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Math Help - Can't Find My Mistake. Area between Function and its Floor Function.

  1. #1
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    Can't Find My Mistake. Area between Function and its Floor Function.

    Greetings everyone. For the following problem, I think I got the idea but I cannot find the mistake that I made. I know there is one because my answer is  k + \frac{5}{6} which is NOT
     k + \frac{11}{6}. I am also trying to get this right before I sum each representative area between function and the floor function.

    Thank you everyone.



    My work

    \text{For }x\in (k,k-1)\text{ on }x\text{-axis, area = }\int_{k-1}^{k}{{{x}^{2}}+3x-1}\text{ }dx-1\cdot {{(k-1)}^{2}}+3(k-1)-1

    = \left[ \frac{{{x}^{3}}}{3}+\frac{3}{2}{{x}^{2}}-x \right]_{k-1}^{k}-1\cdot {{(k-1)}^{2}}+3(k-1)-1

    = \frac{{{k}^{3}}-{{k}^{3}}+3{{k}^{2}}-3k+1}{3}+\frac{3{{k}^{2}}-5{{k}^{2}}+10k-5}{2}-k+(k-1)-3\left( k-1 \right)+1

    = k+\frac{1}{3}-\frac{5}{2}+3=k+\frac{5}{6} <br />
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  2. #2
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    Re: Can't Find My Mistake. Area between Function and its Floor Function.

    You should take interval (k,k+1) because you end up with summing up k+\frac{11}{6} for k=1 to n-1.
    If you take interval (k-1,k) then for k=1 this is interval (0,1) but you need to find integral from 1 to n, not from 0 to n.
    So, take interval (k,k+1).
    A=\int_k^{k+1}(x^2+3x-1)dx-(k^3+3k-1)

    Also notice that your answer is k+\frac56=(k+\frac{11}{6})-1.
    This is because you chose starting point to be 0, not 1.
    Value of integral (using your formula) on interval (1-1=0,1) is 0+\frac56, on interval (2-1=1,2) is 2+\frac56=1+\frac{11}6 which is exactly same answer for interval (1,1+1) using required formula.
    It is just a question of shifting.
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  3. #3
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    Re: Can't Find My Mistake. Area between Function and its Floor Function.

    A really simple solution is just to evaluate

    \int_{1}^{n} x^2 + 3x - 1 \, dx - \sum_{i=1}^{n-1} i^2 + 3i - 1

    Note that for the summation, i ranges from 1 to n-1 (because we are dealing with floor values).

    =(\frac{1}{3}n^3 + \frac{3}{2}n^2 - n) - (\frac{1}{3} + \frac{3}{2} - 1) - \sum_{i=1}^{n-1} i^2 + 3i - 1

     = \frac{1}{3}n^3 + \frac{3}{2}n^2 - n - \frac{5}{6} - \sum_{i=1}^{n-1} i^2 + 3i - 1.

    To evaluate \sum_{i=1}^{n-1} i^2 + 3i - 1 we split it up to \sum_{i=1}^{n-1} i^2 + 3 \sum_{i=1}^{n-1} i - \sum_{i=1}^{n-1} 1

    =\frac{(n-1)(n)(2n-1)}{6} + 3\frac{n(n-1)}{2} - (n-1)

     = \frac{1}{6} (n(n-1)(2n-1) + 9n(n-1) - 6(n-1)) and take it on from there.
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  4. #4
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    Re: Can't Find My Mistake. Area between Function and its Floor Function.

    Thank youv ery much! I got it!
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