- #1

tahayassen

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Bare in mind, my riemann sums is extremely weak.

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- Thread starter tahayassen
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In summary, integration is the reverse process of differentiation, where we find the function whose derivative is the original function. Differentiation can be done using the limit definition or by using shortcuts such as the product rule, while integration can be done using the limit of the Riemann sum or by finding the anti-derivative and using the fundamental theorem of calculus. The anti-derivative is essentially the inverse of differentiation and is used to calculate the definite integral using the fundamental theorem of calculus.

- #1

tahayassen

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Bare in mind, my riemann sums is extremely weak.

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- #2

STEMucator

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tahayassen said:

Bare in mind, my riemann sums is extremely weak.

I find the point of the Riemann sum is to prove a function is integrable. It's nice that they can also give us the value of the integral on that particular interval as n → ∞.

Anti derivatives and methods of finding them merely speed up the long process of integration which we would otherwise have to do with a Riemann sum every time ( May I note how time consuming it is ).

- #3

Vorde

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Why can't we just say that anti-derivatives are the 'short-cut' for Riemann summing?

- #4

Mark44

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We saytahayassen said:It seems like we calculate integration by doing the reverse of derivation.

No, it's not. Differentiation is the action of finding the derivative of some function by any means. Differentiation includes doing so by the limit definition (first principle) or by such shortcuts as the power rule, product rule, quotient rule, etc.tahayassen said:Differentiation is basically just using short-cuts for differentiation by first principles (e.g. power rule).

There's some confusion here, between indefinite integrals and definite integrals. For definite integrals, one way of evaluating them is using a Riemann sum.tahayassen said:If integration by first principles is the Riemann sum, then why don't we use short-cuts of the Riemann sum to do integration? Why do we instead use reverse derivation?

For indefinite integrals, the idea is to find a function whose derivative is the function in the indefinite integral - i.e., finding the antiderivative of the function in the integral.

Zondrina said:I find the point of the Riemann sum is to prove a function is integrable. It's nice that they can also give us the value of the integral on that particular interval as n → ∞.

Anti derivatives and methods of finding them merely speed up the long process of integration which we would otherwise have to do with a Riemann sum every time .

What would the Riemann sum look like for this (indefinite) integral?

$$ \int e^x~dx$$

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- #6

Mark44

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Under Differentiation you have the limit definition of the derivative and the various rules that are specific applications of the limit definition (constant multiple rule, sum and difference rules, product rule, quotient rule, etc.) In other words, these techniques would be leaves that come off the limit definition.

Riemann sums are the basic way of evaluating

Every time we develop a new differentiation formula, we get a integral formula for free. For example, there is the sum rule: d/dx(f(x) + g(x)) = f'(x) + g'(x), which implies that

## \int ( f'(x) + g'(x) )dx = f(x) + g(x) + \text{an arbitrary constant}##

This formula is usually presented as the sum rule for integration:

## \int ( f(x) + g(x) )dx = F(x) + G(x) + \text{an arbitrary constant}##, where F'(x) = f(x) and G'(x) = g(x). IOW, F and G are antiderivatives of f and g, respectively.

The technique of substitution in integrals, is nothing more than the differentiation chain rule, working backwards.

The Fundamental Theorem of Calculus isn't part of differentiation - what it does is show

1. that differentiation and antidifferentiation are essentially inverse processes

$$ \frac{d}{dx} \int_a^x f(t)~dt = f(x)$$

2. how to evaluate a definite integral using the antiderivative of the integrand.

$$\int_a^b f(t)dt = F(b) - F(a)$$

where F is an antiderivative of f (i.e., F' = f).

- #7

rollingstein

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tahayassen said:I think I didn't ask my question properly, so I've made a mindmap to make my question more clear.

I liked your Mind Map.

Can I ask what you used to draw it?

- #8

tahayassen

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I'll try an alternate method. I'll list a list of assumptions that I've made and maybe you can tell me if any are incorrect. Thank you by the way for taking the time to reply.

- Differentiation by using the limit definition is essentially taking the limit of x/y where both x and y approach zero
- A rigorous proof of the product rule involves using the limit definition. Therefore, the product rule is a shortcut of differentiation by first principles.
- Therefore, you can differentiate a function either by: 1) limit definition 2) shortcuts for the limit definition such as the product rule.
- Integration by using the limit of the Riemann sum is essentially multiplying an infinite sum times an infinitesimally small value
- Integration using the limit of the Riemann sum will always give you the definite integral.
- As far as integration using the limit of the Riemann sum is concerned, the indefinite integral is unnecessary.
- Finding the definite integral using the limit of the Riemann sum is limited and not very powerful.
- A better method is to find the indefinite integral (or anti-derivative) and use the indefinite integral to calculate the definite integral using the fundamental theorem of calculus.
- Therefore, you can find the definite integral of a function either by: 1) limit of the Riemann sum 2) finding the anti-derivative and then using fundamental theorem of calculus to find the definite integral using the anti-derivative.
- Anti-derivative is reverse differentiation.
- Both differentiation and integration can be done using first principles. But differentiation has a shortcut for its first principles while integration's "shortcut" relies on reversing differentiation (with the use of anti-derivative).
- The reason why integration does not have a shortcut is because multiplying an infinite sum times an infinitesimally small value is not as easy (or is not as predictable?) as taking the limit of x/y where both x and y approach zero.

rollingstein said:I liked your Mind Map.

