Question: Let $f(x) = 0$ if $x \neq 1$, and $f(x) = 1$ if $x=1$ Let $F(x) = \int_0^xf $. At which points is $F'(x) =f(x)$?

Answer Can I get a check to see if my answer is correct?

I understand that if a function $g$ is integrable on $[a,b]$, and we let $h = g$ except at finitely many $x_i \in [a,b]$, then $h$ is still integrable, and $\int_a^b g = \int_a^b h$.

Thus in the above question $F'(x)=0$ for all $x$, so $F'(x)=f(x)$ for all points except $x=1$.

  • $\begingroup$ The only problematic point is $x=1$ why? What is the domain of $F$? If it is $[0,1]$, then you can only check the left-hand limit of the difference quotient:$\frac{F(1-h)-F(1)}{h}=\frac{0-0}{h}=0$ so $F_-'(1)=0.$ $\endgroup$ Mar 11, 2020 at 0:38
  • 1
    $\begingroup$ Your answer is fine. We even have $F(x) =0$ for all $x$. $\endgroup$
    – Berci
    Mar 11, 2020 at 1:27
  • $\begingroup$ In any case you can have a look at FTC for general functions at math.stackexchange.com/a/2149700/72031 $\endgroup$
    – Paramanand Singh
    Mar 12, 2020 at 1:46


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