What's the demonstration that the antiderivative of a function is the integral?
Why is the integral the antiderivative of a function?
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5It's known as the [fundamental theorem of calculus](http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus) and you can look it up in rigorous treatments of calculus. – 2012-11-11
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0Any calculus text will have a proof. – 2012-11-12
2 Answers
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}
It's too hard to give a proof of the general case but I'll prove that $f(t) = \int_0^t x^2$ is an antiderivative of $x^2$. $\forall x \in \mathbb{R}\forall t \in \mathbb{R}\frac{x + t^3}{3} = \frac{x^3}{3} + x^2t + xt^2 + \frac{t^3}{3}$. The $x^2t$ term tells you that the derivative of $\frac{x^3}{3}$ is $x^2$ and therefore that $\frac{x^3}{3}$ is an antiderivative of $x^2$. Also for any positive real number $t$, $\int_0^t x^2 = \lim_{h\rightarrow0^+}\sum_{i=1}^{\lfloor\frac{x}{h}\rfloor}h^3i^2 = \frac{t^3}{3}$. It's not that hard to show that also for negative real numbers $t$, $\int_0^t x^2 = \frac{x^3}{3}$. Therefore $f(t) = \int_0^t x^2 = \frac{t^3}{3}$ is also an antiderivative of $x^2$.