Find some formulas for the integral of the step function [t]^2

by RoRi

July 18, 2015

For $n \in \mathbb{Z}_{>0}$ , prove
$\int_0^n [t]^2 \, dt = \frac{n(n-1)(2n-1)}{6}.$
For $x \in \mathbb{R}$ , with $x \geq 0$ , define
$f(x) = \int_0^x [t]^2 \, dt.$

Draw the graph of $f$ on the interval $[0,3]$ .
Find all real $x > 0$ such that
$\int_0^x [t]^2 \, dt = 2(x-1).$

Proof. Let $P = \{ 0,1, \ldots, n \}$ . Then $P$ is a partition of $[0,n]$ and $[t]^2$ is constant on the open subintervals of $P$ . Further, $[t]^2 = (k-1)^2$ for $k-1 < t < k$ . So,
$\begin{align*} \int_0^n [t]^2 \, dt &= \sum_{k=1}^n (k-1)^2 \cdot (k-(k-1)) \\ &= \sum_{k=1}^n (k-1)^2 \\ &= \sum_{k=0}^{n-1} k^2 \\ &= \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} \\ &= \frac{n(n-1)(2n-1)}{6}. \end{align*}$

The second to last line follows from this exercise (I.4.7, #6) $. \qquad \blacksquare$
The graph is:
By inspection, we have, $x = 1, \frac{5}{2}$ .

Stumbling Robot