mishin05

I it read for a long time all and I know. Just because I know and I say that in official calculus there are errors. The first significant error - the INTEGRATION CONSTANT "+С". It can't be! I ask you prove that it is. Textbooks were written by other people. I can't ask them a question. And to you to a smog. Time you consider that it is correct, prove to me now here again! I will a little prompt to you: 1. [math]\int adx=a\int dx, a=const.[/math] 2. [math]\frac{dC}{dx}=0[/math]; 3. [math]\displaystyle \int dC=\int 0dx[/math]. 4. According to 1. (a=0): [math]\displaystyle \int dC=0\int dx=0[/math]. 5. According to 4: [math]\displaystyle \int dC=0[/math]; [math]\displaystyle \int 0dx=0[/math]; [math]\displaystyle \int 0dx \not=C[/math]! "... Thus if F(x) is an antiderivative of some function f(x), so is G(x)=F(x)+c..." [math]\frac{dG(x)}{dx}=\frac{F(x)+c}{dx}[/math]; [math]\frac{dG(x)}{dx}=\frac{F(x)}{dx}+\frac{c}{dx}[/math]; [math]\frac{dG(x)}{dx}=\frac{F(x)}{dx}+0[/math]; [math]\frac{dG(x)}{dx}=f(x)[/math]; [math]dG(x)=f(x)dx[/math]; [math]\displaystyle\int dG(x)=\int f(x)dx[/math]; [math]G(x)=F(x)[/math]; [math]c=0[/math]. Now prove that [math]\int dx=x+C, C\not=0[/math]!

That the integral started from any point to it enough to have variable limits: [math]\int\limits_{0}^{l} f(x)dx+\int\limits_{l}^{x}f(x)dx=\int\limits_{0}^{x}f(x)dx, if \int\limits_{l}^{x}f(x)dx=C[/math]. Because in the formula [math]\int f(x)dx=F(x)+C[/math] the right part isn't equal to the left if [math]C\not=0[/math]! If you consider their equal prove here. It is not necessary any references, I know all. Take and prove here! The constant +C sets displacement which depends on value of the integral. This displacement can be set integral limits then it will be correct. The integral where this displacement is issued in its limits otherwise should be set [math]x^2+C=\int\limits_{0}^{\sqrt{x^2+C}}2tdt[/math]. Integral with "dx" has borders (x-0) and you don't receive antiderivative [math]G(x)=x^2+C[/math]. For this purpose it is necessary to expand borders to (t-0)!

At what here derivatives? We speak about integration. Who can prove, what [math] \int dx=x+C [/math]?

This constant isn't present! Prove that it is! If to assume that it is, it should be equal to zero. Look at start-topic! There there is a proof! It yet the most important error!

You don't understand a simple thing. Integration - is process, return to differentiation. Differentiation doesn't have two various processes. There is a process of search of an increment with reduction and without reduction. And - all! And integration doesn't have two various processes. The uncertain integral too has a limit is an argument of integration. And certain differs from uncertain that its limit is less - it is limited by two values of argument. And - all! The rest is a misunderstanding!

Error of an official science that on the basis: [math] \frac {dt} {dx} = \frac {dx} {dx} =1 [/math] the conclusion that [math] x=t [/math], because actually [math] t=x+C [/math] has been drawn. It only one of many errors of an official science. I develop "the Structural analysis" which cleans all errors!

