How does calculus work?

What you are looking for is an explanation from "first principles".

Take a look at the links below.

The key here is that the formula works by successively narrowing the difference between two points so that they "approach" zero i.e. become vanishingly small and therefore, to all intents and purposes a point.

Calculus is part of a larger branch of mathematics called "analysis", which is, by definition, the study of infinity and limits. What you're doing when you take "shortcuts" is taking the behavior of the function and using it to draw conclusions about information at a single point (like the derivative), or over the entire interval (the integral).

For example, let's suppose you want to find out how far a horse at the track has run. Ideally, you would just multiply it's speed by the time it ran. Unfortunately, this doesn't work well in practice, because it runs at many different speeds from the time it leaves the gate to the time it crosses the finish line. What you would have to do then, is multiply each different speed by the amount of time it was running at that speed, then add 'em all up. The smaller the intervals of time you consider, the more accurate your answer will be. So if you think about instantaneous velocity (which is velocity at a single moment in time), your answer will be 100% complete. A mathematical object like this is called an integral, of course.

This, again, is the basic idea of a limit, studying the behavior of the function (distance) over smaller and smaller intervals (time) to derive information about a single point (instantaneous velocity).

You are correct when you explain the way integration finds area.

For explanation, I will use a geometric progression.

Consider the progression,

1/2, 1/4, 1/8, 1/16 .........

If I sum the first 2 terms I get 0.75 (3/4.)

the sum of first 3 terms is 0.875 (7/8)

sum of first 10 terms is 0.9990234375

sum of the first 50 terms is 0.99999999999999911182158029987477

sum of first 100 terms is 0.99999999999999999999999999999921 (approximately)

You might have observed that as we sum more and more terms the value tries to reach one. At a time it may even become 0.999999999999999999999999999999999999999999999999999999999999999999999999999

but it would never be 'equal' to 1. Only the number of nine's would keep on increasing. It will happen only when the number of terms will be infinite, So the sum of infinite terms of an increasing geometric progression is not approximately calculated. The value calculated is 'exact'. If someday we are able to write it to infinite terms and then sum, we would get what we had calculated.

For example, 1\100 even if raised to the highest power known would not give 0. Only when raised to infinite power would it be equal to nothing. so multiplication of 1\100 infinite times is not approximately zero, it exactly is. This is how we get to defining a quantity 'infinite'. We use this number by thinking it's properties.

It is same for calculus. In integration we don't actually split the curve into infinite parts. We study the properties of summation on the curve for finite quantities and then do it for infinite terms.

Similarly in differentiation the distance between two points is not made zero. We observe what happens when the distance is lowered and then we calculate what would happen if we reduce the distance to zero.

This is how Calculus works.

Well, I suppose you know that the fathers of differential calculus were Newton and Leibnitz, actually the great Newton was the one who started noticing a kind of pattern in the results of calculating the limit of a Riemann's Sum when the number of divisions tended to infinite, for the different sorts of functions they had studied... However, those short-cut equations do not always work and sometimes it's necessary to go back and do the whole work from Riemann's Sum. I think that should answer your question.

Good luck!

I think most answerers misunderstand you. The answer is somehow too complicated to be resolved in here but I can refer you to a theorem, pivotal, creative and fundamenatal in calculus.

No wonder it is called the fundamental theorem of calculus.

It proves that differentiating ( finding the derivative) and integration are inverse operations. I believe that you understand how derivatives work. After that theorem the only question to be solved in an individual problem when integrating is thinking of the derivative of what function would yield the function you are trying to find the integral of.

It is good to be confused.

Cheers.

The derivative is basically he slope of a line tangent to the curve at any given point. Using that information, about the rate of change and stuff, you can eventually come up with more complicated things like integrals, or anything that uses derivatives.

Heres an example, from physics, the following are functions:

s(x) is the DISPLACEMENT of an object

v(x) is the VELOCITY of the object, displacement/time

a(x) is the ACCELERATION of the object, displacement/time/time

j(x) is the JERK of an object, displacement/time/time/time

they are all related by derivatives, the derivative of s(x) is v(x), derivative of v(x) is a(x), derivative of a(x) is j(x). Each time its divided by time. The information that you're looking at gets finer and finer in each step, because you look at how that information changes. Think about what a graph about velocity would say about acceleration, if the velocity is increasing, then its accelerating. if the graph of acceleration is increasing, then theres a jerk, velocity should spike. etc.

Does that make sense?

If I understand you correctly, what you're asking is why the area below the graph is equal to the difference of two values of an anti-derivative in to points?

Well, there's another theorem named the Lagrange theorem that states that If for a function f, a derivative is defined between x1 and x2, there's a point c between them such that f(x2)-f(x1)=f'(c)(x2-x1)

Now, let's divide an interval (a,b) into a=x0,,x1,x2,x3,x4,...,x(n)=b

Then for each interval [x(n),x(n+1)] we'll chose a point c(n) such that

f(x(n+1))-f(x(n))=f'(c)(x(n+1)-x(n))

Now, if The sum of all f'(c(k)(x(k+1)-x(n)) when k is between 0 and n-1 is equal to the sum of f(x(k+1)-f(x(k)) = f(x(n+1)-f(xn)), and this is the limit when d, the maximal diameter of a little interval approaches zero.

when u square a number

x^2

the ans is x^2....sheesh

2^2=4

got it?

wen u differentiate it, its doubled wierd is nt it?

d(x^2)=2x

d(2^2)=4

so differential calculus gives u an entirely different solution or function.

integration is the reversal of differentiation

calculus is diving something into infinitesimly small pieces and then adding those pieces up.

Little gnomes.