Can I ask what you used to draw it?

http://www.text2mindmap.com/ :P

- #9

DrewD

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I think you might have been getting at this with one of your mindmap comments, but Riemann integral is not usually described by multiplying an infinite sum by an infinitesimal amount (although I'm pretty sure it can be - I don't find it particularly helpful). I think it is much better thought of as an infinite sum of a function times an infinitesimal amount. I'm not saying you were wrong, but I've found a lot of physics students having trouble by thinking about it the way you stated it.

- #10

Mark44

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Your English is very good, and better than that of some native speakers I have seen. I agree that you might have used some terms incorrectly, but that comes with the territory when you are learning something new.tahayassen said:Mark44, I understand everything in your post. I think my issue is that I'm having trouble communicating my question due to my poor English and because I might have used the wrong terms by accident.

tahayassen said:I'll try an alternate method. I'll list a list of assumptions that I've made and maybe you can tell me if any are incorrect. Thank you by the way for taking the time to reply.

- Differentiation by using the limit definition is essentially taking the limit of x/y where both x and y approach zero
- A rigorous proof of the product rule involves using the limit definition. Therefore, the product rule is a shortcut of differentiation by first principles.
- Therefore, you can differentiate a function either by: 1) limit definition 2) shortcuts for the limit definition such as the product rule.
- Integration by using the limit of the Riemann sum is essentially multiplying an infinite sum times an infinitesimally small value

- That's (above) pretty much it if your partition is divided into subintervals of the same size. It's possible to divide the partition into variable-length subintervals, with smaller subintervals where the function you are integrating increases or decreases rapidly, and larger subintervals where the function is relatively flat.

In this case you would be adding up an infinite number of terms, where each term is the product of a function value and the width of the subinterval.

tahayassen said:[*]Integration using the limit of the Riemann sum will always give you the definite integral.

[*]As far as integration using the limit of the Riemann sum is concerned, the indefinite integral is unnecessary.

[*]Finding the definite integral using the limit of the Riemann sum is limited and not very powerful.

tahayassen said:[*]A better method is to find the indefinite integral (or anti-derivative) and use the indefinite integral to calculate the definite integral using the fundamental theorem of calculus.

tahayassen said:[*]Therefore, you can find the definite integral of a function either by: 1) limit of the Riemann sum 2) finding the anti-derivative and then using fundamental theorem of calculus to find the definite integral using the anti-derivative.

[*]Anti-derivative is reverse differentiation.

[*]Both differentiation and integration can be done using first principles. But differentiation has a shortcut for its first principles while integration's "shortcut" relies on reversing differentiation (with the use of anti-derivative).

[*]The reason why integration does not have a shortcut is because multiplying an infinite sum times an infinitesimally small value is not as easy (or is not as predictable?) as taking the limit of x/y where both x and y approach zero.

tahayassen said:

- #11

mishin050

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Mark44 said:The Fundamental Theorem of Calculus...

1. that differentiation and antidifferentiation are essentially inverse processes

$$ \frac{d}{dx} \int_a^x f(t)~dt = f(x)$$

2. how to evaluate a definite integral using the antiderivative of the integrand.

$$\int_a^b f(t)dt = F(b) - F(a)$$

where F is an antiderivative of f (i.e., F' = f).

I think that on it the correct theory comes to an end.

$$\text{This integral:}~~\int F'(x)dx=F(x)+C~~\text{begins the wrong theory.}$$

Nobody proved it and it doesn't make sense.

$$\text{It would be correct to go on such way:}~~F(x)=\int \limits_{x_0}^x f(t)~dt,~~F(x_0)=0;~~~~~C=\int\limits_0^C~dt.$$

$$\text{If}~~f(t)=\frac{1}{t}:~~~F(x)=\int\limits_1^x f(t)dt=ln(x)=ln\left(\frac{x}{1}\right);~~~\int \limits_{t_1}^{t_2}\frac{dt}{t}=ln\left(\frac{t_2}{t_1}\right);$$

$$F(b)-F(a)=\int\limits_{a}^{b}\frac{dx}{x}=ln\left(\frac{b}{a}\right);~~~~\int \frac{dx}{x}=ln|x|+C -~~\text{is wrong!}.$$

Because:

$$y=F(x),~~ u=F(x)+C,~~u(x,p)=F(x)+F(p)_{p=F^{(-1)}(C)};$$

$$f(x)=F'(x)=\frac{dF(x)}{dx};$$

$$f(x)=\frac {\partial u(x,p)}{\partial x}=\frac{\partial (F(x)+C)}{\partial x};$$

$$\int\limits_{x_0}^x f(t)\partial t=F(x)+C.$$

----------------------------------------------------------------------

"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"

Thomas Ioannes Stiltes. ...

