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Integral question

  1. Nov 27, 2011 #1
    If n is a positive integer, prove that ∫ [t] dt = n(n-1)/2 in [0,n]

    Solution:

    sub intervals: [0,1] , [1,2] , [n-1,1]

    each sub interval has base 1 and height 1, 2 , .... n respectively....

    Thus: ∫ [t] dt = Ʃ [t](base) from i = 1 to n = n is base(=1)(1 + 2 + .... + n)

    = 1 + 2 + .... n =/= n(n-1)/2

    some help please...?
     
  2. jcsd
  3. Nov 27, 2011 #2

    Mark44

    Staff: Mentor

    What function does [t] represent? Greatest integer function - [itex]\lfloor t \rfloor[/itex]? Least integer function - [itex]\lceil t \rceil[/itex]?
     
  4. Nov 27, 2011 #3
    Greatest integer
     
  5. Nov 27, 2011 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Your only mistake is in thinking that 1+ 2+ 3+ ...+ n is NOT equal to n(n+1)/2.

    If n= 3, 1+ 2+ 3= 6 and 3(3+1)/2= 6.
    If n= 4, 1+ 2+ 3+ 4= 10 and 4(4+ 1)/2= 10.


    It can be shown that, for all n, 1+ 2+ 3+ ...+ n= n(n+1)/2.

    Here is a proof attributed to Gauss when he was a child:

    Write
    1+ 2+ 3+ ...+ n and below it
    n+ n-1+ n-2+ ...+ 1

    And add each vertical pair. Each pair adds to n+ 1 and there are, of course, n pairs. That total sum is n(n+ 1). But that clearly adds the original sum twice. The sum itself is half of that, n(n+1)/2.

    It is also fairly easy to prove this by induction. When n= 1 we have 1= 1(1+ 1)/2.

    Assume that, for some k, 1+ 2+ 3+ ...+ k= k(k+1)/2.

    Then 1+ 2+ 3+ ...+ k+ k+1= k(k+1)/2+ (k+1)

    Factor k+ 1 out of the right side: (k+1)(k/2+ 1)= (k+1)(k/2+ 2/2)= (k+ 1)(k+ 2)/2 which is just n(n+1)/2 with n= k+1.
     
  6. Nov 27, 2011 #5

    Dick

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    Homework Helper

    1+2+...+(n-1) is n(n-1)/2. What's the value of [t] for n-1<=t<n? Is [n]=n going to change the integral?
     
  7. Nov 27, 2011 #6
    I know of that proof....
    So i am guessing the author is assuming who ever is reading the book already knows that 1+ 2+ 3+ ...+ n is equal to n(n+1)/2?

    I thought that there might be some way to solve [t] and directly get n(n+1)/2, instead of getting 1+ 2+ 3+ ...+ n then adding n+ n-1+ n-2+ ...+ 1 to it...
     
  8. Nov 27, 2011 #7
    I thought of another way of proving it that seems more natural.... please look at it and tell me what you think....

    I created a step function then sections it off into a rectangle and a triangle.....

    1. area of rectangle = (b)(h) = (n)(1)

    2. area of large triangle = (b)(h)/2 = (n)(n)/2

    3. area of small triangle's = n/2


    Area of ordinate satisfying greatest integer function from [0,n] is the sum of these areas....

    (n)(1) + n^2/2 - n/2 = n(n-1)/2..
     

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  9. Nov 27, 2011 #8
    Never mind I just looked at it and realized it does not equal n(n-1)/2
     
  10. Nov 27, 2011 #9

    Dick

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    Homework Helper

    You did the algebra wrong. n+n^2/2-n/2=n*(n+1)/2. But yes, you work it out like that, but I think your graph is wrong. Isn't [t] the greatest integer less than or equal to t? Shouldn't [1/2] be 0? That's not what your graph shows.
     
  11. Nov 27, 2011 #10
    your correct... I realized that it was equal to n*(n+1)/2., which I replied back to saying that... and yes my graph is wrong... thank you.
     
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