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?

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    $\begingroup$ Are you asking for us to check your proof, or are you asking us to prove it? If it's the first, please put your proof inside your question and tag it as proof-verification. $\endgroup$ – user61527 Sep 23 '13 at 1:38
  • $\begingroup$ This is an demostrations that I want to share. There's another way to ask and answer this? thanks! $\endgroup$ – Tomi Sep 23 '13 at 1:52
  • $\begingroup$ Are you asking a question, or just stating the proof, then? $\endgroup$ – user61527 Sep 23 '13 at 1:53

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 $h \to 0$.


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: $${1\over{-h}}{\int_{x_0+h}^{x_0}\epsilon dt}={\epsilon\over{-h}}{(x_0-x_0-h)} = \epsilon$$


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