mishin05

Well, we will go some other way. Let Structural Analysis - addition to Calculus. Because applying it it is possible to describe all geometry as formulas. it will be so good?

The main reason for that in calculus there were mistakes consists in the following. When Rene Descartes created the Cartesian system of coordinates, he meant that setting one function on the horizontal line, and another on vertical, the width and length of the rectangle which area represents one more function are set. The main mistake of founders of calculus in that that they badly studied: "Descartes's Rules for the Direction of the Mind". Because when horizontal (OX) and vertical (OY) pieces that are postponed the rectangle turns out . And the line of the schedule is only a part of drawing. Actually on graphics of function 9 functions are located but not one! Euler decided to investigate some lines on one schedule that was a mistake. Instead of 9 functions began to investigate only one which was represented in the form of the line. But also it interpreted incorrectly having presented as function which postponed on the vertical line. P.S. I badly know English therefore I transport through the robot.

All right if don't want to read everything at once, I will spread in parts

Well, I think, what this link is not infected: http://www.mathisfun...hp?id=18422&p=3

My link

I already placed such subject on other site. But anybody can tell nothing! I suggest to discuss's theme excerpts from this site

This idiotic definition: " Integral of function — natural analog of the sum of sequence." In the formula [math]f'(x)=\lim_{\Delta x\to 0}\frac{\Delta f(x)}{\Delta x}=\lim_{\Delta x\to 0}\frac{f(x_2)-f(x_1)}{x_2-x_1}[/math] Possible variants of aspiration of an increment of argument to "0" aren't defined. There are three variants: [math]{\bold{1.}~x_1\rightarrow x_2;~~~~~\bold{2.}~x_1\leftarrow x_2;~~~~~\bold{3.}~x_1\rightarrow x \leftarrow x_2.}[/math] Accordingly, there are three variants of a limit [math]\lim_{\Delta x\to 0}:[/math] [math]{\bold{1.}~\lim_{\Delta x\to 0}=\lim_{x_1\to x_2};~~~~~\bold{2.}~\lim_{\Delta x\to 0}=\lim_{x_2\to x_1};~~~~~\bold{3.}~\lim_{\Delta x\to 0}=\lim_{x_1\to x \leftarrow x_2}.}[/math] Therefore three variants of results of application of the formula are possible [math]f'(x)=\lim_{\Delta x\to 0}\frac{\Delta f(x)}{\Delta x}:[/math] [math]{\bold{1.}~f'(x_2)=\lim_{x_1\to x_2}\frac{\Delta f(x)}{\Delta x};~~~~~\bold{2.}~f'(x_1)=\lim_{x_2\to x_1}\frac{\Delta f(x)}{\Delta x};~~~~~\bold{3.}~f'(x)=\lim_{x_1\to x \leftarrow x_2}\frac{\Delta f(x)}{\Delta x}.}[/math] Все три варианта различны и не равны друг другу. Приведу пример на функции [math]y=x^3:[/math] [math]\begin{array}{*{20}{l}}{\bold{1.}~(x_2^3)'}&\!\!\!{=\lim\limits_{x_1\to x_2}\dfrac{x_2^3-x_1^3}{x_2-x_1}= \lim\limits_{x_1\to x_2}\dfrac{(x_2-x_1)(x_2^2+x_1x_2+x_1^2)}{x_2-x_1}=}\\[18pt]{}&\!\!\!{=\lim\limits_{x_1\to x_2}(x_2^2+x_1x_2+x_1^2)=x_2^2+x_2\cdot x_2+x_2^2=3x_2^2;}\end{array}[/math] [math]\begin{array}{*{20}{l}}{\bold{2.}~(x_1^3)'}&\!\!\!{=\lim\limits_{x_1 \leftarrow x_2}\dfrac{x_2^3-x_1^3}{x_2-x_1}=\lim\limits_{x_1\leftarrow x_2}\dfrac{(x_2-x_1)(x_2^2+x_1x_2+x_1^2)}{x_2-x_1}=}\\[18pt]{}&\!\!\!{=\lim\limits_{x_1 \leftarrow x_2}(x_2^2+x_1x_2+x_1^2)=x_1^2+x_1\cdot x_1+x_1^2=3x_1^2;}\end{array}[/math] [math]\begin{array}{*{20}{l}}{\bold{3.}~(x^3)'}&\!\!\!{=\lim\limits_{x_1\rightarrow x \leftarrow x_2}\dfrac{x_2^3-x_1^3}{x_2-x_1}= \lim\limits_{x_1\rightarrow x \leftarrow x_2}\dfrac{(x_2-x_1)(x_2^2+x_1x_2+x_1^2)}{x_2-x_1}=}\\[18pt]{}&\!\!\!{=\lim\limits_{x_1\rightarrow x \leftarrow x_2}(x_2^2+x_1x_2+x_1^2)=x^2+x\cdot x+x^2=3x^2.}\end{array}[/math]

