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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!

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  • $\begingroup$ Existence of antiderivative does not imply continuity, which is the condition for FTC 1 to hold $\endgroup$ – Poon Levi May 6 '18 at 18:27
  • $\begingroup$ "a function $f$ might be of the form $g'$ even if $f$ is not continuous." Note that the second FTC does not require $f$ to be continuous. But the first FTC requires $f$ to be continuous at least at one point. $\endgroup$ – user144410 May 6 '18 at 18:49
  • $\begingroup$ So FTC 1 says if a function is integrable and is continuous over some interval [a,b], f is a derivative of some function, namely the area function F. Corollary 1 - As a consequence, I can compute the area under the curve if I know some function g with derivative f also, meaning that F = g + c, for some constant c, so the integral is computed by g(b) - g(a). The confusion is why I need mean value theorem to prove this in FTC 2 when I have this corollary and since it also says "f is of the form g' even though f is not continuous". $\endgroup$ – john bishop May 7 '18 at 3:07

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