0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I've never heard of an "Archimedes-Riemann" theorem - I suspect the terminology is unique to your textbook. What exactly does it say? $\endgroup$ – jmerry Mar 3 at 6:11
  • $\begingroup$ @jmerry . Ditto for me. Riemann and Archimedes were about 21 centuries apart. $\endgroup$ – DanielWainfleet Mar 3 at 7:36
0
$\begingroup$

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.

Well, actually, that's exactly what's going on. That's the function we have.
Theorem: Given $a<b<c$, a function $f$ is (Riemann) integrable on $[a,c]$ if and only if it is integrable on both $[a,b]$ and $[b,c]$, and $\int_a^c f = \int_a^b f+\int_b^c f$.

And when we have a piecewise defined function, with multiple formulas? The standard practice to evaluate it is to carve up the interval so as to deal with one formula at a time. Is it any surprise we can do this even when we go down to the definitions?

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.