0--------[math]x[/math]----------A-----------[math]C[/math]------------B>[math]t[/math] 0->[math]t[/math] - Axis of abscisses, {A} - Mobile point, therefore [math]x=|OA|[/math] - a variable, [math]|AB|=C=const.[/math] ' the Structural analysis ' approves, that record of the mathematical analysis [math]\displaystyle\int dx=x+C[/math] makes no sense, since in it actually three are laid various integral: 1. If [math]\displaystyle t=x+C[/math], [math]\displaystyle\frac{dt}{dt} = \frac {d (x+C)}{d (x+C)}=1.[/math] [math]|OB|=\displaystyle\int \limits _{0}^{t}dt=t = \int\limits_{0}^{x+C}d(x+C) = \int\limits_{0}^{x+C}dt=\int d(x+C)=x+C.[/math] 2. Special case [math]\displaystyle t=x+C[/math] at [math]\displaystyle C=0 [/math] [math]|OA|=|OB|=\displaystyle\frac{dt}{dt} = \frac{dx}{dx}=1.[/math] [math]\displaystyle\int\limits_{0}^{t}dt[/math] [math] (t=x) [/math] [math]\displaystyle = \int \limits_{0}^{x}dx= \int dx=x.[/math] 3. [math]\displaystyle\frac{dt}{d(t-C)}=\frac{dt}{dx} = \frac{d (x+C)}{dx} = \frac {dx}{dx} =1.[/math] [math]|OA|=\displaystyle\int\limits_{0}^{t-C} dt= \int \limits_{0}^{x}dt=x[/math]. [math]\displaystyle\int dx\not=x+C[/math], because [math]\displaystyle\int dx=|OA|[/math], [math]\displaystyle x+C=|OB|[/math]. The uncertain integral is limited by an integration variable. The certain integral is limited by two values of a variable of integration. Geometrical interpretation of formula [math]\displaystyle U\cdot V=\int UdV+\int VdU[/math] for the elementary functions shows that both uncertain integrals are limited by arguments [math]V[/math] and [math]U[/math], therefore sum [math]\displaystyle\int UdV+\int VdU[/math] is equal to the area of rectangle [math]\displaystyle U\cdot V.[/math]

And you can prove it? Give, prove on a formula example: [math] \displaystyle U\cdot V =\int UdV +\int VdU [/math]!!! ' the Structural analysis ' approves, that record of the mathematical analysis [math]\displaystyle\int dx=x+C[/math] makes no sense, since in it actually three are laid various integral of Reimann! 1. If [math]\displaystyle t=x+C[/math], [math]\displaystyle\frac{dt}{dt} = \frac {d (x+C)}{d (x+C)}=1.[/math] [math]\displaystyle\int \limits _{0}^{t}dt=t = \int\limits_{0}^{x+C}d(x+C) = \int\limits_{0}^{x+C}dt=x+C ====== \int d(x+C)=x+C.[/math] 2. Special case [math]\displaystyle t=x+C[/math] at [math]\displaystyle C=0 [/math] [math]\displaystyle\frac{dt}{dt} = \frac{dx}{dx}=1.[/math] [math]\displaystyle\int\limits_{0}^{t}dt[/math] [math] (t=x) [/math] [math]\displaystyle = \int \limits_{0}^{x}dx=x ================ \int dx=x.[/math] 3. [math]\displaystyle\frac{dt}{d(t-C)}=\frac{dt}{dx} = \frac{d (x+C)}{dx} = \frac {dx}{dx} =1.[/math] [math]\displaystyle\int\limits_{0}^{t-C} dt=x ======================== \int \limits_{0}^{x}dt=x[/math].

If nobody wants to communicate with me prompt other forum where there are more than people.

[math]\displaystyle \int\limits_{0}^{x} f(a+x)dx=\int f(a+x)dx[/math] [math]\displaystyle \int\limits_{0}^{x} f(a+x)dx=ax+\frac{x^2}{2}[/math] [math]\displaystyle \int f(a+x)dx=ax+\frac{x^2}{2}[/math] [math]\displaystyle \int f(a+x)dx\not=ax+\frac{x^2}{2}+C[/math] [math]\displaystyle \int\limits_{a}^{\sqrt{(a+x)+2C}} ftdt=ax+\frac{x^2}{2}+C[/math] I want to show that the uncertain integral doesn't mean - boundless! The uncertain integral is limited by an integration variable, and certain is limited by values of this variable. Here in what at them a difference. Only that is known authentically about uncertain integral is a formula [math] \displaystyle U\cdot V =\int UdV +\int VdU [/math]. Geometrical interpretation of this formula of primitive functions show that the sum of two uncertain integrals is limited by the rectangle area [math] U\cdot V [/math]. Understand? It is limited [math] U [/math] and [math] V [/math]. Therefore [math] \displaystyle\int UdV=U _ {ANTIDERIVATIVE} (V) [/math] also can't be [math] \displaystyle\int UdV=U _ {ANTIDERIVATIVE} (V) +C [/math].

