I'm stuck trying to calculate the area of the region defined by:
$\begin{Bmatrix} {(x,y)\in\mathbb{R^2}:0\leq x\wedge 0\leq y\leq 2^{-\lfloor x \rfloor}} \end{Bmatrix}$
I'm just starting studying Calculus I and I don't have much information to give: I could do this exercise if it wasn't for $x\leq0$, however I can't calculate the improper integral because I can't integrate the floor function and I though about using an infinite sum (I don't know if that doesn't make sense) but we haven't started learning it yet so I assume that ain't it.
If we plot y against x, $[x]$ is the floor function with steps going upward. However the function $y = 2^{-[x]}$ has steps going downward with each step half of the height before it. Hence if $N <= x < N + 1$ we have area equal to
$$\int2^{-[x]}dx = 1 + \frac{1}{2} + \frac{1}{2^2} + ... + \frac{1}{2^{(N - 1)}} + \frac{1}{2^N}(x - N)$$
$$= \frac{1 - \frac{1}{2^N}}{1 - \frac{1}{2}} + \frac{1}{2^N}(x - N) = 2 - \frac{1}{2^N}(2 - (x - N))$$
$$= 2 - \frac{1}{2^{[x]}}(2 - x + [x])$$
Since N = [x].