Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

Thanks

Hi

I am very happy you rated it as a clear material )

**peter010**- Replies: 4

Hello,

In my high school, i was told to add a constant C to the results generated from Indefinite Integration! but I did not know why should I do so!! This video illustrates the reason behind this:

**peter010**- Replies: 4

Hello,

I have finished recently this video for presenting the essential meaning of Integration:

https://www.youtube.com/watch?v=5Y-LGfbS6JA

What do you think?

bob bundy wrote:

hi markosheehan

Use

a = -g, Write the two equations for h with t1 and t2. Equate to get an expression for u in terms of g, t1 and t2. Substitute back into either to eliminate u.

Bob

ps. Why did I do it this way? The problem involves height, gravity, initial velocity and time so that forces which equation to use. With two equations you can eliminate one variable. I chose h because it looked easiest but then had to rework to eliminate u. Maybe I should have made u the subject of the two and equated those to get the answer straight off. I leave it to the reader to try that if desired.

pps. Tried it myself and it comes out easily this way in four lines.

Hi,

I followed that, but we still have then h1 (at t1) and h2 (at t2)..

[t1.h2-t2.h1]/t1.t2=0.5g[t1-t2]

what do you think ?

final t is comprised of t1 of t2 times .. in other words -> t_final=t1*t2 (considering that t2 is just a unit less number)

-- Applying motion equation:

v= u-gt // minus because its going up

but v when particle reaches the top = 0

0=u-gt

u=gt..

lets integrate this over time..

h=0.5gt^2

//substitute: t=t1*t2

h=0.5g[t1*t2]^2

//arranging the equation..

[t1*t2]^2 = 2h/g // but i know its not correct since units are not compatible on both sides-> unless considering the whole set of t1*t2 is of second's unit but not second square, as pre my assumption t=t1*t2 !

then i can say:

t1*t2 = 2h/g

I thought person like you dreams even in mathematics

bobbym wrote:

Okay, good to talk to you. I am going to have dinner, see you later.

Hi,

And this is the outcome ))

https://www.youtube.com/watch?v=5Y-LGfbS6JA

regards.

bobbym wrote:

You can make particular formulas for any function but it appears there will always be some that will not like your formula.

I see

So, what an exciting night

Thanks indeed again..

bobbym wrote:

Did you get anything close to 2 / 3 = .66666666...?

= 0.266

but this still does not mean that i cannot customizing the solution/code to fit this kind of functions/

bobbym wrote:

Many integrals are what we call pathogenic ( diseased or sick ) in numerical work it means they do not respond well to numerical integration. One of these is

It is easy to do using calculus but not so easy using numerical methods. You might like to try it when you have some spare time.

hahaha I just did..

And I found yes it is diseased

bobbym wrote:

That is fine! You are making discoveries for yourself and thinking about how to get your computer to do them. The epitome of EM. I was just giving you some of what I do in that field. My favorites are numerical integration and curve fitting.

That is a lovely field.

And I appreciate and welcoming your advices, which is the reason why I come to this forum

bobbym wrote:

Hi;

Any function precisely? No amount of rectangles can do that unless it is an infinite number of them. There are even many more precise methods of numerical integration ( exactly what you are doing ) but each except for limited cases is just an approximation.

ok

i just wanted to share "my code" of series "approximation" method with ppl.

noth. more,,

Regards.

bobbym wrote:

Did you put a lot of very thin rectangles under the curve?

This code is to meet any function precisely (not specifically the function of 10 ) with a rectangular width=0.0001 (it can be definitely adjusted by the user)

Here I put my code for calculating the integration by approximation method (the series),,,

//Integrating Functions

double function;

double width = .0001;

double sum = 0, j = 1;

int interval = 5;

for (double i = width; j <= (interval / width); j++, i += width)

{

function = 10;

sum += (width * function);

}

Console.WriteLine(sum);//50

Console.WriteLine("\nDone!");

///

thickhead wrote:

I cant describe this case better than this

zetafunc wrote:

He might do, yes, but the opening line of his post suggested that he was using that definition of an integral for a general function .

Thanks

bobbym wrote:

He might do, yes, but the opening line of his post suggested that he was using that definition of an integral for a general function f.

That is correct and why I answered in the way I did.

Thanks Indeed

bobbym wrote:

As I said summation and integration can be used as approximations of each other. The discretizing trick will not always produce exact same answers.

For instance:

Thanks

I totally agree. Its here a special matter of integrating a constant, thus I presume exact answer could be achieved.

thickhead wrote:

Thanks

So, what if I considered it like: 0 to 1 represents: 1, 1 to 2 represents: 2, 2 to 3 represents: 3, and so on, thus I can presume the following:

5

∫10 = (10*5) - (10*0) = 50

0

What do you think ?

bobbym wrote:

Hi;

Integration can be defined as:

b b

∫ f(x).dx = ∑ f(x)

a aWho said integration and summation were the same thing? Sometimes, we can approximate one with the other but they are not the same thing.

http://www.mathcentre.ac.uk/resources/Engineering%20maths%20first%20aid%20kit/latexsource%20and%20diagrams/8_12.pdf

Pages: **1**