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More integrals of step functions

  1. Compute

        \[ \int_0^9 \left[ \sqrt{t} \right] \, dt. \]

  2. For n \in \mathbb{Z}_{>0} prove

        \[ \int_0^n \left[ \sqrt{t} \right] \, dt = \frac{n(n-1)(4n+1)}{6}. \]


  1. We compute,

        \[ \int_0^9 \left[ \sqrt{t} \right] \, dt = (0 \cdot 1) + (1 \cdot (4-1)) + (2 \cdot (9-4)) = 13.  \]

  2. Proof. Let P = \{ 0,1,4,9, \ldots, n^2 \}. Then P is a partition of [0,n^2] and \left[ \sqrt{t} \right] is constant on the open subintervals of P. Further, for (k-1)^2 < t < k^2 we have \left[ \sqrt{t} \right] = (k-1). Thus, we have (using this exercise and this exercise to evaluate some of the sums),

        \begin{align*}  \int_0^{n^2} \left[ \sqrt{t} \right] \, dt &= \sum_{k=1}^n (k-1) \cdot (k^2 - (k-1)^2) \\ &= \sum_{k=1}^n (k-1) \cdot (2k-1) \\ &= \sum_{k=1}^n (2k^2 - 3k + 1 ) \\ &= 2 \sum_{k=1}^n k^2 - 3 \sum_{k=1}^n k + \sum_{k=1}^n 1 \\ &= \frac{2n^3}{3} + n^2 + \frac{n}{3} - \frac{3n^2}{2} - \frac{3n}{2} + n \\ &= \frac{4n^3 - 3n^2 - n}{6} \\ &= \frac{n(n-1)(4n+1)}{6}. \qquad \blacksquare \end{align*}

One comment

  1. mrseb says:

    You are helping me a lot with the “proof writing”, thanks a lot.

    Just one little minor thing: in point b there is a typo , is n**2 instead of n.
    Thanks again! :)

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