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Finding integral using sum of limits?

  1. Dec 21, 2011 #1
    1. The problem statement, all variables and given/known data
    Find [itex]\int^{2}_{1}x^{-2}[/itex]dx. Hint: Choose x[itex]^{*}_{i}[/itex] to be the geometric mean of x[itex]_{i-1}[/itex] and x[itex]_{i}[/itex] (that is, x[itex]^{*}_{i}[/itex] = [itex]\sqrt{x_{i-1}x_{i}}[/itex]) and use the identity

    [itex]\frac{1}{m(m+1)}[/itex] = [itex]\frac{1}{m}[/itex] - [itex]\frac{1}{m+1}[/itex]


    3. The attempt at a solution
    First off I am very confused by x[itex]^{*}_{i}[/itex] and the term geometric mean. Of the 70 questions prior to this question x[itex]^{*}_{i}[/itex] = x[itex]_{i}[/itex] but in this question x[itex]^{*}_{i}[/itex] is the square root of.... Yeah I am confused by that too. Could someone please explain how to go about solving this question? I know I have to use the sum of limits as suggested by my notes, but I am not sure where to start.
     
  2. jcsd
  3. Dec 21, 2011 #2

    Deveno

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    well, first of all, it should be clear that:

    [tex]x_i^* = \sqrt{x_{i-1}x_i} \leq \sqrt{x_i^2} = x_i[/tex] and

    [tex]x_i^* = \sqrt{x_{i-1}x_i} \geq \sqrt{x_{i-1}^2} = x_{i-1}[/tex]

    since everything in sight is greater than or equal to 1. so [itex]x_i^*[/itex] is indeed "some" point in [itex][x_{i-1},x_i][/itex], as taking a riemann sum requires.

    this particular [itex]x_i^*[/itex] is chosen for you as a kindness, to make it easier to evaluate [itex]f(x_i^*)[/itex]. so evaluate [itex]f(x_i^*)[/itex], and use the identity provided.
     
  4. Dec 21, 2011 #3
    Oh I see, so you don't have to take the right, left or mid points to find the integral you can use any point?
     
  5. Dec 21, 2011 #4

    Deveno

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    if a function is integrable, than ANY riemann sum will converge to the value of the integral. this is basic.

    now, proving that a given function is actually integrable, can be an arduous process. however, in your class, you should have at least been told (and hopefully proved) that continuous functions, at least, are integrable (there are also a lot of non-continuous functions that are integrable, too...most notably "step functions").

    since f(x) = 1/x2 is indeed continuous on the interval [1,2], you're good to go.

    (the basic idea, is that no matter which point in our subinterval we pick, and no matter how we come up with a partition of [a,b], when the mesh size (the length of the largest subinterval) approaches 0, all the points in that subinterval get "closer together". for a continuous function, this means the values of f(x) on that subinterval also get closer together, so "getting the wrong one" isn't going to affect the sum very much. as a matter of practicality, if one actually has to (for some reason) find a bona-fide riemann sum, it is often convenient to choose a REGULAR partition, and choose either

    a) an endpoint
    b) the mid-point

    of the sub-intervals. but if some other point in the (sub-)interval is more convenient for a particular function, yes, we can use that one instead. that's why there's a star on xi, to indicate that it is arbitrarily chosen (so any method of choosing one, is as good as any other...but maybe the calculations aren't so nice).).
     
  6. Nov 22, 2012 #5
    I was refreshing myself on calculus earlier and came across this problem in a textbook. If we consider that x0 = 1 and x1* = √[1 * (1 + 1/n)] or equivalently √[(1 + 0/n)(1 + 1/n)], this may help you find that, in general, xi* = √[(1 + (i-1)/n)(1 + i/n)].

    Expand it and simplify the result and you will eventually reach a point where you can apply that identity and divide the resultant summation into a difference of summations. Then carefully analyze the only distinction between the summations. Write out the summations explicitly if required. You will find that almost everything cancels except two terms. This will be your final answer.
     
    Last edited: Nov 22, 2012
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