The Structural analysis
- Thread starter mishin05
- Start date
And that is exactly what the OP continues to do.
More than people? Like robots?
You've spoken to at least 3 PhDs in maths or physics in this thread and convinced none of them. You're making a mistake someone aged 16 shouldn't make.
Watch this:thescienceforum.com/Mathematics-forum-6f.php
Are you, perchance, using Google translate (or another automated translation service) for your posts? Because, frankly, this gibberish is painful to 'read'.
int dx =/= x+C
0--------$$x$$----------A-----------$$C$$------------B>$$t$$
0->$$t$$ - Axis of abscisses,
{A} - Mobile point, therefore $$x=|OA|$$ - a variable,
$$|AB|=C=const.$$
' the Structural analysis ' approves, that record of the mathematical analysis $$\displaystyle\int dx=x+C$$ makes no sense, since in it actually three are laid various integral:
1. If $$\displaystyle t=x+C$$,
$$\displaystyle\frac{dt}{dt} = \frac {d (x+C)}{d (x+C)}=1.$$
$$|OB|=\displaystyle\int \limits _{0}^{t}dt=t = \int\limits_{0}^{x+C}d(x+C) = \int\limits_{0}^{x+C}dt=\int d(x+C)=x+C.$$
2. Special case $$\displaystyle t=x+C$$ at $$\displaystyle C=0 $$
$$|OA|=|OB|=\displaystyle\frac{dt}{dt} = \frac{dx}{dx}=1.$$
$$\displaystyle\int\limits_{0}^{t}dt$$ $$ (t=x) $$ $$\displaystyle = \int \limits_{0}^{x}dx= \int dx=x.$$
3. $$\displaystyle\frac{dt}{d(t-C)}=\frac{dt}{dx} = \frac{d (x+C)}{dx} = \frac {dx}{dx} =1.$$
$$|OA|=\displaystyle\int\limits_{0}^{t-C} dt= \int \limits_{0}^{x}dt=x$$.
$$\displaystyle\int dx\not=x+C$$, because $$\displaystyle\int dx=|OA|$$, $$\displaystyle x+C=|OB|$$.
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 $$\displaystyle U\cdot V=\int UdV+\int VdU$$ for the elementary functions shows that both uncertain integrals are limited by arguments $$V$$ and $$U$$, therefore sum $$\displaystyle\int UdV+\int VdU$$ is equal to the area of rectangle $$\displaystyle U\cdot V.$$
0--------$$x$$----------A-----------$$C$$------------B>$$t$$
0->$$t$$ - Axis of abscisses,
{A} - Mobile point, therefore $$x=|OA|$$ - a variable,
$$|AB|=C=const.$$
' the Structural analysis ' approves, that record of the mathematical analysis $$\displaystyle\int dx=x+C$$ makes no sense, since in it actually three are laid various integral:
1. If $$\displaystyle t=x+C$$,
$$\displaystyle\frac{dt}{dt} = \frac {d (x+C)}{d (x+C)}=1.$$
$$|OB|=\displaystyle\int \limits _{0}^{t}dt=t = \int\limits_{0}^{x+C}d(x+C) = \int\limits_{0}^{x+C}dt=\int d(x+C)=x+C.$$
2. Special case $$\displaystyle t=x+C$$ at $$\displaystyle C=0 $$
$$|OA|=|OB|=\displaystyle\frac{dt}{dt} = \frac{dx}{dx}=1.$$
$$\displaystyle\int\limits_{0}^{t}dt$$ $$ (t=x) $$ $$\displaystyle = \int \limits_{0}^{x}dx= \int dx=x.$$
3. $$\displaystyle\frac{dt}{d(t-C)}=\frac{dt}{dx} = \frac{d (x+C)}{dx} = \frac {dx}{dx} =1.$$
$$|OA|=\displaystyle\int\limits_{0}^{t-C} dt= \int \limits_{0}^{x}dt=x$$.
$$\displaystyle\int dx\not=x+C$$, because $$\displaystyle\int dx=|OA|$$, $$\displaystyle x+C=|OB|$$.
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 $$\displaystyle U\cdot V=\int UdV+\int VdU$$ for the elementary functions shows that both uncertain integrals are limited by arguments $$V$$ and $$U$$, therefore sum $$\displaystyle\int UdV+\int VdU$$ is equal to the area of rectangle $$\displaystyle U\cdot V.$$
This guys a waste of bandwidth. If he wants to fail high school algebra let him do it somewhere else.
My "mishin05 : Powered by Google Translate" hypothesis is growing stronger with each post.
Error of an official science that on the basis: $$ \frac {dt} {dx} = \frac {dx} {dx} =1 $$ the conclusion that $$ x=t $$, because actually $$ t=x+C $$ has been drawn.
It only one of many errors of an official science. I develop "the Structural analysis" which cleans all errors!
It only one of many errors of an official science. I develop "the Structural analysis" which cleans all errors!
You are frightening , indeed. This is how they teach calculus now in the former soviet republics? Which one are you ? Russia? Chechnia?
Leibnitz says you fail.
I know it better, than you! Once again I repeat a question: who here can independently prove: $$\int dx=x+C$$?