# Show that $f$ is integrable using Archimedes-Riemann theorem

Define $$f(x)=\begin{cases} x, 2 \leq x \leq 3 \\ 2, 3 < x \leq 4 \end{cases}$$ Show that $$f$$ is integrable on $$[2,4]$$ using Archimedes-Riemann theorem (Hint: find an Archimedian sequence of partitions that all have 3 as a partition point).

My attempt: I tried to use the regular partition, but it skips the point 3, so I'm thinking of a partition with $$x_i-x_{i-1}=\frac{1}{n}.$$ Then $$m_i=\inf\{f(x)|x in [x_{i-1},x_i]\}=x_{i-1}, i =1,..., n, m_i = 2, i = n+1,...,2n.$$ $$M_i=\sup\{f(x)|x in [x_{i-1},x_i]\}=x_{i}, i =1,..., n, M_i = 2, i = n+1,...,2n.$$ Therefore, $$L(f,P_n)=\sum_{i=1}^{2n}m_i(x_i-x_{i-1})=2*\frac{1}{n}+(2+\frac{1}{n})*\frac{1}{n}+...+(2+\frac{n-1}{n})*\frac{1}{n}+2*\frac{1}{n}+...+2*\frac{1}{n}=\frac{9n-1}{2n}.$$ Similarly, $$U(f,P_n)=\sum_{i=1}^{2n}M_i(x_i-x_{i-1})=\frac{9n+3}{2n}.$$

$$\lim_{n \rightarrow \infty} L(f,P_n)=\lim_{n \rightarrow \infty} U(f,P_n)=\frac{9}{2},$$ so $$f$$ is integrable.

Is it correct? I noticed that what I've done actually is showing that $$x$$ is integrable on subinterval $$[2,3]$$ and 2 is integrable on $$[3,4]$$, so I'm kind of confused and I suppose this is not the right approach. Maybe I should've chosen another sequence of partitions... Correct me if I'm wrong.

• I've never heard of an "Archimedes-Riemann" theorem - I suspect the terminology is unique to your textbook. What exactly does it say? Mar 3, 2019 at 6:11
• @jmerry . Ditto for me. Riemann and Archimedes were about 21 centuries apart. Mar 3, 2019 at 7:36

I noticed that what I've done actually is showing that $$x$$ is integrable on subinterval $$[2,3]$$ and $$2$$ is integrable on $$[3,4]$$, so I'm kind of confused and I suppose this is not the right approach.
Theorem: Given $$a, a function $$f$$ is (Riemann) integrable on $$[a,c]$$ if and only if it is integrable on both $$[a,b]$$ and $$[b,c]$$, and $$\int_a^c f = \int_a^b f+\int_b^c f$$.
One quibble: $$M_{n+1}=\sup\{f(x)|x\in [x_n,x_{n+1}]\}$$ is equal to $$3$$, not $$2$$. After all, the lower endpoint of that interval is $$3$$, and $$f(3)=3$$. Fortunately, that only applies to a single subinterval, with a width that goes to zero.