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Prove an identity for the sum of integrals of particular step functions

Prove

    \[ \int_a^b [x] \, dx + \int_a^b [-x] \, dx = a-b. \]


Proof. From this exercise (1.11 #4, part b) we know

    \[ [-x] = \begin{cases} -[x] & \text{if } x \in \mathbb{Z} \\ -[x]-1 & \text{if } x \notin \mathbb{Z}. \end{cases} \]

But, since [x] is constant on the open subintervals of the partition

    \[ P = \{ a, [a] + 1, \ldots, [a] + [b-a], b \} \]

which contains every integer between a and b, we have [-x] = -[x]-1 on the open subintervals of P (since there are no integers in the open subintervals). Hence, \int_a^b [-x] \, dx = \int_a^b (-[x] -1)\, dx. Thus,

    \begin{align*}  \int_a^b [x] \, dx + \int_a^b [-x] \, dx &= \int_a^b [x] \, dx + \int_a^b -[x]-1 \, dx \\ &= \int_a^b -1 \, dx \\ &= \int_b^a 1 \, dx & (\text{let } P = \{ a,b \})\\ &= a-b. \qquad \blacksquare \end{align*}

One comment

  1. Trevor Garrity says:

    The partition does not contain all integers. For example, let a = 1.9 and b =3.8. [a] = 1, b-a = 1.9, and [b-a] = 1. Therefore, your proposed partition is 1.9, 2, 3.8. This is missing the integer 3.

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