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Integration from First Principles

  1. Oct 24, 2015 #1
    Hello,

    This is not a homework problem, but a worked example I encountered from Hobson and Riley 3e pg 60, if anyone has the book. If not I took a screen shot. [I actually just decided to post the photos on a blog so no one has to download anything]. http://justinphysicsforums.blogspot.com/2015/10/integration-from-first-principals.html

    Okay here's the problem, they go ahead and give a formula for integration from first principals, and they make each rectangle an equal width h, and then they go to say without explanation that the Kth rectangle has an area of
    (kh)^2h = k^2h^3

    This is the first thing that makes no sense to me.

    The second part that doesn't make sense to me is how they change the index in the summation and then I haven't thought too much about anything that follows. They give this formula for the Sum (before taking the limit as n tends towards infinity)

    I am having a hard time writing the formula, I am just going to include a picture.

    So essentially they are using this formula in the specific example shown in the picture and then I'm just lost as to what they actually did mathematically. I could go back to an easier calculus book and understand this but it would be nice to understand what mathematics they are actually using, because this book gets more and more advanced (I am using it to get ready for Lagrangian and Hamiltonian mechanics as well as quantum mechanics next semester, studying independently).

    Any help would be appreciated very much, I am just going to go on, I hope someone can sort of break this down for me because I know calculus and this isn't the easiest way to write this at all, it's making me question this book, I just wanted to review calculus quickly before getting into other things.

    Alright thanks,
    Justin
     
  2. jcsd
  3. Oct 24, 2015 #2

    SammyS

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    What is the x-coordinate for the kth rectangle?

    The height of the rectangle is simply f(x) for that x value.

    What is the width of that rectangle?
     
  4. Oct 24, 2015 #3
    I'll upload a descriptive graph as to how they get the area of the kth block tomorrow when i have time to draw it out. What you have to know is, say b=3, so your integrating from x=0 to 3. Say you divide it into 12 rectangles, so n=12 and h=0.25. Area=height*width.....in this case the width is "h" and the height is the solution to y=x2 at the kth rectangle. The reason the height is equal to kh is because you have to multiply the solution at k by the scaling factor of h. for example at the 4th rectangle, the height would be 1 because the lower end of the 4th rectangle lies at x=1. But if you plug k=4 (for the 4th height) then y(4)=42 which is 16. So, you have to multiply by the integer division of h, then y(4)=(kh)2 = (4*0.25)2 = 1. Now it is correct. So with the height given as y evaluated at x=kh, then area becomes width*height, where the height is given by making x=kh. Substituting, you get A=(kh)2h = k2h3. Its easier to understand if you draw out the graph of the equation and draw the dividing rectangles you can more easily see how x=kh. If the width was 1, where h=1, then the width would be equal to the unit size of x, and the area would simply be x2h. I hope this helps some, draw out the graph and I believe you will be able to see how this was set up more easily.

    To me when trying to solve a general case of all variables, its easier for me to make up numbers for each variable, do the equations with the numbers and make sure i get the right answer. Then, i can transition into leaving it arbitrary and using just variables. You know area equals width * height, all you need to see is how x is represented by kh as the height
     
  5. Oct 24, 2015 #4
    It is a great example you've given me. I tried to draw out your graph exactly as well as h equaling 0.25, and I'm still not getting it. I've got to put it away for now, this is very frustrating.

    Considering it more, I understand the need to multiply by h in order to get the correct value for y at the left end point of that rectangle, I'm just not fully following it as I go into a generalization. Perhaps it is because I haven't taken intro to analysis yet. I don't ever even recall doing a problem like this in my calculus class. I do not fully understand it but you've convinced me it has to be right. Fine showing.

    I am going to move onto the next part and I already know I do not know why the summation index switches and how I can understand that, but I am going to work on as much as I can. Would it be alright to get back to you after I mess around with it a little bit? I'm not very good at math which was why I chose this book that is a lot harder than my calculus, linear algebra, or other math methods books, to sort of try to understand some of the deeper math. This book doesn't offer much by way of explanation sometimes though.
     
  6. Oct 25, 2015 #5

    SammyS

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    Maybe this link will help. https://en.wikipedia.org/wiki/Riemann_sum

    You might find it helpful to look up Riemann Sum in most any standard Calculus textbook.
     
  7. Oct 25, 2015 #6
    In reality, this is just showing you how an integral works and is the proof behind it as well. The summation is adding all small rectangles to approximate the area under the curve. Not fully understanding the proof will not hinder you in higher up math courses, because you will never see this type of setup again. Once you pass this and get into actual integration; it takes the place of all this crap. It is very confusing, but integration is much easier (not to say that it's easy, but easier to understand than these proofs). When you take n to infinity, the summation becomes an integral. Once you reach integration, you no longer have to do this kind of setup. There are a few basic rules to learn, same as differentiation, and it's more simple. Proof's are always confusing, I never understood why in math they show you the most complicated way to do something, then a week later you're shown the short, easy way of doing the same thing and never use the long way again lol. But, I guess its necessary to know where the theories come from.

    The book you are looking at seems to be a bit out of the ordinary in the way of introducing this proof, it is a different approach than my calculus book. I always found summations to be confusing anyway. Integration is easier to understand to me.
     
  8. Oct 25, 2015 #7
    Right. I understand integration intuitively, and I'm sure if we were sitting down together you could break it down for me. Your example was really solid and I drew your graph exactly and convinced myself it had to be what it is. The explanation in the book is definitely a bit strange, and I went back to my intro calculus book and it's much more simple in there. I'm not too sure about this book, the reason I liked it is because it's succinct, and right now I'm not interested in the existence of limits or derivatives and such, I'm just looking for a quick review involving 'physical' functions (continuous and differentiable, that you can take for granted things exist). This book is a bit tough though, the linear algebra part gets pretty wild. I had to just move past this part because I need to review single and multi variable calculus, vector calculus, linear algebra, Fourier analysis, and ordinary and partial differential equations for the quantum mechanics and Lagrangian mechanics I am taking next semester.

    Thanks a lot for the help, much appreciated, and Sammy, you as well.

    I'm sure this book will become even more convoluted as I go so you guys have not heard the last of me! I am probably going to switch over to a different math methods book because I only have until 12-1-15 to get through everything I mentioned above, I need to move quickly. Fortunately I've done all the stuff before so it should come back fairly quickly.

    Alright thanks again!
     
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