=\int_{a}^{b}f(t)\,dt$, it follows that $\int_{a}^{b}f(t)\,dt=(F(b)+C)-(F(a)+C)=F(b)-F(a)$. The $t$ can now be changed to an $x$ because it is a dummy variable. Thus, $\int_{a}^{b}f(x)\,dx=F(b)-F(a)$. 
A: Intuitively, this theorem says that "the total change is the sum of all the little changes", and this can be made into a rigorous proof.
Here is some intuition.
Chop the interval $[a,b]$ up into tiny pieces:
$a = x_0 < x_1 < \cdots < x_N = b$.
Let $\Delta F_i$ be the change in $F$ as its input changes from $x_i$ to $x_{i+1}$.
Of course, $F(b) - F(a) = \sum_i \Delta F_i$.  ("The total change is the sum of all the little changes.")
But $\Delta F_i \approx F'(\xi_i) \Delta x_i$, where $\xi_i$ is any point in $[x_i,x_{i+1}]$.  Hence
\begin{align*}
F(b) - F(a) &= \sum_i \Delta F_i \\
& \approx \sum_i F'(\xi_i) \Delta x_i \\
& \approx \int_a^b F'(x) \, dx.
\end{align*}
This intuitive argument can be made into a rigorous proof by using the Mean Value Theorem to write $\Delta F_i = F'(\xi_i) \Delta x_i$ with exact equality, for some $\xi_i \in (x_i, x_{i+1})$.
Assume $F$ is continuous on $[a,b]$, differentiable on $(a,b)$,
and $F' = f$ is integrable. Let $\epsilon > 0$.  There exists $\delta > 0$ such that if
$a = x_0 < x_1 < \ldots < x_N = b$ is a partition of $[a,b]$ with mesh less than $\delta$, and $\xi_i \in [x_i,x_{i+1}]$ for $i = 0, \ldots, N - 1$, then the Riemann sum $\sum_{i=0}^{N-1} f(\xi_i) \Delta x_i$ is within $\epsilon$ of $\int_a^b f(x) \, dx$.
So, let $a = x_0 < x_1 < \ldots < x_N = b$ be a partition of $[a,b]$ with mesh less than $\delta$, and select $\xi_i \in (x_i,x_{i+1})$ so that $\Delta F_i = F'(\xi_i) \Delta x_i$.  (This is possible by the Mean Value Theorem.)  Then we have that
\begin{align}
F(b) - F(a) &= \sum_i \Delta F_i \\
&= \sum_i F'(\xi_i) \Delta x_i 
\end{align}
is within $\epsilon$ of $\int_a^b F'(x) \, dx$.
Because $\epsilon > 0$ was arbitary, we see that $F(b) - F(a) = \int_a^b F'(x) \, dx$.
A: What about using the first part of the Fundamental Theorem of Calculus? It seems you have no problem with that part. The first part of FTC says that the function $G(x)=\int_a^x f(t) dt$ is the unique antiderivative of $f$ (so $G'(x)=f(x)$ for any $x$) satisfying $G(a)=0$.
Remark also that $\int_a^b f(t) dt=G(b)=G(b)-G(a)$.
Now, you now that if $F$ is another antiderivative of $f$ on $[a,b]$, then $F$ and $G$ differ from a constant. So there exists a constant $C$ such that $F(x)=G(x)+C$ for any $x$.
So $F(b)-F(a)=G(b)-G(a)=\int_a^b f(t) dt$ for any antiderivative of $f$ on $[a,b]$.
A: If you are asking for the proof of $\int f(x)\,dx=F(x)\implies F'(x)=f(x)$
$$F(b)-F(a)=\lim_{n\to\infty}\frac1{n}\sum_{r=na}^{nb}f\left(\frac{r}{n}\right)$$
Letting $b=a+h$ with $h=\frac{1}{n}$
$$\lim_{h\to0}F(a+h)-F(a)=\lim_{n\to\infty}\frac{1}{n}\sum_{r=na}^{na+1}f\left(\frac{r}{n}\right)$$
$$\lim_{h\to0}F(a+h)-F(a)=\lim_{n\to\infty}\frac1{n}f(a)$$
$$\lim_{h\to0}\frac{F(a+h)-F(a)}{h}=f(a)$$
$$F'(a)=f(a)$$
A: If you will forgive me for linking to my own site, I wrote a blog post for my students about understanding the fundamental ideas of one variable calculus. 
The proof the the second fundamental theorem of calculus takes place before what I called definition 4 (defining integrals as areas) and theorem 5 (the second fundamental theorem).
