Answer
Write a Riemann sum and know that the first term blows up if we choose $ x_ {1} ^ {*} $ arbitrarily close to the original.
Work Step by Step
We need to show that the given function $\mathrm{f}$ isn't integrable on the interval [0,1. Use Def. 5.5.1
\[
f(x)=\left\{\begin{array}{ll}
0 & \text { if } x=0\\
\frac{1}{x} & \text { if } x \neq 0
\end{array}\right.
\]
We start with verifying the fact by using Theorem $5.5 .8,$ which says that this function is actually not an integral because it is unlimited at $0=x \in[0,1]$.
Assume we have partitioned the interval [0,1] using $x_{k}$ such that $x_{0}=0$ and $x_{n}=1$
The Riemann sum is given by
\[
A=\lim _{\max \Delta x_{k} \rightarrow 0} \sum_{k=1}^{n} \Delta x_{k} f\left(x_{k}^{*}\right)
\]
where, $x_{k}^{*}$ can be chosen by choice in $\left[x_{k}-1, x_{k}\right]$
Definition 5.5.1 says that this sum, $A$, should be defined (the limit should exist $)$ Consider the first partition $\left[x_{0}, x_{1}\right]=\left[0, \frac{1}{n}\right]$ The area of the corresponding rectangle would be $\frac{1}{n x_{1}^{*}}, \quad x_{1}^{*} \in[0,1]$ It's clear from here that this area is not defined (tends to infinity) if we choose $x_{1}^{*}=0$ no matter how large $n$ becomes. So, the limit and hence $A$ isn't defined. $f$ isn't integrable in [0,1]
