Rate of change of area under curve f(x) = f(x)

[rajeshmarndi]

What lead to the equality of the, rate of change of area under curve f(x) = f(x).

Was it, they were just compared(OR believed to be equal) and mathematically found to be equal. Or when one was integrated or differentiated the other appeared.

Also I knew, integration was being used since Archimeded and Greeks times.

Thanks.

 

[PeroK]

What lead to the equality of the, rate of change of area under curve f(x) = f(x).

Was it, they were just compared(OR believed to be equal) and mathematically found to be equal. Or when one was integrated or differentiated the other appeared.

Also I knew, integration was being used since Archimeded and Greeks times.

Thanks.

If you take a function at a point $f(x_0)$. That's the height of the curve at that point. The area between this and another point a short distance to the right is approximately a rectangle of width $\Delta x$, hence of area $\Delta A \approx f(x_0) \Delta x$.

And the rate of change of area is approximately $\frac{\Delta A}{\Delta x} \approx f(x_0)$

Using calculus makes this exact, but intuitively the rate of change of area under the curve is simply the height of the curve at that point.

 

[rajeshmarndi]

It look so easy, how is it that only calculus make it possible later , the relation between integration and differentiation.

 

[PeroK]

It look so easy, how is it that only calculus make it possible later , the relation between integration and differentiation.

I think it is relatively easy to see. But, calculus is needed to prove it formally and also to put conditions on the function involved - in this case, for example, that the function $f$ is continuous is sufficient. It's not so easy to see intuitively that it works for every continuous function, as some continuous functions are very strange indeed.