Prove the Fundamental Theorem of Calculus Prove the Fundamental Theorem of Calculus with this hypothesis:
If $f$ is integrable over $[a,b]$, if $g:[a,b]\rightarrow\Bbb R$ given by $g(x)=\int_{a}^{x}f(t)dt$ and $f$ is continuous in $x_0 \in [a,b] \implies g'(x_0)=f(x_0)\Rightarrow g'(x_0)-f(x_0) = 0 $
How can this be proven?
 A: I argue like this:
$g(x)
=\int_a^x f(t)\ dt
$,
so
$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
$
so
$\dfrac{g(x+h)-g(x)}{h}
=\dfrac1{h}\int_x^{x+h} f(t)\ dt
$.
Here is where I get
purposely sloppy.
Since $f$ is continuous,
it does not vary much
from $x$ to $x+h$,
and as $h$ gets small,
$f$ varies less and less.
Therefore,
$f(t) \sim f(x)$
for $t$ from $x$ to $x+h$,
so
$\dfrac1{h}\int_x^{x+h} f(t)\ dt
\sim \dfrac1{h}\int_x^{x+h} f(x)\ dt
= \dfrac1{h}f(x)\int_x^{x+h}  dt
= \dfrac{f(x)}{h}(h )
= f(x)
$.
Therefore,
as $h$ gets small,
$\dfrac{g(x+h)-g(x)}{h}
\sim f(x)
$
and the left side
is just the definition of the
derivative of $g$
at $x$ as $h \to 0$.
If you want to be more rigorous,
you can write
$\begin{align}
\dfrac{g(x+h)-g(x)}{h}
&=\dfrac1{h}\int_x^{x+h} f(t)\ dt\\
&=\dfrac1{h}\int_x^{x+h} (f(x) +(f(t)-f(x)))\ dt\\
&=\dfrac1{h}\int_x^{x+h} f(x)\ dt +\dfrac1{h}\int_x^{x+h}(f(t)-f(x))\ dt\\
&=f(x) +\dfrac1{h}\int_x^{x+h}(f(t)-f(x))\ dt\\
\end{align}
$
and use the $\delta-\epsilon$
definition of continuity
to show that
$\dfrac1{h}\int_x^{x+h}(f(t)-f(x))\ dt
\to 0
$
as $h \to 0$.
Similarly,
you can understand
the definition of derivative
as "sneaking up" on a function
using two points on the function
that get closer and closer
to see how the
slope of the line 
through the two points
becomes
(or gets as close as you want to)
the tangent.
Just as the preceding derivation of
the fundamental theorem
involves a remainder term
that goes to $0$ as $h \to 0$,
the definition of derivative
involves a term that
goes to $0$ as $h \to 0$:
$g$ is the derivative of $f$
at $x$ if
$\big|\dfrac{f(x+h)-f(x)}{h}- g(x)\big|
\to 0
$
as $h \to 0$.
For example,
if $f(x) = x^2$
(the canonical example),
$\dfrac{f(x+h)-f(x)}{h}
=\dfrac{(x+h)^2-x^2}{h}
=\dfrac{(x^2+2xh+h^2)-x^2}{h}
=\dfrac{2xh+h^2}{h}
=2x+h
$,
so that,
if $g(x) = 2x$,
$\big|\dfrac{f(x+h)-f(x)}{h}-g(x)\big|
=|(2x+h)-2x|
=|h|
$,
and this obviously
(and fortunately)
goes to $0$ as $h \to 0$.
A: Let $x_0 \in [a,b], f$ is continuous in $x_0 \implies (\forall \epsilon>0) (|x-x_0|<\delta \implies |f(x)-f(x_0)|<\epsilon) $ then:
$$\left|{g(x_0 +h) - g(x_0)\over{h}} -f(x_0)\right| = \left|{\int_a^{x_0+h}f(t)dt - \int_a^{x_0}f(b)dt\over{h}} -f(x_0)\right| = \left|{\int_a^{x_0+h}f(t)dt +\int_{x_0}^{a}f(t)dt\over{h}} -f(x_0)\right| = \left|{\int_{x_0}^{x_0+h}f(t)dt \over{h}} -{f(x_0)\over{h}}\right|=\left|{\int_{x_0}^{x_0+h}f(t)dt  -f(x_0) \int_{x_0}^{x_0+h}dt}\over{h}\right|=\left|{\int_{x_0}^{x_0+h}f(t)dt - \int_{x_0}^{x_0+h}f(x_0)dt}\over{h}\right| = \left|{1\over{h}}{\int_{x_0}^{x_0+h}[f(t) -f(x_0)]dt}\right| = {{1\over{|h|}}\left|{\int_{x_0}^{x_0+h}[f(t)-f(x_0)]dt}\right|\le}{{1\over{|h|}}{\int_{x_0}^{x_0+h}|f(t)-f(x_0)|dt}}\lt{{1\over{h}}{\int_{x_0}^{x_0+h}\epsilon dt}}={{\epsilon\over{h}}{\int_{x_0}^{x_0+h}dt}} = {{\epsilon\over{h}}{(x_0+h-x_0)}}=\epsilon$$
Analogously is for $-\delta<h<0$, taking limit:
$$\lim_{h\to0}{g(x_0+h)-g(x_0)\over{h}} \rightarrow g'(x_0)=f(x_0)$$
Then:
$${1\over{-h}}{\int_{x_0+h}^{x_0}\epsilon dt}={\epsilon\over{-h}}{(x_0-x_0-h)} = \epsilon$$
