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Compute integrals of some more step functions

  1. Prove

        \[ \int_0^2 [t^2] \, dt = 5 - \sqrt{2} - \sqrt{3}. \]

  2. Compute

        \[ \int_{-3}^3 [t^2] dt. \]


  1. Proof. Let P = \{ 0, 1, \sqrt{2}, \sqrt{3}, 2 \} be a partition of [0,2]. Then [t^2] is constant on the open subintervals of P, so,

        \begin{align*}  \int_0^2 [t^2] \, dt &= \sum_{k=1}^5 s_k \cdot (x_k - x_{k-1}) \\  &= (0 \cdot 1) + (1)(\sqrt{2}-1) + (2)(\sqrt{3}- \sqrt{2}) + (3)(2 - \sqrt{3}) \\  &= 5 - \sqrt{2} - \sqrt{3}. \qquad \blacksquare \end{align*}

  2. We compute,

        \begin{align*}  \int_{-3}^3 [t^2] \, dt &= 2 \cdot \int_0^3 [t^2] \, dt \\  &= 2 \cdot \left( \int_0^2 [t^2] \, dt + \int_2^3 [t^2] \, dt \right) \\  &= 2 \cdot \left( 5 - \sqrt{2} - \sqrt{3} + 4(\sqrt{5}-2) + 5(\sqrt{6} - \sqrt{5}) + 6(\sqrt{7}-\sqrt{6}) \right.\\ & \qquad \quad \left.+ 7(\sqrt{8} - \sqrt{7}) + 8(3 - \sqrt{8}) \right) \\ &= 2 (21 - 3\sqrt{2} - \sqrt{3} - \sqrt{5} - \sqrt{6} - \sqrt{7}). \end{align*}

4 comments

  1. M.d.S says:

    How come does the integral from -3 to 0 equals the one from 0 to 3?
    The function from -1 to 0 equals to -1, while the function from 0 to 1 equals 0 != -(-1)
    from -2 to -1 it is -2, while the positive part is 1 from 1 to 2
    i guess i just didn’t understand

    • Anonymous says:

      Actually the function from 0 to -1 is also equal to 0 ( 0 being the smallest integer between the interval). \int_{-3}^{3} [t^{2}] \ dx = \int_{0}^{3} [t^{2}] \ dx + \int_{-3}^{0} [t^{2}] \ dx = \int_{0}^{3} [t^{2}] \ dx + \int_{0}^{3} [(-t)^{2}] \ dx = 2\int_{0}^{3} [t^{2}] \ dx

    • Anonymous says:

      Don’t forget about the exponent 2, the smallest integer between [(-1)^2,0] is 0. \int_{-3}^{3} [t^{2}] \ dx = \int_{0}^{3} [t^{2}] \ dx + \int_{-3}^{0} [t^{2}] \ dx = \int_{0}^{3} [t^{2}] \ dx + \int_{0}^{3} [(-t)^{2}] \ dx = 2\int_{0}^{3} [t^{2}] \ dx

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