Thorham Posted June 27, 2015 Share Posted June 27, 2015 In this integral (error function), what is d and what is t? This stuff is hard to find for math noobs 1 Link to comment Share on other sites More sharing options...
ajb Posted June 27, 2015 Share Posted June 27, 2015 t looks to be a real number. dt is a formal symbol that tells you what you are integrating with respect to. For simple integrals like that there is no more in it really. You can use Riemann's definition. 1 Link to comment Share on other sites More sharing options...
Unity+ Posted June 27, 2015 Share Posted June 27, 2015 (edited) In this integral (error function), what is d and what is t? This stuff is hard to find for math noobs What do you mean what is d and what is t? That is part of integral notation. EDIT: I thumbed up the wrong post. I meant just to thumb up ajb. Oops? Edited June 27, 2015 by Unity+ 1 Link to comment Share on other sites More sharing options...
Thorham Posted June 27, 2015 Author Share Posted June 27, 2015 (edited) t looks to be a real number. dt is a formal symbol that tells you what you are integrating with respect to. For simple integrals like that there is no more in it really. You can use Riemann's definition. I have no Idea what that means. Could someone explain this as if I'm a four year old? Edited June 27, 2015 by Thorham Link to comment Share on other sites More sharing options...
Unity+ Posted June 27, 2015 Share Posted June 27, 2015 (edited) I have no Idea what that means. Could someone explain this as if I'm a four year old? It means that you are taking the reverse derivative of an equation with respect to a particular variable. For example, let's say you have ax+b and the point is to integrat this equation with respect to x, or apply the integral to that particular variable. You would treat every other variable like a constant and simply integrate x. ax^2/2 + bx Edited June 27, 2015 by Unity+ Link to comment Share on other sites More sharing options...
Thorham Posted June 27, 2015 Author Share Posted June 27, 2015 It means that you are taking the reverse derivative of an equation with respect to a particular variable. For example, let's say you have ax+b and the point is to integrat this equation with respect to x, or apply the integral to that particular variable. You would treat every other variable like a constant and simply integrate x. ax^2/2 + bx I don't understand that, either. The gaps in my maths knowledge are just to large. Perhaps it's time to go and do something about that. Link to comment Share on other sites More sharing options...
Unity+ Posted June 27, 2015 Share Posted June 27, 2015 I don't understand that, either. The gaps in my maths knowledge are just to large. Perhaps it's time to go and do something about that. What about it do you not understand? Link to comment Share on other sites More sharing options...
Thorham Posted June 27, 2015 Author Share Posted June 27, 2015 What about it do you not understand? Everything. I took a quick look at integrals and thought it wasn't so hard, but I don't get any of it. I should probably just take my time to learn integrals properly. Link to comment Share on other sites More sharing options...
mathematic Posted June 28, 2015 Share Posted June 28, 2015 Everything. I took a quick look at integrals and thought it wasn't so hard, but I don't get any of it. I should probably just take my time to learn integrals properly. Good idea!!!!! You need to understand the basic notation. 1 Link to comment Share on other sites More sharing options...
Unity+ Posted June 28, 2015 Share Posted June 28, 2015 Everything. I took a quick look at integrals and thought it wasn't so hard, but I don't get any of it. I should probably just take my time to learn integrals properly. Before learning integrals, I think you should start with derivatives first. If that is confusing, I would start with limits of functions then. Link to comment Share on other sites More sharing options...
LaurieAG Posted June 28, 2015 Share Posted June 28, 2015 Everything. I took a quick look at integrals and thought it wasn't so hard, but I don't get any of it. I should probably just take my time to learn integrals properly. The integral is just the opposite of the derivative so the tutorials on this forum should give you a helping hand. http://www.scienceforums.net/topic/29473-introduction-to-calculus-differentiation/ http://www.scienceforums.net/topic/4108-calculus-i-lesson-1-a-background-to-differentation/ http://www.scienceforums.net/topic/4182-calculus-i-lesson-2-a-continuation-from-first-principles/ 1 Link to comment Share on other sites More sharing options...
ajb Posted June 28, 2015 Share Posted June 28, 2015 Or you you can think in terms of the area under the graph. If you ploy y = f(x) then the integral over x1 to x2 is the area under the graph between these two points. 1 Link to comment Share on other sites More sharing options...
Strange Posted June 28, 2015 Share Posted June 28, 2015 Everything. I took a quick look at integrals and thought it wasn't so hard, but I don't get any of it. I should probably just take my time to learn integrals properly. This is a good intro to calculus: https://www.coursera.org/learn/calculus1 Link to comment Share on other sites More sharing options...
Thorham Posted June 29, 2015 Author Share Posted June 29, 2015 The integral is just the opposite of the derivative so the tutorials on this forum should give you a helping hand. http://www.scienceforums.net/topic/29473-introduction-to-calculus-differentiation/ http://www.scienceforums.net/topic/4108-calculus-i-lesson-1-a-background-to-differentation/ http://www.scienceforums.net/topic/4182-calculus-i-lesson-2-a-continuation-from-first-principles/ This is a good intro to calculus: https://www.coursera.org/learn/calculus1 Thanks for those links Link to comment Share on other sites More sharing options...
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