# Thread: Real Analysis: Riemann Integrable

1. ## Real Analysis: Riemann Integrable

Here is the question:

How do I do this problem?? Thank you!!

2. Originally Posted by Phyxius117
Here is the question:

How do I do this problem?? Thank you!!

Is $k$ also a natural number?

Is $n$ a fixed natural number, or does ${k\over n}$ represent all rational numbers on the interval?

3. Originally Posted by Phyxius117
Here is the question:

How do I do this problem?? Thank you!!
There is a very general theorem which does away with this, but assuming that you don't know this theorem here's what I would do:

Start by considering $\varepsilon<1$. Consider then the intervals

$I_k=\begin{cases}\left[\frac{k}{n}-\frac{\varepsilon}{64n},\frac{k}{n}+\frac{\varepsi lon}{64n}\right] & \mbox{if}\quad k=1,\cdots,2n-1\\ \left[2-\frac{\varepsilon}{64n},2] & \mbox{if}\quad k=2n\end{cases}$

Notice then that if

$J_k=\begin{cases}[0,\frac{1}{n}-\frac{\varepsilon}{64n}] & \mbox{if}\quad k=1\\ [\frac{k-1}{n}+\frac{\varepsilon}{64n},\frac{k}{n}-\frac{\varepsilon}{64n}] & \mbox{if}\quad k=2,\cdots,2n\end{cases}$

Then $[0,2]=I_1\cup\cdots\cup I_{2n}\cup J_1\cup\cdots\cup J_{2n}$. Note though that $f_{\mid J_k}=x^3,\quad k=1,\cdots,2n$ and so by continuity of $x^3$ on $J_1,\cdots,J_{2n}$ we may find partitions $P_1,\cdots,P_{2n}$ such that $U(P_k,f_{\mid J_k})-L(P_k,f_{\mid J_k})<\frac{\varepsilon}{4n}$. Note next that $P=P_1\cup\cdots\cup P_{2n}$ is a partition for $[0,2]$ and

$\displaystyle U(P,f)-L(P,f)=\underbrace{\sum_{k=1}^{2n}\left(U(P_k,f_{\ mid J_k})-L(P_k,f_{\mid J_k})\right)}_{\mathbf{(1)}}+\underbrace{\frac{\va repsilon}{32n}\sum_{k=1}^{n}\left(\sup_{x\in I_k}f(x)-\inf_{x\in I_k}f(x)\right)}_{\mathbf{(2)}}$

Note though that by construction $\mathbf{(1)}=2n\cdot \frac{\varepsilon}{4n}=\frac{\varepsilon}{2}$ and ,since $\displaystyle \inf_{x\in I_k}f(x)=0$,

$\displaystyle \mathbf{(2)}=\frac{\varepsilon}{32n}\sum_{k=1}^{n} \sup_{x\in I_k}f(x)\leqslant \frac{\varepsilon}{32n}\sum_{k=1}^{2n}8=\frac{\var epsilon}{2}$

And thus $U(P,f)-L(P,f)=\mathbf{(1)}+\mathbf{(2)}<\frac{\varepsilon }{2}+\frac{\varepsilon}{2}=\varepsilon$ and since $\varepsilon$ (after taking the min of $1$ and $\varepsilon$ or any $\varepsilon$) was arbitrary it follows that[ math]f[/tex] is integrable. Thus,

$\displaystyle \int_0^2 f(x)\text{ }dx=\overline{\int_0^2}f(x)dx=\sup_{P\in\mathcal{P }}U(P,f)=\sup_{p\in\mathcal{P}}U(P,x^3)=\int_0^2 x^3\text{ }dx=4$

where we've used the obvious fact that $U(P,f)=U(P,x^3)$ for any $P\in\mathcal{P}$ and since $x^3\in\mathcal{R}([0,2])$ that $\displaystyle \sup_{P\in\mathcal{P}}U(P,x^3)=\int_0^2 x^3\text{ }dx$.

4. Assuming that n is a fixed natural number and k must be an integer, then k/n is in [0, 2] only for k= 0, 1, 2, ..., 2n. That is, f(x) is continuous every where except on the finite set {0, 1/n , 2/n, ..., 2}. Every finite set has Lebesque measure 0 so f is continuous except on a set of measure 0 and so is Riemann integrable. its integral is exactly $\int_0^2 x^3dx= 4$.