Integral of a floor function

Question

Let $f(x) = \lfloor 1-x^2 \rfloor$ with $x \in [-2,2]$. Calculate:

$$F(x) = \int_{-2}^{x}f(t)dt$$

I know that:

$$f(x) = \begin{cases} -3 & : x \in [-2,-\sqrt{3})\\ -2 & : x \in [-\sqrt{3},-\sqrt{2})\\ -1 & : x \in [-\sqrt{2},-1)\\ 0 & : x\in [-1,1) \\ 1 & : x \in [1,\sqrt{2} )\\ 2 & : x \in [\sqrt{2}, \sqrt{3})\\ 3 & : x \in [\sqrt{3}, 2)\\ 4 & : x = 2 \end{cases}$$

But I don't remember how to integrate that function.

I remember that the result is a stepwise function, not of this kind.

See this technique. Do not forget to up vote the answers if you benefit from them. — Mhenni Benghorbal, Apr 27 '13 at 04:07

score 5 · Accepted Answer · answered Apr 27 '13 at 01:42

You’re just adding signed areas of rectangles.

Example: If $x\in[-\sqrt2,-1)$, then

$$F(x)=(-3)\Big(-\sqrt3-(-2)\Big)+(-2)\Big(-\sqrt2-(-\sqrt3)\Big)+(-1)\Big(x-(-\sqrt2)\Big)\;.$$

Simplify, and do the analogous thing for the entire interval on which $F$ is defined.

Integral of a floor function

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