[math]f(x)=\lim_{\Delta x\rightarrow 0}\frac{\Delta F(x)}{\Delta x}[/math] - Differentiation process. Where in this process you see y=F (x)? She doesn't participate in this process. [math]y_1=F(x_1); y_2=F(x_2); \Delta y=y_2-y_1[/math] is the preparatory period and differentiation process doesn't concern! [math]\frac{\Delta F(x)}{\Delta x}[/math] - The process beginning, the his first stage. [math](x^2)'=\lim_{x_1\rightarrow x\leftarrow x_2}\frac{x_2^2-x_1^2}{x_2-x_1}=\lim_{x_1\rightarrow x\leftarrow x_2}\frac{(x_2-x_1)(x_2+x_1)}{x_2-x_1}=\lim_{x_1\rightarrow x\leftarrow x_2}(x_2+x_1)=x+x=2x[/math]. [math]\int\limits_{x_1}^{x_2} 2xdx=a\lim_{x_1\leftarrow x\rightarrow x_2}(2xdx)=a\lim_{x_1\leftarrow x\rightarrow x_2}2x\cdot\Delta x=(x_1+x_2)(x_2-x_1)=x_2^2-x_1^2.[/math] [math]\int 2xdx=a\lim_{x_1\leftarrow x\rightarrow x_2}(2xd(x-0))=a\lim_{x_1=0\leftarrow x\rightarrow x_2=x}2x\cdot\Delta x=(0+x)(x-0)=x^2-0^2.[/math] At school learn nonsenses. For example: [math]\int 0dx=C[/math]. It is necessary to create school where people don't deceive!

Excuse, if I have offended you. Also I am sorry about other normal people who was offended by my phrase! I meant such as Mr Skeptic which makes mistakes not pardonable for kids, but also criticizes others! You inattentively read! The point {0} is motionless. Points {B} and {F} - are mobile. On the rest, except a rectangle yet we do not pay attention. If you name variable |OB | - "x" or any other letter while I won't object. But, I think that we clever people and we understand: properties of size doesn't depend on its name! And from the rename don't change! On Lession 1 it is necessary to show the beginning of integration. When [math] \int tdx=tx [/math] What is [math]d|OB|?[/math] This is [math]d|OB|=\lim_{\Delta |OB|\rightarrow 0}\Delta |OB|=\lim_{B_1\rightarrow B\leftarrow B_2}(|OB_2|-|OB_1|)[/math].

Integration - process, return to differentiation. Differentiation process - unique. Integration process too the unique. The certain and uncertain integral - two versions of the unique process. In process differentiation participate: 1. A function increment. 2. An argument increment. 3. A limit. Function doesn't participate in this process. Therefore the function increment can be result of integration only! No function!

