# How did Newton (or whoever invented the process) come up with the idea that the anti-derivative can be used to calculate the area under a curve?

I don't know if it has anything directly to do with the fundamental theorem of calculus, but if it does I can't seem to see the connection. Or maybe I just don't understand the theorem.

I can understand how the definition of derivative works... you plug in any $f(x)$ and you get out $f\prime(x)$.

$$f\prime(x)=\lim_{h\to \infty}\frac{f(x+h)-f(x)}{h}$$

But I can't understand the connection between the limit process of finding the area under a curve and the definite integral.

$$\lim_{n\to \infty}\sum_{i=1}^{n}y_{i}\Delta x\ on\ interval\ [a,b]=\int_{a}^{b}f(x)\ dx$$

Why is it that the anti-derivative of a function can serve as a shortcut to solving this problem?

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Barrow is often given credit for the Fundamental Theorem, at least by people writing in English. Derivative is not mentioned by him, he works very geometrically with tangents to what in hindsight can be called the area function. –  André Nicolas Jan 23 '13 at 13:58
Similar question: math.stackexchange.com/q/228018 –  Martin Jan 23 '13 at 14:00

The wikipedia article on Fundamental theorem of calculus (which is the statement you are looking for) the has a good intuitive explanation of why the two are equal. In essence, if you take the area $A$ up to a point $x+h$, then: $$A(x+h)-A(x)\approx f(x)h$$ So that: $$\frac{A(x+h)-A(x)}{h}\approx f(x)$$ Now take $h\to0$, showing that the derivative of the area is actually the function $f(x)$.

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Both of these equations aren't really equations. The point is that if you take $h$ small, the left hand sides can be approximated by the right hand sides. So you should have $A(x+h)-A(x)\approx f(x)h$ for example. –  Michael Albanese Jan 23 '13 at 14:06
@MichaelAlbanese - of course. I was just trying to explain the intuition behind it. –  nbubis Jan 23 '13 at 14:22

Newton was a physicist, and studied Mechanics heavily. My guess is:

It was well-known, understood, and somehow intuitive that the area below a function of velocity would be the distance traveled. (Because if you divide it in "tiny intervals", you have "constant velocity", and can make $d=vt$. That would also justify the way integral got developed by Riemann.) And it was well-known that the derivative (rate of change) of the distance was the velocity. Then, you had the derivative of $d(t)$ was equal to $v(t)$, and the area below $v(t)$ was equal to $d(t)$. There is the relation you wanted know.

Physical ideas help developments in mathematics, and vice-versa. I think this was the case.

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