If $f(x)=|x|$, then we know $f(x)$ has a sharp point at $x=0$, and now I want to calculate the derivative of $f(x)$ on interval $[0,1]$, i.e., $$\int_{0}^{1} f '(x)\,dx$$ but $f'(x)$ is not defined on $0$, so I want to know whether the function of $f'(x)$ has been restricted on domain $[0,1]$ as we write down the definite integral, so the derivative of $f'(x)$ at zero just has to consider its derivative from right direction limit, thanks for answers~!
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Formatting tips here. – Em. Aug 17 '16 at 04:34
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You mean "I want to calculate the area under the graph of $f'(x)$ on the interval [0.1]" instead of "I want to calculate the derivative of f(x) on interval [0,1]" – Our Aug 17 '16 at 05:07
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By the way, I not sure whether I understand your question.Could you be more clear about what exactly you are asking ? – Our Aug 17 '16 at 05:11
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If we let $f(x) = |x|$, we have that $$f'(x) = \begin{cases} - 1 & x<0 \\ c & x =0 \\ 1 & x>0\end{cases}$$
Note that the value of $c$ here doesn't matter. You can say that it's undefined at $0$ just as well.
It's known that the Riemann integral doesn't change when finitely many points are changed (this is formalized better through measure theory). This is the case here, so $$\int_0^1 f'(x) \mathrm{d} x = 1$$
Mark Schultz-Wu
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If the integral is not defined at a point in the interval then a limit approach needs to be taken as the integral approaches that value. If the limit exists then the value can be obtained. In this case the limit is x -> 0 from the right.
John
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