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The second part of the fundamental theorem of calculus states that if $$ f(x)=g'(x) $$ then $$ \int^b_a f(x)\,dx = g(b)-g(a)$$

And so I was wondering that if you solved $\int^b_a f(x)\,dx$ using some other method like the limit definition and showed to to be equal to some $g(b)-g(a)$, then does that imply that $g'(x)=f(x)$?

This question came to me when I was solving for $\int^b_a x\,dx$ using the limit definition of the integral and I showed it to be equal to $\frac{b^2-a^2}2$, and so I was just wondering if I could do this with any integral that could be solved using alternate methods.

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    $\begingroup$ Do you mean that $\int_a^b f(x)\,dx = g(b)-g(a)$ for all $a,b$ or just one particular choice of $a,b$? $\endgroup$ – JimmyK4542 Sep 15 '14 at 2:25
  • $\begingroup$ If you mean for all $a,b$, then yes it is true. This is more or less the other fundamental theorem. If you mean for specific $a,b$, then no it is not necessarily true. $\endgroup$ – Cameron Williams Sep 15 '14 at 2:25
  • $\begingroup$ I meant for all $a,b$ $\endgroup$ – ASKASK Sep 15 '14 at 2:30
  • $\begingroup$ I won't post a full answer, but the "graduate level" answer to your question is the Lebesgue Differentiation Theorem. $\endgroup$ – icurays1 Sep 15 '14 at 2:39
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You do need some additional conditions on $f(x)$ to make it work.

Consider the function $f(x) = \begin{cases}1 & x \ \text{is an integer,} \\ 0 & x \ \text{is not an integer.}\end{cases}$

For any reals $a,b$, we have $\displaystyle\int_a^b f(x)\,dx = 0 = g(b)-g(a)$, where $g(x) = 0$.

But $g'(x) = 0 \neq f(x)$ for integer values of $x$.

If we add the condition that $f(x)$ is continuous, then your claim works.

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  • $\begingroup$ Thanks, this didn't occur to me $\endgroup$ – ASKASK Sep 15 '14 at 3:14
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If $f(x)$ is continuous and, leaving $a$ fixed (for example, $a = 0$), the equation $$ \int^b_a f(x) \, dx = g(b)-g(a)$$ is true for all choices of $b$, then yes, $g'(x) = f(x)$.

Another way to say exactly the same thing is that if $f(x)$ is continuous and you define a function $g(x)$ by $g(x) = \int^x_a f(t) \, dt$, then we have $g'(x) = f(x)$.

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If it holds for all $a,b \in [A,B]$ that $$ \int_a^b f(x) \, dx = g(b) - g(a) $$ for some function $g : [A,B] \to \mathbb R$, (and added : $f$ is continuous), then in particular $$ g(b) = \int_A^b f(x)\, dx + g(A) $$ so without loss of generality we can assume $g(A) = 0$ (otherwise you could define $\widetilde g(b) \overset{def}= g(b) - g(A)$). Also, for $x_0 \in ]A,B[$, we have $$ \lim_{x \to x_0} \frac{g(x) - g(x_0)}{x-x_0} = \lim_{x \to x_0} \frac 1{x-x_0} \int_{x_0}^x f(x) \, dx = f(x_0) $$ by the first fundamental theorem of calculus. Therefore $g(x)$ is differentiable on $]A,B[$ and $g'(x) = f(x)$. (We said "without loss of generality" for $g$ by changing $g$ up to a constant ; note that $g'(x) = f(x)$ does not depend on this constant.)

The continuity assumption is used when computing the last limit. You can either convince yourself with the formal $\varepsilon-\delta$ proof or with the idea that if $x \approx x_0$, then $f(x) \approx f(x_0)$ by continuity and thus $\int_{x_0}^x f(x) \,dx \approx (x-x_0)f(x_0)$.

Hope that helps,

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  • $\begingroup$ You need some additional assumptions on $f$ in order to guarantee that your last limit exists everywhere. If you only demand almost everywhere, you need at least integrability of $f$. $\endgroup$ – icurays1 Sep 15 '14 at 2:38
  • $\begingroup$ @icurays1 : Sure, let's say I assumed continuous. $\endgroup$ – Patrick Da Silva Sep 15 '14 at 4:36

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