Let's say I have a floor function that I want to integrate, $$\int_{-1}^{3} \left[ x + \dfrac{1}{2} \right] \; dx$$

and I graphed it as shown below, enter image description here

Would it be correct to evaluate it saying that for example from $-1$ to $-0.5$ the step function is $-1$ looking at the $y$-axis? and from $2.5$ to $3$ it is equal to $3$? I have tried that and I got a correct answer, but I don't know if this is definitely correct or it was just a coincidence for this particular floor function. I know I can solve this by finding the area under each graph, but I just want to know if this method that I mentioned above is correct or not?

Solving this another method, I got for example that from $-1$ to $0$, the step function is equal to $-\dfrac{1}{2}$, which I also noticed that from the graph from $-1$ to $0$, the open circle lies at $-0.5$, is that a correct way of solving? by looking at each open circle and where it lies?


1 Answer 1


Yes, this is correct. You are using $$\int_{-1}^{3} f(x) \; dx = \int_{-1}^{-\frac{1}{2}} f(x) \; dx + \int_{-\frac{1}{2}}^{\frac{1}{2}}f(x) \; dx + \int_{\frac{1}{2}}^{\frac{3}{2}} f(x) \; dx + \int_{\frac{3}{2}}^{\frac{5}{2}} f(x) \; dx + \int_{\frac{5}{2}}^{3} f(x) \; dx$$ and the fact that your function is constant on these intervals.

  • $\begingroup$ And is it also correct to say that from $-1$ to $0$ the step function is $-\dfrac{1}{2}$? Does that has to do with the open circle in the graph? $\endgroup$
    – Friedrich
    Feb 1, 2019 at 11:17
  • 1
    $\begingroup$ No, the step function is $-1$ on $[-1,\frac{1}{2})$ and $0$ on $[-\frac{1}{2},0]$. You are thinking of the average, which is a different function, but integrates (by definition) to the same value. The open circle indicates that the point is excluded, e.g. $f(-\frac{1}{2}) = 0$ and not $-1$. $\endgroup$
    – Klaus
    Feb 1, 2019 at 11:22
  • $\begingroup$ I used $m \leq x+0.5 < m+1$, and I got $-1 \leq x < 0$ and $m=-\dfrac{1}{2}$... so that is what is meant by the average? Thank you for clarifying $\endgroup$
    – Friedrich
    Feb 1, 2019 at 11:27

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