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What's the demonstration that the antiderivative of a function is the integral?

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    $\begingroup$ It's known as the fundamental theorem of calculus and you can look it up in rigorous treatments of calculus. $\endgroup$ – Michael Greinecker Nov 11 '12 at 23:11
  • $\begingroup$ Any calculus text will have a proof. $\endgroup$ – Stefan Smith Nov 12 '12 at 16:11
  • $\begingroup$ I upvoted this question because in my opinion, it's a real question because some mathematicians have a demand for rigour ad just like the Jordan curve theorem, it seems so obviously true that an integral is an antiderivative but is probably actually quite difficult to give a rigorous proof of. I actually was specifically looking for a question like this so I did a Google search to find it. $\endgroup$ – Timothy May 27 at 3:29
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Here's the intuition. Suppose $f$ is continuous, and let \begin{equation} F(x) = \int_a^x f(t) \, dt. \end{equation} Let $\Delta x > 0$ be tiny. Then \begin{align*} F(x + \Delta x) - F(x) &= \int_x^{x+\Delta x} f(t) \,dt. \end{align*} But since $f$ is continuous, $f$ is approximately constant over the tiny interval $[x,x + \Delta x]$. Thus \begin{align*} \int_x^{x+\Delta x} f(t) \,dt &\approx \int_x^{x + \Delta x} f(x) \, dt \\ &= f(x) \int_x^{x + \Delta x} 1 \, dt \\ &= f(x) \Delta x. \end{align*}

So we see that \begin{align*} & F(x + \Delta x) - F(x) \approx f(x) \Delta x \\ \implies& \frac{F(x + \Delta x) - F(x)}{\Delta x} \approx f(x). \end{align*}

As $\Delta x \to 0$, the approximation gets better and better, so we conclude that \begin{equation} F'(x) = \lim_{\Delta x \to 0} \frac{F(x + \Delta x) - F(x)}{\Delta x} = f(x). \end{equation}

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