# Doubt From Notes Regarding Fundamental Theorem Of Calculus

I am reading a notes in which following is written:

Let $f$ be continuous periodic function defined on $\mathbb R$ with period 2l. Furthermore assume that $f’$ is piecewise continuous then $\int_ l^x f’(t) dt= f(x)$

I think the author is applying Fundamental Theorem of Calculus but $f’$ is just piecewise continuous and I know that Fundamental theorem of calculus is not valid for piecewise continuous function.Could you please help me verifying the above claim?

• If it really says "$\int_ l^x f’(t) dt= f(x) dx$" you may want to find another set of notes to read - that makes no sense. A revised version $\int_ l^x f’(t) dt= f(x)$ makes sense but isn't true. (Exactly what do the notes say here?) But anyway, if $f'$ is piecewise continuous then it's certainly true that $f(x)-f(y)=\int_y^x f'(t)\,dt$; who told you this was not so? – David C. Ullrich Mar 2 '18 at 16:03
• @DavidC.Ullrich: Sorry for the typo. Could you please say how does the claim you mentioned in the last holds ? I know the proof of this only for continuous function. – Math Lover Mar 2 '18 at 16:27
• Sorry, I should have said that if $f$ is continuous and $f'$ is piecewise continuous then $f(x)-f(y)=\int_y^x f'(t)\,dt$. Here we're given that $f$ is continuous. (In any case, what you say it says, as corrected, now makes sense but it's still false - that intergral is $f(x)-f(l)$, not $f(x)$.) – David C. Ullrich Mar 2 '18 at 16:40
• @DavidC.Ullrich: I only know the “ordinary” fundamental theorem of Calculus which assumes continuity of f and shows that the function F defined by $F(x)= \int_ a^x f(t) dt$ is differentiable and $F’(x)= f(x)$. Could you please give the reference for the result you mentioned? – Math Lover Mar 2 '18 at 16:54
• I just posted a proof in an Answer... – David C. Ullrich Mar 2 '18 at 16:57

I tend to suspect that what's actually written in the notes is not the same as what's written above. For example, the integral is actually $f(x)-f(l)$, not $f(x)$. Also it's assumed that $f$ is periodic, but this is totally irrelevant to the conclusion.
If $f$ is continuous, $f$ is differentiable except at finitely many points, and $f'$ is piecewise continuous then it's certainly true that $$f(y)-f(x)=\int_x^yf'(t)\,dt.$$
Suppose for example that $f'$ is continuous on $[x,a)$, continuous on $(a,y]$, and has one-sided limits at $a$. (Note that the existence of one-sided limits is part of the definition of "piecewise continuous".) For small $\delta>0$ we have $$(\int_x^{a-\delta}+\int_{a+\delta}^y)f'(t)\,dt=f(y)-f(a+\delta)+f(a-\delta)-f(x).$$Now let $\delta\to0$: The fact that $f'$ has one-sided limits at $a$ shows that $$(\int_x^{a-\delta}+\int_{a+\delta}^y)f'(t)\,dt\to\int_x^y f'(t)\,dt,$$while the continuity of $f$ shows that $$-f(a+\delta)+f(a-\delta)\to0.$$
Similarly if $f'$ has more than one (but only finitely many) points of discontinuity.
• @MathLover That's the same as saying that $\int_{a-\delta}^{a+\delta}f'(t)\,dt\to0$. Since $f$ has one-sided limits it is bounded near $a$; now if $|f'|\le M$ then $\left|\int_{a-\delta}^{a+\delta}f'(t)\,dt\right|\le 2\delta M$. – David C. Ullrich Mar 2 '18 at 19:31