mishin05 3 Posted December 15, 2010 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 0 Share this post Link to post Share on other sites

the tree 222 Posted December 15, 2010 [math]\displaystyle\int(a+x)dx=\frac{(a+x)^2}{2}-\frac{a^2}{2}[/math]lulwut. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 15, 2010 lulwut. 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]. 0 Share this post Link to post Share on other sites

uncool 219 Posted December 15, 2010 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]. There are several problems with this. First problem is that you have not defined f. Second problem is that you do not have limits when you introduce the integral that you say is equal to x^2/2. You need to have limits from the beginning. Third problem is that you don't have limits on the integral of f at any point, so you cannot do step 8 at all. =Uncool- 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 15, 2010 There are several problems with this. First problem is that you have not defined f. Second problem is that you do not have limits when you introduce the integral that you say is equal to x^2/2. You need to have limits from the beginning. Third problem is that you don't have limits on the integral of f at any point, so you cannot do step 8 at all. =Uncool- 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]. 0 Share this post Link to post Share on other sites

uncool 219 Posted December 16, 2010 (edited) 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]? I thought it was, but your post is sufficiently unclear that I did not want to assume the answer. Additionally, the fact that you used f at all usually implies that there is some meaning to it, which is why I wanted to make sure I knew what you meant by it. It is defined by a sign "=" in the formula! [math] f(a+x)=a+x[/math]! OK, so we now have that f(a + x) = a + x. In other words, f(y) = y for any y. Thank you for already substituting everything later. 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]; Not quite. The correct equation is: 6. [math]\displaystyle \int (a+x)dx=\left(\frac{(a+x)^2}{2}-\frac{a^2}{2}+\int_{0}^{x} xdx\right)-\int xdx[/math]; To make this more clear, I will do a substitution of variables. 6. [math]\displaystyle \int (a+x)dx=\left(\frac{(a+x)^2}{2}-\frac{a^2}{2}+\int_{0}^{x} ydy\right)-\int xdx[/math]; I will put corrections on number 7. 7. [math]\displaystyle \int (a+x)dx=\left(\frac{(a+x)^2}{2}-\frac{a^2}{2}\right)+\left(\int_0^x xdx-\int xdx\right)[/math]; 8. [math]\displaystyle \int (a+x)dx=\frac{(a+x)^2}{2}-\frac{a^2}{2}[/math]; Due to the corrections, we can now see that step 8 does not follow from step 7 unless the boundaries of the rightmost integral are 0 and x. 1. and 8. [math]\rightarrow[/math][math]\displaystyle \int (a+x)dx=\int\limits_{a}^{a+x} f(t)dt[/math]. Again, you are canceling two integrals without checking that their boundaries are the same. The boundary of the integral that you added in is 0 and x, while the boundary of the right-most integral is undefined. You cannot cancel them unless their boundaries are the same. Once you do that, your final step becomes: 1. and 8. [math]\rightarrow[/math][math]\displaystyle \int_{0}^{x} (a+y)dy=\int\limits_{a}^{a+x} t dt[/math]. Which is entirely correct. =Uncool- Edited December 16, 2010 by uncool 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 16, 2010 [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]. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 17, 2010 If nobody wants to communicate with me prompt other forum where there are more than people. 0 Share this post Link to post Share on other sites

Bignose 946 Posted December 17, 2010 If nobody wants to communicate with me prompt other forum where there are more than people. mishin, it is tough to communicate when you make mistakes like [math] \displaystyle \int\limits_{0}^{x} f(a+x)dx=\int f(a+x)dx [/math] This equation is wrong. One side has limits, the other doesn't. This is not an equality. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 17, 2010 (edited) mishin, it is tough to communicate when you make mistakes like [math] \displaystyle \int\limits_{0}^{x} f(a+x)dx=\int f(a+x)dx [/math] This equation is wrong. One side has limits, the other doesn't. This is not an equality. 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]. Edited December 17, 2010 by mishin05 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 18, 2010 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] 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 18, 2010 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] 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 Share this post Link to post Share on other sites

John Cuthber 3646 Posted December 18, 2010 makes no sense, 0 Share this post Link to post Share on other sites

uncool 219 Posted December 18, 2010 Mishin, you are again mixing up integrals that have boundaries and integrals that don't. Integrals without boundaries, such at the integral of dx, have a +C term because they can start at any point, with any constant. Integrals with boundaries, on the other hand, do not have a +C because they are the difference of a function at two points. =Uncool- 0 Share this post Link to post Share on other sites

uncool 219 Posted December 18, 2010 Mishin, you are again mixing up integrals that have boundaries and integrals that don't. Integrals without boundaries, such at the integral of dx, have a +C term because they can start at any point, with any constant. Integrals with boundaries, on the other hand, do not have a +C because they are the difference of a function at two points. =Uncool- 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 18, 2010 Mishin, you are again mixing up integrals that have boundaries and integrals that don't. Integrals without boundaries, such at the integral of dx, have a +C term because they can start at any point, with any constant. Integrals with boundaries, on the other hand, do not have a +C because they are the difference of a function at two points. =Uncool- 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! 0 Share this post Link to post Share on other sites

D H 1371 Posted December 18, 2010 The rest is a misunderstanding! The only misunderstanding here is yours. You are spamming multiple forums with this nonsense. Many people have told you many, many times over that you are omitting the constant of integration in the indefinite integral. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 18, 2010 The only misunderstanding here is yours. You are spamming multiple forums with this nonsense. Many people have told you many, many times over that you are omitting the constant of integration in the indefinite integral. 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! 0 Share this post Link to post Share on other sites

insane_alien 840 Posted December 18, 2010 okay, if the derivative of a function is equal to 2x then what is its integral? at x=2 the integrated function has a value of 5. nothing above is mathematically impossible. explain how your posts are consistent with the above. 0 Share this post Link to post Share on other sites

D H 1371 Posted December 18, 2010 This constant isn't present! Prove that it is! The derivative of a constant function is zero, and the derivative of sum of two functions is the sum of the derivatives. Thus if F(x) is an antiderivative of some function f(x), so is G(x)=F(x)+c. This is basic, basic stuff. Have you taken even one course in calculus? 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 18, 2010 The derivative of a constant function is zero, and the derivative of sum of two functions is the sum of the derivatives. Thus if F(x) is an antiderivative of some function f(x), so is G(x)=F(x)+c. This is basic, basic stuff. Have you taken even one course in calculus? At what here derivatives? We speak about integration. Who can prove, what [math] \int dx=x+C [/math]? 0 Share this post Link to post Share on other sites

D H 1371 Posted December 18, 2010 (edited) Just. Read. This. http://en.wikipedia..../Antiderivative Russian wikipedia entry: Неопределённый интеграл — Википедия Edited December 18, 2010 by D H 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 19, 2010 Integrals without boundaries, such at the integral of dx, have a +C term because they can start at any point, with any constant. =Uncool- 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)! 0 Share this post Link to post Share on other sites

insane_alien 840 Posted December 19, 2010 At what here derivatives? We speak about integration. Who can prove, what [math] \int dx=x+C [/math]? answer my post and it will become apparent 0 Share this post Link to post Share on other sites

mississippichem 456 Posted December 19, 2010 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)! Constants disappear when you take a derivative. The C in the solution to an indefinite integral is just there because an indefinite integral is the reverse of a derivative and there is no way to know if the function contained a constant or what that constant was. I don't see what is so hard about that. [math] \frac{d}{dx}x^{2}+13=2x[/math] [math]\frac{d}{dx}x^{2}+500=2x[/math] You may know how to find integrals, but I doubt you know what an integral is. 0 Share this post Link to post Share on other sites