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\title{Math 165: Honors Calculus I\\
Assignment 15 Solution}

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Graph $F(x) = \int_0^x [t]\,dt$ on $[-4,4]$.

\bigskip
\emph{Solution.}
The graph of $[t]$ for $-4 \le t \le 4$:

\[
\resizebox{!}{1.5in}{\includegraphics{floor.eps}}
\]

First, let us assume $x \ge 0$.  Using additivity of intervals and
letting $n=[x]$:
  \begin{eqnarray*}
   F(x) &= &\int_0^x [t]\,dt = \int_0^1 0 \,dt + \int_1^2 1\,dt + \int_2^3 2\,dt
   + \dots + \int_{n-1}^n(n-1)\,dt + \int_n^x n\,dt\\
   &=& 0 +  1 + 2 + \dots + (n-1) + n(x-n)\\
   &=& (n-1)n/2 + nx - n^2 = n(x - (n+1)/2)
  \end{eqnarray*}
Therefore, $F(x) = [x](x-([x]+1)/2)$. Of course, $[t]$ is a step function
and this result could also be obtained using the partition 
$\{0,1,\dots,n,x\}$ and the usual formulas for step functions.

If $x < 0$:
  \begin{eqnarray*}
   \int_0^x [t]\,dt &=& \int_0^{-1}(-1)\,dt + \int_{-1}^{-2}(-2)\,dt +
   \int_{-2}^{-3}(-3)\,dt + \dots + \int_{-(n-2)}^{-(n-1)}(-(n-1))\,dt +
   \int_{-(n-1)}^x(-n)\,dt\\
   &=& (-1)(-1) + (-2)(-1) + (-3)(-1) \dots + (-(n-1))(-1) + (-n)(x+(n-1))\\
   &=& (n-1)n/2 - nx - n(n-1) = n(-x - (n-1)/2)
  \end{eqnarray*}
In this case, $n = -[x]$, so $F(x) = -[x](-x - (-[x]-1)/2 =
[x](x-([x]+1)/2)$ which is the same expression as for the case $x>0$.
Here is the graph of $F(x) = [x](x - ([x]+1)/2)$ for $-4 \le x \le 4$:

\[
\resizebox{!}{2.5in}{\includegraphics{int_floor.eps}}
\]

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