I also was afraid of it most of all! You that couldn't define [math] "f"[/math] under the formula [math]\displaystyle \int f(a+x)dx=(a+x)x-\int xd(a+x)[/math]? It is defined by a sign "=" in the formula! [math] f(a+x)=a+x[/math]! Naturally, it. As a result all will look so [math]\displaystyle \int (a+x)dx=(a+x)x-\int xd(a+x)[/math]. Now I understand, why on this site it is a lot of prevews, but it is not enough posts 1. [math]\displaystyle \int\limits_{a}^{a+x}tdt=\frac{(a+x)^2}{2}-\frac{a^2}{2}[/math]; 2. [math]\displaystyle \int (a+x)dx=(a+x)x-\int xd(a+x)[/math]; 3. [math]\displaystyle \int (a+x)dx=(a+x)x-\int xda-\int xdx[/math]; 4. [math]\displaystyle \int (a+x)dx=(a+x)x-\int xdx[/math]; 5. [math]\displaystyle \int (a+x)dx=\left(\frac{(a+x)^2}{2}-\frac{a^2}{2}+\frac{x^2}{2}\right)-\int xdx[/math]; 6. [math]\displaystyle \int (a+x)dx=\left(\frac{(a+x)^2}{2}-\frac{a^2}{2}+\int xdx\right)-\int xdx[/math]; 7. [math]\displaystyle \int (a+x)dx=\left(\frac{(a+x)^2}{2}-\frac{a^2}{2}\right)+\left(\int xdx-\int xdx\right)[/math]; 8. [math]\displaystyle \int (a+x)dx=\frac{(a+x)^2}{2}-\frac{a^2}{2}[/math]; 1. and 8. [math]\rightarrow[/math][math]\displaystyle \int (a+x)dx=\int\limits_{a}^{a+x} f(t)dt[/math].

1. [math]\displaystyle \int\limits_{a}^{a+x}tdt=\frac{(a+x)^2}{2}-\frac{a^2}{2}[/math]; 2. [math]\displaystyle \int f(a+x)dx=(a+x)x-\int xd(a+x)[/math]; 3. [math]\displaystyle \int f(a+x)dx=(a+x)x-\int xda-\int xdx[/math]; 4. [math]\displaystyle \int f(a+x)dx=(a+x)x-\int xdx[/math]; 5. [math]\displaystyle \int f(a+x)dx=\left(\frac{(a+x)^2}{2}-\frac{a^2}{2}+\frac{x^2}{2}\right)-\int xdx[/math]; 6. [math]\displaystyle \int f(a+x)dx=\left(\frac{(a+x)^2}{2}-\frac{a^2}{2}+\int xdx\right)-\int xdx[/math]; 7. [math]\displaystyle \int f(a+x)dx=\left(\frac{(a+x)^2}{2}-\frac{a^2}{2}\right)+\left(\int xdx-\int xdx\right)[/math]; 8. [math]\displaystyle \int f(a+x)dx=\frac{(a+x)^2}{2}-\frac{a^2}{2}[/math]; 1. and 8. [math]\rightarrow[/math][math]\displaystyle \int f(a+x)dx=\int\limits_{a}^{a+x} f(t)dt[/math].

The classical analysis has errors. That to eliminate them I have thought up the structural analysis. The basic formula:[math]\displaystyle \int f(a+x)dx=\int\limits_{a}^{a+x} f(t)dt[/math]. It leads to following contradictions for example: the Structural analysis: [math]\displaystyle\int(a+x)dx=\frac{(a+x)^2}{2}-\frac{a^2}{2}\not=\int(a+x)d(a+x)[/math]; [math]\displaystyle\int\limits_{0}^{x}2tdt=\int2xdx=x^2[/math]; [math]\displaystyle\int\limits_{0}^{\sqrt{x^{2}+C}}2tdt=x^2+C[/math]. The historical background: HERE