Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
5  
It's known as the fundamental theorem of calculus and you can look it up in rigorous treatments of calculus. –  Michael Greinecker Nov 11 '12 at 23:11
    
Any calculus text will have a proof. –  Stefan Smith Nov 12 '12 at 16:11

1 Answer 1

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}

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.