I know that the fundamental theorem of calculus states.

If f is integrable over $[a,b]$ and if we define $F(x) = \int_{a}^{x}f(t)dt$.

If f is continuous at $c\in[a,b]$ then $F$ is differentiable at $c$ and $F′(c)=f(c)$

But what if we say:

If $F$ is a continuous function defined on $[a,b]$ such that $F′(c)=f(c)$ for all $c\in [a,b]$

And if $f$ is continous on $[a,b]$ then for all $x\in[a,b]$ $F(x) = \int_{a}^{x}f(t)dt +c$

I would try to prove this using the definition of the Darboux integral. But I'm not sure if this is just another way of writing the fundamental therem of calculus. I don't know if writing it that way makes sense.

  • $\begingroup$ Just a technicality but you should say F is differentiable for that line F'(c) to make sense. $\endgroup$ – CJD Mar 21 '18 at 5:45

Just define $G(x)=\int_a^xf(t)dt$ and observe that $G'(x)=f(x)$ for all $x\in[a,b]$ by the fundamental theorem of calculus. Since $F'(x)=f(x)$ for all $x\in[a,b]$ as well, this means $F$ and $G$ differ by a constant, which is exactly what you want.


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