# Fundamental Theorem of Calculus at Discontinuities

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 $$x=1$$.

• 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 $F_-'(1)=0.$ Mar 11, 2020 at 0:38
• Your answer is fine. We even have $F(x) =0$ for all $x$. Mar 11, 2020 at 1:27
• In any case you can have a look at FTC for general functions at math.stackexchange.com/a/2149700/72031 Mar 12, 2020 at 1:46