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?

Question

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?

Answer 1

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 $f(x)$.

Answer 2

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.