It puzzled me greatly when going over the Newtonian theory, so I know F = ma, so if F is known, m is known, at any instant, one can calculate the distance, velocity, and acceleration rate of an object precisely.

Distance may be fine to understand, but velocity, acceleration rate, these are metrics defined to be measured by end status minus initial status, right? if one can deduce a precise value at a single moment, that means, an infinitesimal value is commutable!

Even in this case, I take it as because F and m are measurable, so I can get acceleration rate a, so this part is easy, but it is from the fundamental Newton’s law: F = ma, it’s more of a declared statement than a rigorous mathematical proving.

Then from acceleration rate a, I can find the relationship/formula between a and velocity c, a is the differentiation of v relative to time t. a = [v(t2) – v(t1)]/(t2-t1), of which, I don’t know v yet, so in the same vein, I need to find out v first, v = [s(t2) – s(t1)]/(t2-t1). Now most people stuck, how to solve?

Maybe many people can solve it, without invoking the genius of Newton Leibniz, or Riemann. It took me a while to figure out the solution that Newton Leibniz, or Riemann already figured out.

They applied rigorous approach to prove the below fundamental theorem of calculus. The conclusion seems squarely straightforward, yet, hard to truly grasp. It’s basically saying the derivative of the anti-derivation of a function f relative to x is equal to f(x). Within a limited scope from a to b, the area underneath, integral of f(x)dx, is equal to the anti- derivation of f(x) at point b minus the anti- derivation of f(x) at point a. (from wiki)

Beautifully, it convert a problem of a micro, infinitesimal value computation to a macro, finite value deduction, and vice versa.

The tricky part, and vital importantly, is that one needs to find anti-derivation of “any” functions(f(x)). which, through years of efforts of our ancestors, there are existing formulas and calculating skills to achieve. For instance,

Equipped with this math tool, many difficult, seemingly impossible math problems can be solved easily. For example, find the area of a circle:

To push one step further, what is the surface area of a sphere: