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April 30, 2025, 01:52:46 am

Author Topic: Fundamental Theorem of Calculus (I)  (Read 1736 times)  Share 

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Mao

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Fundamental Theorem of Calculus (I)
« on: September 25, 2008, 04:49:47 pm »
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REINVENTING THE WHEEL

I guess many of you have not seen this (which is the Fundamental Theory of Calculus). For public benefits, I will make a fool of myself and post this.

Calculus involves infinitesimals, where differential calculus concerns itself with instantaneous rates of change and integral calculus concerns itself with area under curve.

For a positive function , the area under the curve on the interval [a,b] can be represented by

for example, , its area on the interval [0,x] can be represented by the function F(x). When x=3, F(x)'s value is the area of the shaded area in the image below.


Now, as we increase x, the area gets larger. For instance, when we increase x to 4,


now notice that the increase in area can be approximated by a rectangle of width dx (in this case, dx = 4-3 = 1), and the height is the average height of f(t) for that interval of increase (mean value theorem, ignore this if you have not heard of that before). We denote this as


As we shrink the change in x, we notice that the rectangle becomes a better and better estimate for this change in area. That is, for smaller and smaller dx. We find that, for an infinitesimal change in x, the corresponding change in F (the area) is equal to f(x)dx.
(massive jump to conclusion here)

That is, the rate of change in Area, i.e. the change in F(x), is f(x). Or, the derivative of the integral is the integrand (function inside the integral). Formally, <--- thats the theorem, but is useless for you guys at this moment :P
(now, I have made many many jumps, and have completely ignored all the correct things such as continuity and etc)

This is the underlying principle of using fnInt to solve differential equations. It is common knowledge that . Or, we are saying that the antiderivative is the area function of the derivative.
Visualising that, it means that we do not know what F(x) is, as we do not know where the left end of the area started. In fact, that can be anywhere (hence C can start from anywhere). However, we do know that for a given change in x, we can work out the change in F by finding the area under f for that interval. That is, if we know the initial value, and the derivative, we can find any value by adding increments of area under the derivative.

so, we are saying that,

. So, the area function is the area it already have (the y0) plus any increments (the integral).

« Last Edit: September 25, 2008, 04:57:36 pm by Mao »
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polky

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Re: Fundamental Theorem of Calculus (I)
« Reply #1 on: September 25, 2008, 04:51:21 pm »
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THANKS MAO.

Now I shall read it :D
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/0

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Re: Fundamental Theorem of Calculus (I)
« Reply #2 on: September 25, 2008, 05:37:46 pm »
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Neat, thanks for that Mao. I find it funny how they don't explicitly teach us this even though it is an integral part of calculus. Lol pun  ;D

cara.mel

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Re: Fundamental Theorem of Calculus (I)
« Reply #3 on: September 25, 2008, 08:33:05 pm »
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what part is the Fundamental Theorem

i have heard this mentioned so many times and dont know what it is

dcc

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Re: Fundamental Theorem of Calculus (I)
« Reply #4 on: September 25, 2008, 08:36:24 pm »
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what part is the Fundamental Theorem

i have heard this mentioned so many times and dont know what it is

Basically just says that where
« Last Edit: September 25, 2008, 08:38:12 pm by dcc »

Glockmeister

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Re: Fundamental Theorem of Calculus (I)
« Reply #5 on: September 26, 2008, 12:11:37 am »
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You use it most often with when you do that box thing when you definitely integrate a function If you know that box thing that you do when you have definite integrals.
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Re: Fundamental Theorem of Calculus (I)
« Reply #6 on: September 26, 2008, 03:45:35 pm »
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http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus


Mmmm Wikipedia. The generalisations are quite interesting, actually - the conditions on the function are merely that it is locally Lebesgue integrable.
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