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In the exercise section of the integration chapter of Bartles's book it says that if we define the function

$$f(x)=\begin{cases}0&x\in [0,1) \\ 1 & x\in [1,2]\;,\end{cases}$$

this Riemann-integrable function is not the derivative of any function. But

$$F(x)=\begin{cases}1 & x\in [0,1)\;\\x &x\in [1,2]\end{cases}$$ has f as a derivative, right? Then am I wrong? This is in the context of verifying the hypothesis for the Fundamental Theorem of Calculus.

Thanks in advance for the answers.

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That function does not have a derivative at $x=1$. –  anon Sep 16 '11 at 18:11
    
yes, you are completely right didn't see that gap. –  Ivan3.14 Sep 16 '11 at 18:16
    
@anon: Could you post that as an answer so it can get accepted? –  joriki Sep 16 '11 at 18:21
    
I have an old answer that seems relevant. –  Dylan Moreland Sep 16 '11 at 21:23
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up vote 2 down vote accepted

The function $F$ defined by $1$ when $x\in(0,1)$ and $x$ when $x\in[1,2]$ does not have a derivative at the point $x=1$. To see this analytically (it should be obvious graphically!), consider the definition of the derivative with the limit $$\lim_{h\to0}\frac{F(x+h)-F(x)}{h}.$$ Note that by definition this limit doesn't exist if it isn't the same from both directions. At the point $x=1$, from the negative direction, i.e. $h<0$, we have $\Delta F=1-1=0$, but from the positive direction ($h>0$) we have $\Delta F = (x+h)-x=h$. Thus the limit from the left is $0/h\to0$ and from the right is $h/h\to1$. But $0\ne1$.

This problem will arise in whatever function you try to have a derivative of $f(x)$ for the simple reason that $F$ would be constant from the left of $x=1$ and a line segment with slope $m=1$ on the right, from which the same side discrepancy occurs in evaluating the limit definition of the derivative. In other words, the fundamental theorem of calculus gives $$\lim_{h\to0+}\frac{1}{h}\int_{1-h}^1 f(x)dx=F'(1^-)=0$$ but $$\lim_{h\to0+}\frac{1}{h}\int_1^{1+h}f(x)dx=F'(1^+)=1$$ hence $F'(1)$ cannot exist.

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