Consider the function $f$ defined on $[0, 1]$, where $f(0) = 0$ and, $f(x) = 2^n$, $x \in (2^{−2n}, 2^{−2n+2}], n = 1,2,....$
Compute the area under the curve, from $x = 0$ to $x = 1$. Does this Riemann integral of f exist formally?
I understand that the Riemann Integral can be defined by the Riemann Sum $\sum_{i=0}^n f(c_i)\Delta x_i$ where $c_i$ is any point in the interval $[x_{i-1}, x_i]$ and $\Delta x_i = \dfrac{b - a}{n}$ represents the magnitude of the subintervals.
I also understand that in order for a function $f$ to be integrable it must be continuous over the closed interval $[a, b]$ or have finitely many discontinuities.
However, I do not know how to proceed in solving the problem. I would greatly appreciate it if people could please take the time to assist me in solving this problem and understanding the reasoning behind how to solve it.
Since the length of the intervall $(2^{-2n}, 2^{-2n+2}]$ is $3 \cdot2^{-2n}$ the area under the graph of $f$ on this interval is $2^n \times$ this expression which is $\frac{3}{2^n}$, which implies that the area under the graph is the sum over these expressions ($n=1\ldots\infty$). I would assume you have learned how to calculate this sum, if not do some research on the term geometric series.
If you have learned that, in order to be Riemann integrable (using the definition), the function has to be bounded, which it is not, since $2^n\rightarrow \infty$, which is another reason why this cannot be Riemann integrable. As pointed out in a comment the improper integral will exist, however.