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Integration of step functions

  1. Jan 1, 2015 #1
    1. The problem statement, all variables and given/known data

    This is from Apostol's Calculus Vol. 1. Exercise 1.15, problem 6.(c)

    Find all x>0 for which the integral of [t]2 dt from 0 to x = 2(x-1)

    2. Relevant equations

    [t] represents the greatest integer function of t.

    3. The attempt at a solution

    Integral of [t]2 dt from 0 to x = {x(x-1)(2x-1)}/6, which on equating with 2(x-1) gives x(2x-1) = 12 which does not have a rational solution. The answers given at the back of the textbook are 1 and 5/2. Can someone please give me a hint as to where I'm going wrong?
     
  2. jcsd
  3. Jan 1, 2015 #2

    Mark44

    Staff: Mentor

    Your notation really threw me off for a while. I think this is what you're trying to convey:
    $$ \int_0^x \lfloor t \rfloor ^2 dt = 2(x - 1)$$
    No, it isn't. You can't just integrate ##\lfloor t ^2 \rfloor## as if it were the same thing as ##t^2##. Draw a sketch of ##y = \lfloor t ^2 \rfloor## and y = 2(t - 1) and compare the cumulative area under the graph of the integrand with the y values on the straight line graph.
     
  4. Jan 1, 2015 #3
    Yes, that was exactly what I was trying to say :) Sorry, I do not know how to use the definite integral symbol.

    But but I'm supposed to be integrating [t]2 and not [t2] as you pointed out. Won't those two be different?
     
  5. Jan 1, 2015 #4

    Ray Vickson

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    (1) You don't need to know how to do a definite integral sign; you can just say something like int_{t=0..x} [t]^2 dt or int([t]^2 dt, t=0..x). That would be perfectly clear and would avoid the kind of confusion I was victim to when I first tried to read your message. Or you could have said that "the integral of [t]^2 from 0 to x is equal to 2x-1"---that would also have been clear. Saying "...x = 2x-1" is maximally confusing.
    (2) Yes, [t^2] and [t]^2 are very different.
     
  6. Jan 1, 2015 #5

    HallsofIvy

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    This is, as you say, a step function. It's graph is just a series of horizontal lines and its integral, the area under the graph, is just the sum of areas of rectangles. The first thing you should do is draw the graph. If x is from 0 to 1, [x]= 0 so [x]2= 0. If x is from 1 to 2, [x]= 1 so [x]2= 1. If x is from 2 to 3, [x]= 2 so [x]2= 4, etc.
     
  7. Jan 1, 2015 #6

    Stephen Tashi

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    In this problem, the formula for summing the squares of integers is only relevant when [itex] x [/itex] is an integer. And when [itex] x [/itex] is an integer, I think you are "off by 1". For example, as I understand the symbol "[t]^2" in this problem, when [itex] 4 < t < 5 [/itex] , [itex] [t]^2 = [4]^2 = 16 [/itex]. So the area involved in [itex] \int_0^5 [t]^2 dt [/itex] does not include a rectangle with height 25. The areas to be summed are [itex] (1)(0^2)+(1)(1^2) + (1)(2^2) + (1)(3^3) + (1)(4^2) [/itex].

    If the upper limit of integration is between two integers you must sum the squares of the whole rectangles involved in the are and then add the area of the fraction of the rectangle at the end to the sum.
     
    Last edited: Jan 1, 2015
  8. Jan 1, 2015 #7

    Mark44

    Staff: Mentor

    Absolutely, as is the one below.
     
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