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Part One of the Fundamental Theorem of Calculus states that if $g$ is a continuous function on $[a,b]$ that is differentiable on $(a,b)$, and if $g'$ is integrable on $[a,b]$ then $$\int_{a}^{b}g'=g(b)-g(a)$$ What I am wondering is: what does it mean that $g'$ is integrable on $[a,b]$? I reckon that it is necessary for the Theorem to hold, but I am not sure what the statement means by itself.

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    $\begingroup$ A function is Riemann integrable if the lower sums and the upper sums coincide. $\endgroup$ – Valerin Dec 10 '13 at 19:32
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You know the definition of the Riemann integral of a function on an interval: take a partition, form the upper and lower sums, repeat to get the set of all possible upper sums and all possible lower sums. If there's a unique number $s$ that's no less than any lower sum and no greater than any upper sum, then $s$ is called the integral, and the function's said to be integrable on the interval.

But in some cases (for example, the function that's one on the rational numbers and 0 on the irrationals), the lower sums and upper sums are all far apart -- the number "$s$" above is not unique. In that case, the function's not integrable.

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