[math]\int f(x)dx=F(x)-F(0)![/math]

Это полный пиздец! Ku-ku! I have got on a mathematical forum or to a psychiatric clinic? 1. [math]\frac{d}{dx} f(x) =f'(x)(!!!!!!!)= 2x[/math]; [math] f(x) =\int 2xdx[/math]. 2. [math] F(x) =\int f(x)dx[/math]; [math] F(x) =\int\left(\int 2xdx\right)dx[/math]; [math]\int\left(\int 2xdx\right)dx=x^2+C???????????!!!!!!!!!!!!!!![/math]

I got it. You are that person who possesses independent thinking. Unlike other self-assured silly persons. I am ready to spend for the sake of you the time. Lesson 1. In the beginning we will look, integration whence undertook. As it is connected with arithmetics and algebra. We look the attached drawing: If to take a rectangle its area will be equal to multiplication of length with the width. In this case [math] S _ {OFGB} = |OF |\cdot|OB | [/math]. Condition: the length from width doesn't depend. They are variables. To fill with itself all area of a rectangle length [math] |OF | [/math] which is set of points on one line we will increase by distance between two being nearby points [math] |OB | (d|OB |) [/math]. The area of very thin plate [math] |OF |\cdot d|OB | [/math] As a result will turn out area of a rectangle/ when these thin plates it is stacked densely to each other from a point {0} to a point {B}. We designate it so: [math] S _ {OFGB} = |OF |\cdot|OB | =\int\limits _ {0} ^ {|OB |} |OF|d|OB | [/math]. The break line.pdf

Certainly, you are right, for this reason my reasonings have gone further: 8/3 is not equal to 5, but not have stopped here. I thought that to any fool it is clear that there there should be two words: "... Incorrectly, therefore..." Such people as you there is a jackal of Tabaki from "Maugli". Than is more I read posts of those people which write about me awfully, I am more convinced that is worthless people, but with a lot of ambitions because have read another's thoughts in textbooks and think that it now their own thoughts. Suddenly they read the person whose thoughts don't coincide with their thoughts. What to do? The clever person will tell: "Give we will look, give me the proofs!" The idiot will tell: "I don't want to hear it. Shut up him", because except someone's thoughts he don't have.

What isn't pleasant to you? When I tell something, I bring to the formula proof. That is I communicate in a mathematical language. Why, when opponents speak, they communicate in philosophical language? What, really it is not clear that the mathematical language is unequivocal, and philosophy language is multiple-valued? Answer me in a mathematical language, than you are dissatisfied? Integration was invented by the nature. The problem of the person was to think up corresponding for the general understanding of this phenomenon badges. I have a little altered badges for clearness. What from this? It not new process, is additional words of a mathematical language. I have expanded it! This is bad?

Look here. It is written in Russian, but, I think that you will understand. by forum.pdf

You simply don't understand that such integral. And at school of it almost don't explain. There very well explain that such a derivative (though and it is not absolutely correct), and the integral is a return action of a derivative. Now I to you will a little open eyes. You know that such a derivative? It is a limit of the relation of an increment of function to an argument increment when the argument increment aspires to zero. Look attentively this definition: differentiation process begins with a function increment. Integration process as return, comes to an end the with the same. A function increment. Understand? Not function, but its increment. These are all were persuaded to consider that uncertain integral - antiderivative function. But actually result of integral not function but its increment. Whence the increment can have a constant? The constant derivative therefore also is equal to zero that the difference of constants is equal to zero! The integral also comes to an end with the same difference!

