1
$\begingroup$

The second part of the fundamental theorem of calculus states that

If $f:[a,b]\to\mathbb{R}$ is a Lebesgue-Integrable function and $F$ is a primitive of $f$, then $\int_{a}^{b}f(x)dx=F(b)-F(a)$.

I have been unable to prove this result by myself or find any reference for a proof of it.

My attempt

Lets consider the function $G(t):=f(a)+\int_{a}^{t}f(x)dx$, which is absolutely continuous, so $G$ is differentiable almost everywhere, $G'$ is Lebesgue-Integrable and $\int_{a}^{b}G'(x)dx=G(b)-G(a)=\int_{a}^{b}f(x)dx$. So we have the desired conclusion for the function $G$ instead of a primitive $F$. We also know that $G$ is a primitive of $f$ almost everywhere, but I really don't know how to go further proving the equility for a primitive $F$.

$\endgroup$
  • $\begingroup$ Rudin's RCA has a proof of this. $\endgroup$ – Kavi Rama Murthy Jun 1 at 8:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.