This is a passage in Spivak's calculus just before the second fundamental theorem of calculus is introduced.
" A function $f$ may be integrable without being the derivative of another function. For example if $f(x) = 0$ for $x≠1$ and $f(1) = 1$, then $f$ is integrable, but $f$ cannot be a derivative. If $f$ is continuous, then we know $f = g'$ for some function $g$ and we know this only because of the first fundamental theorem of calculus. The function $f(x) = \frac{1}{x}$ provides an excellent illustration: if $x>0$, then $f(x) = g'(x)$, where
$$g(x) =\int_{1}^{x} 1/x dx$$ and we know of no simpler function with this property. So, a function $f$ might be of the form $g'$ even if $f$ is not continuous. If $f$ is integrable, then it is still true that
$$\int_{a}^{b} f = g(b) - g(a)$$
The proof however is entirely different - we cannot use FTC 1"
Then he proves FTC 2 by the mean value theorem by proving the statement " If $f$ is integrable on $[a,b]$ and $f = g'$ for some function $g$, then $$\int_{a}^{b} f = g(b) - g(a)$$
My question is
why is it necessary to use mean value theorem to prove this, since its already assumed $f = g'$ for some function $g$ in the first part. In other words, I don't seem to get why FTC 1 cant be used if its assumed that $f = g'$ for some $g$.
Any help is appreciated. Thanks!