mishin05 3 Posted December 19, 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? 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]! 0 Share this post Link to post Share on other sites

insane_alien 840 Posted December 19, 2010 you can't prove that because 0 is a possible solution for C if you answer my post then you would see what you want to know. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 19, 2010 okay, if the derivative of a function is equal to 2x then what is its integral? [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... at x=2 the integrated function has a value of 5. Which integrated function has a value of 5? nothing above is mathematically impossible. explain how your posts are consistent with the above. answer my post and it will become apparent Learn to formulate questions competently mathematically!!! 0 Share this post Link to post Share on other sites

insane_alien 840 Posted December 19, 2010 wow, i just gave you an integration problem any highschool student would be able to solve and you failed miserably at it. go learn integration before you start complaining about how it is done. for instance, where the hell did the t and the limits come from? the weren't supplied. and what about the constant that goes after x^2? i clearly said that the integrated function had the value of 5 at x=2 now what bit are you not getting? 0 Share this post Link to post Share on other sites

Cap'n Refsmmat 1351 Posted December 19, 2010 To express the problem mathematically: we have an unknown function [math]f(x)[/math]. [math]\frac{d}{dx} f(x) = 2x[/math] [math]\left. \int f(x) \, dx \right|_{x=2} = 5[/math] Find [math]\int f(x) \, dx[/math]. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 19, 2010 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. 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]!!! 0 Share this post Link to post Share on other sites

insane_alien 840 Posted December 19, 2010 ahh thats why all our calculus based designs and simulations don't work! </sarcasm> 1 Share this post Link to post Share on other sites

mississippichem 456 Posted December 19, 2010 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]!!! Take the derivative of any function that contains a constant. The constant always becomes 0 in the derivative. Now integrate the same function you just derived without integral limits. There is no way you can recover the original constant with an indefinite integral. Hopefully we can agree that: [math] \int \frac{d}{dx}f(x)dx=f(x) [/math] So one must add +C to an indefinite integral to account for any constant that could have been present in the original function. Try it for yourself, that is, if you even know how to integrate or differentiate at all. Also, as insane alien just mentioned and others mentioned before, you can't just arbitrarily stick limits on the integral operator. Seriously, what is your level of mathematics education? You do realize that if we've been doing indefinite integrals wrong since the 1700's that all the physics, chemistry, mathematics, and modeling done since then is completely wrong, right? Are you really that confident in your idea? I'd be willing to bet that everyone who has posted in this thread uses the same math you're disputing on at least a weekly basis in school or at work. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 19, 2010 (edited) ahh thats why all our calculus based designs and simulations don't work! </sarcasm> 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? Take the derivative of any function that contains a constant. The constant always becomes 0 in the derivative. Now integrate the same function you just derived without integral limits. There is no way you can recover the original constant with an indefinite integral. Hopefully we can agree that: [math] \int \frac{d}{dx}f(x)dx=f(x) [/math] So one must add +C to an indefinite integral to account for any constant that could have been present in the original function. Try it for yourself, that is, if you even know how to integrate or differentiate at all. Also, as insane alien just mentioned and others mentioned before, you can't just arbitrarily stick limits on the integral operator. Seriously, what is your level of mathematics education? You do realize that if we've been doing indefinite integrals wrong since the 1700's that all the physics, chemistry, mathematics, and modeling done since then is completely wrong, right? Are you really that confident in your idea? I'd be willing to bet that everyone who has posted in this thread uses the same math you're disputing on at least a weekly basis in school or at work. 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. Edited December 19, 2010 by mishin05 0 Share this post Link to post Share on other sites

Cap'n Refsmmat 1351 Posted December 19, 2010 Please do solve the problem I posed in post #30. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 19, 2010 Please do solve the problem I posed in post #30. Look post #34 0 Share this post Link to post Share on other sites

Cap'n Refsmmat 1351 Posted December 19, 2010 I did. Please solve the problem I posed in post #30. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 20, 2010 (edited) I did. Please solve the problem I posed in post #30. [math]\displaystyle\int\limits_{2}^{\sqrt[3]{23}}x^2dx=5[/math]. Edited December 20, 2010 by mishin05 0 Share this post Link to post Share on other sites

