Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tell me about these basic powefull rules of calculas.

  1. Sep 3, 2011 #1
    There are several rules in calculus which have far reaching applications in field of mathematics, physics etc. unfortunately there is no derivation or visualisation of those rules in my maths book. SO SHAME!!.
    one of these is that definite integral of function from x1 to x2 gives geometrical area under the curve. There is no well proof of this statement in any book i read. Now I am asking this to you tell me where does it came. why the following statement is true
    f(x1)+f(x1+dx)+f(x1+2*dx)+f(x1+3*dx)+...............f(x2)=integral from x1 to x2 of function f(x).


    On more thing is integration by parts this rule is something like this
    integral_of {f(x).g(x)}=f(x)* integral_of {g(x)} - integral_of{ (differentiation_of{g(x)} integral_of{f(x)} }.dx
    let me say integral_of g(x) = h(x)+C
    When we solve any question with help f this rule (ex tan inverse x) then we put integral_of(g(x))= h(x) we simply did not think about C. But why..




    these are two basic question i want to ask hope that unlike other places i will get answer here.
     
  2. jcsd
  3. Sep 3, 2011 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The Riemann sums don't equal the area, they converge to the area as they get finer. The actual proof uses concepts that are likely to be a bit advanced for you given that you are in high school. Here's a link if you wish to see what is involved:

    http://www.google.com/url?sa=t&sour...XP04hLrtw&sig2=2wCWcDU_yWxr_IvygojKGw&cad=rja

    If you try putting the constant of integration during the appropriate step in integration by parts, you don't get a different answer than if you just wait to add the constant at the end. Try it with the integral of xex
     
  4. Sep 3, 2011 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I'm surprised at what vkash is saying. Every Calculus text I have ever seen has included proofs that the area under the curve y= f(x), from x= a to x= b, is equal to an anti-derivative of f, evaluated at b, minus an anti-derivative evaluated at x= a.
     
  5. Sep 3, 2011 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, I will take that back. Once (and only once) I taught a course title "Calculus for Business Administration" for the Business Admin school, using a text they had chosen (titled "Calculus for Economics and Business Admininstration") which was really horrible! That text did not give good proofs of anything!

    As an example, on one page they stated (without proof) the basic "laws of limits":
    a) if lim f= F and lim g= G, then lim (f+ g)= F+ G.
    b) if lim f= F and lim g= G, then lim(fg)= FG.
    c) if lim f= F and lim g= G, and g does not go to 0, then lim(f/g)= F/G.

    On the very next page they defined the derivative as [itex]\lim_{h\to 0} (f(x+h)- f(x))/h[/itex] completely ignoring the fact that those laws do NOT apply because the denominator necessarily goes to 0.
     
  6. Sep 3, 2011 #5
    I am 100% sure that you are not from my Land INDIA. here in starting when calculus is started in schools no such proofs.

    one more thing you write here is also in my question category, that is sometimes these rules of limit (you mentioned in last post) does not work why???????
    when i ask this to my maths teacher he says that, you can't apply these rules when form of limit is changed like if 0/0 or 0*inf form converted into some other form then we can't apply these rules. I did not able to got it's reason can you tell me it's reason.

    what you say in last two lines. an you plz explain that once again.

    I have read first 2 pages in LCKurtz's pdf file. still not got answer of first one.
     
    Last edited: Sep 3, 2011
  7. Sep 3, 2011 #6
    I think I have found think like proof on this page. I understood all the thing at the third point.that goona sandwiched in x1 and x1+dx.

    however this proof uses definite integral as area to proof that definite integral is area.
    Let me say that this is wrong then we are using wrong thing and it gives one more wrong thing. that rule is just after point 1.
    hope you understood.
     
  8. Sep 3, 2011 #7

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The proof you are looking for is on pages 186-190. But as I said before, it is a bit much to be expect to be able to understand that proof if you have only had high-school mathematics courses.
     
  9. Sep 4, 2011 #8

    Stephen Tashi

    User Avatar
    Science Advisor

    One way to look at the matter is to say that the area under the curve is defined as the definite integral rather than say we can prove the definite integral is the area.

    Suppose we assume the formulas for the areas for simple figures such as rectangles and triangles are correct. If we are given a figure which can be exactly divided up into such figures, there is no doubt in our minds that we can compute its area. The total area is the sum of the areas of the simple figures. However, suppose we consider the area of a circle. It is not exactly divisible into non overlapping rectangles and triangles. The argument that the area of a circle if [itex] \pi r^2 [/itex] is based on looking at how the area of a large number of triangles that approximately conform to the shape of the circle becomes nearer and nearer [itex] \pi r^2 [/itex] as the number of triangles increases.

    One may take the view that the area of the circle existed before we discovered how to use the triangles to compute it. Or one may take take the view that it is the process of using the many triangles that defines what the area must be and proves that the area of a circle can be defined.

    You left out a factor of dx in that expression. Even with that factor, the two sides of that equation aren't exactly equal.

