user friendly proof of fundamental theorem of calculus Silly question. Can someone show me a nice easy to follow proof on the fundamental theorem of calculus. More specifically,
$\displaystyle\int_{a}^{b}f(x)dx = F(b) - F(a)$
I know that by just googling fundamental theorem of calculus, one can get all sorts of answers, but for some odd reason I have a hard time following the arguments. 

The key fact is that, if $f$ is continuous, the function $G(x)=\int_a^xf(t)\,dt$ is an antiderivative for $f$. For this,
$$
\frac1h\,\left(\int_a^{x+h}f(t)\,dt-\int_a^xf(t)\,dt\right)
=\frac1h\,\int_x^{x+h}f(t)\,dt\xrightarrow[h\to0]{}f(x).
$$
The justification of the limit basically plays on the fact that $f$ is continuous. A formal proof requires dealing with the formal definition of continuity. Namely, given $\varepsilon>0$ by definition of continuity there exists $\delta>0$ such that $|f(x)-f(t)|<\varepsilon$ whenever $|x-y|<\delta$. If we choose $h<\delta$, then $|f(t)-f(x)|<\varepsilon$ for all $t\in [x,x+h]$. Then
\begin{align}
f(x)&=\frac1h\int_x^{x+h}f(x)\,dt\leq\frac1h\int_x^{x+h}(\varepsilon +f(t))\,dt=\varepsilon + \frac1h\int_x^{x+h}f(t)\,dt\\[0.3cm]
&\leq2\varepsilon + \frac1h\int_x^{x+h}f(t)\,dt=2\varepsilon+f(x). 
\end{align}
Thus
$$
f(x)-\varepsilon\leq \frac1h\int_x^{x+h}f(t)\,dt\leq f(x)+\varepsilon, 
$$
showing the convergence.
Now $G(a)=0$, and $G(b)-G(a)=G(b)=\int_a^bf(t)\,dt$. If $F$ is any other antiderivative of $f$, we have $F'=f=G'$, so $(G-F)'=G'-F'=0$, i .e. $G(x)-F(x)=c$ for some constant. That is, $F(x)=G(x)-c$. Then
$$
F(b)-F(a)=(G(b)-c)-(G(a)-c)=G(b)-G(a)=\int_a^bf(t)\,dt. 
$$