1. [math]\frac{d}{dx} f(x) = 2x[/math]; [math]d f(x) = 2xdx[/math]; [math]\displaystyle\int d f(x) = \int 2xdx[/math]; [math]\displaystyle f(x) = x^2[/math]. 2. [math]\left. \int f(x) \, dx \right|_{x=2} = 5[/math]; [math]\left. \int x^2 \, dx \right|_{x=2} = 5[/math]; [math]\left.\frac{x^3}{3}\right|_{x=2} = 5[/math]; [math]\frac{t^3}{3}-\frac{x^3}{3} = 5-\frac{x^3}{3}(x=2)[/math]; [math]\frac{t^3}{3}-\frac{x^3}{3} = 5 -\frac{2^3}{3} [/math]; [math]\frac{t^3}{3} =\frac{x^3}{3}+ \frac{7}{3}[/math]; [math]t^3 = x^3+7[/math]; [math]t =\sqrt[3]{x^3+7}[/math]; 3. [math]F(x)=\displaystyle\int\limits_{0}^{\sqrt[3]{x^3+7}} t^2dt[/math]. Well, I will repeat one more time: the Uncertain integral differs from defined so: 1. The uncertain integral is limited by argument of integration [math] (d (x-0)) [/math]. 2. The certain integral is limited by values of this argument [math] (x_1, x_2) [/math]. And - all! More anything. No constants of integration are present! Turn to a zero at differentiation a constant can. And to rise from zero at integration - isn't present!

[math]\displaystyle\int\limits_{0}^{\sqrt[3]{x^3+7}} t^2dt=\frac{x^3+7}{3}[/math]. The formula [math] \int dx=x+C [/math] is erroneous, because for its conclusion it is required here: [math] 0 =\frac {dC} {dx} [/math]; [math] \int 0dx =\int dC [/math]; [math] \int 0dx=C [/math]. And it isn't correct, because: [math] \int 0dx=0\int dx=0x=0 [/math]. Therefore all over the world it is necessary to copy textbooks. It is necessary so: [math] x+C =\int\limits_{0}^{x+C}dt=\int d(x-0) +\int\limits _ {0} ^ {C} dx =\int\limits _ {0} ^ {x} dx +\int\limits _ {0} ^ {C} dx =\int\limits _ {0} ^ {x} dt +\int\limits_{x} ^ {x+C} dt [/math].

[math]F(x)=\frac{x^3+7}{3}[/math].

[math]\displaystyle\int\limits_{2}^{\sqrt[3]{23}}x^2dx=5[/math].

Look post #34

Works. But not all! And where errors of didn't show. [math] \displaystyle\int mvdv =\frac {mv^2} {2} +C [/math] - the perpetuum mobile formula. Where it works? You not absolutely correctly have understood me. When calculate it think that apply this formula: [math]\displaystyle\int dx=x+C[/math]. Actually apply this formula: [math]\displaystyle\int\limits_{0}^{x+C} dt=x+C[/math]. Only don't understand it. Because this formula [math]\displaystyle\int dx=x+C[/math] - wrong. At it the right and left parties various also aren't equal each other. Therefore in the textbook have written: [math]\displaystyle\int 0dx=C[/math] that is wrong, because it [math]\displaystyle\int 0dx=0[/math], [math]\displaystyle\int\limits_{0}^{C} dt=C[/math] - are correct. Modern calculus is a freak with legs of a chicken. All use only legs. I have developed the Structural Analisys and now calculus becomes a normal chicken and will be used both legs, and wings, and others. Now it becomes the presents. Also it begins to be used more. For example so: "The current strength is structured by conductivity of a conductor on potential: [math]\displaystyle I=\int\limits_{\phi_1}^{\phi_2}\frac{1}{R}d\phi[/math]", "The circle area is structured by length of a circle on radius: [math]\displaystyle \pi r^2=\int 2\pi rdr[/math] ". And much, many other things. The mathematics becomes even developer to all other sciences.

And who has told, what it is correct: [math]\frac{d(x^2+C)}{dx}=2x[/math]??? Correctly it: [math]\frac{d(x^2+C)}{d\sqrt{x^2+C}}=2\sqrt{x^2+C}[/math]!!!

[math]\int 2xdx=x^2[/math]; [math]\int 2xdt=2xt[/math]; [math]\displaystyle\int\limits_{0}^{x}2xdt=2x^2[/math]; and more, more, more... Which integrated function has a value of 5? Learn to formulate questions competently mathematically!!!