    On a graph of the curve, divide the interval from [itex] x_1 [/itex] to [itex] x_2 [/itex] in N equal intervals of size [itex] dx = \frac{(x_2 - x_1)}{N} [/itex]

    The points that are the endpoints of these intervals are [itex]\{x1. x1+dx,x1+2dx,...x_2\}[/itex]. Let each one of the these intervals be the base of the rectangle. Let the height of the rectangle be the function [itex] f [/itex] evaluated at the left endpoint of the interval. The sum of the areas of these rectangles is:
    [tex] f(x_1)dx + f(x+dx)dx + f(s + 2 dx)dx + ....f(x + (n-1)dx) dx [/tex] (So the sum doesn't include [itex] f(x_2) dx [/itex] like yours did.)

    You can factor out [itex] dx [/itex] from all the terms and get an expression resembling the one you wrote. The sum is not the exact area under the curve. The sum merely indicates an approximation to the area. The assertion of calculus is that as we use more and more intervals (make N larger and thus make dx smaller) the sum approaches the area under the curve. In the notation for the integral [itex] \int_{x_1}^{x_2} f(x) dx [/itex], the [itex] dx [/itex] no longer represents a number. It is just a reminder that [itex] x [/itex] is the independent variable.

    A more technical discussion of the above sum is found by looking up "Riemann Sum", for example: http://en.wikipedia.org/wiki/Riemann_sum
     
  10. Sep 4, 2011 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Those rules of limits cannot be used to find the derivative because, as I said, the quotient rule does not apply if the denominator is going to 0- which always happens for the derivative.

    You need one more property. Remember that the definition of limit says
    "[itex]\lim{x\to a}f(x)= L[/itex] if and only if, given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]0< |x- a|< \delta[/itex], [itex]|f(x)- L|< \epsilon[/itex]".
    Do you see that "[itex]0< |x- a|[/itex]"? The value of the function at x= a is irrelevant to the limit! If f(x)= g(x) for all x except x= a, the [itex]\lim_{x\to a}f(x)= \lim_{x\to b}g(x)[/itex] no matter what f(a) and g(a) happen to be.

    Yes, that is that "additional rule" I mention above. For example, to find
    [tex]\lim_{x\to 2}\frac{x^2- 4}{x- 2}[/tex]
    we cannot just let x= 2 because we get 0/0 (the function is not continuous at x= 2). But we can say that, for all [itex]x\ne 2[/itex],
    [tex]\frac{x^2- 4}{x- 2}= \frac{(x- 2)(x+ 2)}{x- 2}= x+ 2[/tex]
    (Notice at x= 2, we would be dividing by 0 which is invalid. That is why I say that is true for [itex]x\ne 2[/itex].)

    But the two functions, [itex](x^2- 4)/(x- 2)[/itex] and [itex]x+ 2[/itex], have the same value for all x except x= 2 so they have the same limit as x goes to 2.

    The definition of derivative of f, at x= a is
    [tex]\lim_{h\to 0}\frac{f(a+h)- f(a)}{h}[/tex]
    My point was that, although that is a quotient, we cannot use the quotient rule:
    "If [itex]\lim_{x\to a}f(x)= F[/itex] and [itex]\lim_{x\to a}g(x)= G[/itex], and G is not 0, then [itex]\lim_{x\to a} f(x)/g(x)= F/G[/itex]".
    We cannot use that because it violates the condtion "G is not 0".

    Suppose we have [itex](x)= x^2[/itex], to find the derivative at x= a, we would need to calculate
    [tex]\lim_{h\to 0}\frac{(a+ h)^2- a^2}{h}= \lim_{h\to 0}\frac{a^2+ 2ah+ h^2- a^2}{h}= \lim_{h\to 0}\frac{2ah+ h^2}{h}[/tex]
    and we cannot apply the "quotient rule" precisely because the denominator, h, is going to 0.

    However, we can apply my "additional" rule. For all h except 0,
    [tex]\frac{2ah+ h^2}{h}= \frac{h(2a+ h)}{h}= 2a+ h[/tex]
    And so the limit, as x goes to 0, is 2a.
     
  11. Sep 4, 2011 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    These are questions about Calculus, NOT "Differential Equations" so I am moving this thread.
     
  12. Sep 4, 2011 #11
    The Sum f(x1)+f(x1+dx)+f(x1+2*dx)+f(x1+3*dx)+...............f(x2)=integral from x1 to x2 of function f(x).
    is a way to tell you that as you reduce the the change [\Dellta x] to zero we refine the interval [$x_{1}<x<x_{2}$] and with the requirement that f is continuous or possibly smooth enough, the sum is equal to the area under f. What is more, it is as if we sum up the areas of tiny rectangles over the interval [$(x_{1},x_{2})$].

    In the second question, the integration constant need not be repeated, you only need one. This is because sum of arbitrary constants is still an arbitrary constant.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Tell me about these basic powefull rules of calculas.
Loading...