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- #12

DrewD

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- #13

mishin050

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DrewD said:

$$ \frac{d}{dx} \int_a^x f(t)~dt = f(x);$$

$$ dx \cdot\frac{d}{dx} \int_a^x f(t)~dt = f(x)\cdot dx;$$

$$\int d \int_a^x f(t)~dt =\int f(x)~dx;$$

$$ \int_a^x f(t)~dt =\int f(x)~dx;$$

$$\int f(x)~dx= F(x)-F(a)!$$

- #14

Vorde

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mishin050 said:$$ \frac{d}{dx} \int_a^x f(t)~dt = f(x);$$

$$ dx \cdot\frac{d}{dx} \int_a^x f(t)~dt = f(x)\cdot dx;$$

$$\int d \int_a^x f(t)~dt =\int f(x)~dx;$$

$$ \int_a^x f(t)~dt =\int f(x)~dx;$$

$$\int f(x)~dx= F(x)-F(a)!$$

This doesn't work, not the least of the reasons being that you are treating ##\frac{d}{dx}## as a fraction.

- #15

mishin050

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Vorde said:This doesn't work, not the least of the reasons being that you are treating ##\frac{d}{dx}## as a fraction.

##\displaystyle\frac{d}{dx}=\lim_{\Delta x\to 0}\frac{\Delta}{\Delta x}##

- #16

pwsnafu

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mishin050 said:##\displaystyle\frac{d}{dx}=\lim_{\Delta x\to 0}\frac{\Delta}{\Delta x}##

The right hand side is not mathematically well-defined, and furthermore that not the definition of d/dx. As Vorde said, you can't treat d/dx as a fraction.

- #17

Vorde

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pwsnafu said:The right hand side is not mathematically well-defined, and furthermore that not the definition of d/dx. As Vorde said, you can't treat d/dx as a fraction.

And just to add on to this, ##\frac{d}{dx}## is just notation for a 'derivative operator', i.e. ##\frac{d}{dx}## is saying "take the derivative of this:".

##\frac{dy}{dy}##, which is

- #18

Mark44

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mishin050 said:##\displaystyle\frac{d}{dx}=\lim_{\Delta x\to 0}\frac{\Delta}{\Delta x}##

pwsnafu said:The right hand side is not mathematically well-defined, and furthermore that not the definition of d/dx. As Vorde said, you can't treat d/dx as a fraction.

And the right side isn't defined, either, as Δ by itself doesn't mean anything in this context.

- #19

DrewD

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mishin050 said:$$ \frac{d}{dx} \int_a^x f(t)~dt = f(x);$$

$$ dx \cdot\frac{d}{dx} \int_a^x f(t)~dt = f(x)\cdot dx;$$

$$\int d \int_a^x f(t)~dt =\int f(x)~dx;$$

$$ \int_a^x f(t)~dt =\int f(x)~dx;$$

$$\int f(x)~dx= F(x)-F(a)!$$

In your own (incorrect anyway) proof, you used the fact that [itex]\int f(x) dx=F(x)[/itex] between the third and fourth lines. You weren't correct getting to that line, but even if you were, you assume the exact fact that you are trying to prove wrong. Here's my proof...

Let the operator [itex]\int dx[/itex] denote the operation that sends a function [itex]f[/itex] to the equivalence class of antiderivatives [itex]F+C[/itex] where [itex]F[/itex] is any antiderivative and [itex]C[/itex] can be any element of the range of the function F.

Therefore,

$$\int f(x)dx=F(x)$$

because it is the definition! There's no math involved. The fundamental theorem makes this notation meaningful, but this is merely notation that was chosen because it makes sense.

Definite integrals are used to calculate the area under a curve between two given points. However, sometimes it is not possible to find the exact value of an integral using basic integration techniques. This is where anti-derivatives come in. They allow us to find the exact value of a definite integral by using the fundamental theorem of calculus, which relates integrals to anti-derivatives.

While numerical methods like the trapezoidal rule or Simpson's rule can give us an approximate value of a definite integral, they may not always be accurate or efficient. Anti-derivatives, on the other hand, give us the exact value of the integral and are often easier to compute.

Anti-derivatives are functions that are the reverse of derivatives. They allow us to work backwards from the derivative of a function to find the original function. This is useful when calculating definite integrals, as we can use the fundamental theorem of calculus to relate the integral to an anti-derivative and find its exact value.

No, they are not the same thing. Indefinite integrals give us a family of functions that have the same derivative, while anti-derivatives give us a specific function that has a given derivative. However, they are closely related, as the anti-derivative of a function is the same as the indefinite integral of that function.

Not always. Some integrals may be too complex to solve using anti-derivatives or may not have an anti-derivative that can be expressed in terms of elementary functions. In these cases, numerical methods may be the best option for finding the value of the integral.

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