Integration (Theory)

Username99

So I am starting Integration at the moment.

What I am struggling to understand is this.

If I integrate y' = 2x I get back out the original function x^2, but how graphically can I work out what is going on, with Differentiation you could see by Differentiating from first principals how to find the tangent at any point, but how does knowing the derivative of a function and multiplying it by an infinitesimally small change in x 'build' the original function?

Any help greatly appreciated.

Yakuza

Try looking at integration as the summation of a series of slices under the graph of the function.

srsly78

Don't forget about the constant. You get back a function with the same "shape" but not the exact original.

TheBody

There should be lots of video on youtube to help you out. For example:

https://www.youtube.com/watch?v=he9DkW897eQ

Ompala

Its probably overused to death at this point, but try thinking of displacement (s) and velocity (v).
At an infinitesimally small gap in time(dt) the infinitesimally small displacement is going to be ds = v dt, and since the whole displacement is the sum of the infinitesimally small displacements, by adding them all up you can "build up" the original from the derivative. The "whole adding them all up bit" is where integration comes in. Hope that helps

Davidius

Graphically I don't think there's one that's quite as clean as the tangent line representation of the derivative. Integration is in general a lot 'harder' than differentiation

This is a relatively clean visualisation


Here you can see that the change of the area is roughly the function times the 'change' in input h (which intuitively is infinitesimal). Dividing by h and taking the limit you get A'(x) = f(x). Hence the instantaneous rate of the area at x is f(x). Hence if you sum (hence the elongated S) up these infinitesimal areas f(x)dx you get the overall area.

di11on

I think the nature of the op's confusion is that you are trying to interpret what the integral of a derivative is... well it's just the original function, which isn't that interesting.

Perhaps a better way to think of it is this...

Imagine a function f(t) told you how many widgets you had at time t

Say f(t) = 5t

at time t=0, f(t) = 0 (I have zero widgets at time t=0)
at time t=1 f(t) = 5 (I have 5 widgets at time t = 1).

If I wanted to know how many widgets I had a time t=100 we can find this out by integrating.

integral ( f(t) ) dt = 5/2 t^2 + c

Assume I have zero widgets at time t=0. At time t=100 I have:

5/2 * (10,000)

If I wanted to know how fast I am accumulating widgets, that's the derivative... in this case;

d( f(t))/dt = 5

So my rate of change of widgets is 5 widgets a second.

Username99

Hi guys,

Thanks for all the replies, I have been working on this problem all week and I think I am very close.

I am currently preparing a Microsoft Word document which outlines the problem with my understanding of the Fundamental Theorem of Calculus and I will attach it to a post here hopefully by Sunday evening.

Anyone please feel free to go through it when I post the attachment and finds holes etc...

I feel I could be close, but I am just one small slip up to finding myself back at square one.

Thanks for all the help so far.