Cap'n Refsmmat 1351 Posted December 20, 2010 [math]\displaystyle\int\limits_{2}^{\sqrt[3]{23}}x^2dx=5[/math]. I asked for [math]\int f(x) \, dx[/math], which is a function, not a constant value. What is the function? 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 20, 2010 I asked for [math]\int f(x) \, dx[/math], which is a function, not a constant value. What is the function? [math]F(x)=\frac{x^3+7}{3}[/math]. 0 Share this post Link to post Share on other sites

Cap'n Refsmmat 1351 Posted December 20, 2010 [math]F(x)=\frac{x^3+7}{3}[/math]. Now, please show your work. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 20, 2010 (edited) Now, please show your work. [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]. Edited December 20, 2010 by mishin05 -1 Share this post Link to post Share on other sites

Cap'n Refsmmat 1351 Posted December 20, 2010 [math]\displaystyle\int\limits_{0}^{\sqrt[3]{x^3+7}} t^2dt=\frac{x^3+7}{3}[/math]. No, show all of your work, from the beginning. 0 Share this post Link to post Share on other sites

Mr Skeptic 1154 Posted December 20, 2010 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]. This isn't correct either. You can't get a definite answer from an indefinite integral; [math] \int 0dx =\int dC [/math] [math] 0 + C_2 = C + C_3 [/math] To find C2 and C3 you need to know the limits of integration. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 20, 2010 (edited) To express the problem mathematically: we have an unknown function [math]f(x)[/math]. [math]\frac{d}{dx} f(x) = 2x[/math] [math]\left. \int f(x) \, dx \right|_{x=2} = 5[/math] Find [math]\int f(x) \, dx[/math]. 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]. This isn't correct either. You can't get a definite answer from an indefinite integral; [math] \int 0dx =\int dC [/math] [math] 0 + C_2 = C + C_3 [/math] To find C2 and C3 you need to know the limits of integration. 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! Edited December 20, 2010 by mishin05 0 Share this post Link to post Share on other sites

Mr Skeptic 1154 Posted December 20, 2010 Well you're wrong. Integration and derivatives are almost opposite, such that the derivative of an integral of a function always gives you the original function. Moreover, the integral of a derivative can give you back the original function, but not necessarily. Note that some information is lost by taking the derivative, because the derivative of a constant is zero. This means that the integral of zero must be a constant, but it could be any constant and hence is unknown. For example, if f(x) = x + 3, then [math]\frac{d}{dx} f(x) = 1[/math] But [math]\displaystyle\int d f(x) = \int 1 dx ==> f(x) = x + C[/math], but without more info than that you can't show that C=3. Because you don't realize this, you're saying that [math]\frac{2^3}{3}=5[/math], which clearly isn't true. 1 Share this post Link to post Share on other sites

mishin05 3 Posted December 20, 2010 (edited) the derivative of a constant is zero. This means that the integral of zero must be a constant, but it could be any constant and hence is unknown. 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! Edited December 20, 2010 by mishin05 0 Share this post Link to post Share on other sites

DJBruce 143 Posted December 21, 2010 Would you mind writing your definitions out symbolically. I really don't understand what you mean when you say things like, differentiation process begins with a function increment or But actually result of integral not function but its increment. Also just to be clear when you talking about integrals you are meaning the riemann integral of a continuous function correct? 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 21, 2010 Would you mind writing your definitions out symbolically. I really don't understand what you mean when you say things like, or Also just to be clear when you talking about integrals you are meaning the riemann integral of a continuous function correct? Look here. It is written in Russian, but, I think that you will understand. by forum.pdf 0 Share this post Link to post Share on other sites

Mr Skeptic 1154 Posted December 21, 2010 Well, when you invent something new you need a new name for it. You can't call it an integral if it isn't an integral, which your idea isn't. I'm not really too interested in the maths of someone who derives equations from 8/3=5. 0 Share this post Link to post Share on other sites