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 $\displaystyle \varepsilon<1$. Consider then the intervals
$\displaystyle I_k=\begin{cases}\left[\frac{k}{n}-\frac{\varepsilon}{64n},\frac{k}{n}+\frac{\varepsi lon}{64n}\right] & \mbox{if}\quad k=1,\cdots,2n-1\\ \left[2-\frac{\varepsilon}{64n},2] & \mbox{if}\quad k=2n\end{cases}$
Notice then that if
$\displaystyle J_k=\begin{cases}[0,\frac{1}{n}-\frac{\varepsilon}{64n}] & \mbox{if}\quad k=1\\ [\frac{k-1}{n}+\frac{\varepsilon}{64n},\frac{k}{n}-\frac{\varepsilon}{64n}] & \mbox{if}\quad k=2,\cdots,2n\end{cases}$
Then $\displaystyle [0,2]=I_1\cup\cdots\cup I_{2n}\cup J_1\cup\cdots\cup J_{2n}$. Note though that $\displaystyle f_{\mid J_k}=x^3,\quad k=1,\cdots,2n$ and so by continuity of $\displaystyle x^3$ on $\displaystyle J_1,\cdots,J_{2n}$ we may find partitions $\displaystyle P_1,\cdots,P_{2n}$ such that $\displaystyle U(P_k,f_{\mid J_k})-L(P_k,f_{\mid J_k})<\frac{\varepsilon}{4n}$. Note next that $\displaystyle P=P_1\cup\cdots\cup P_{2n}$ is a partition for $\displaystyle [0,2]$ and
$\displaystyle \displaystyle U(P,f)-L(P,f)=\underbrace{\sum_{k=1}^{2n}\left(U(P_k,f_{\ mid J_k})-L(P_k,f_{\mid J_k})\right)}_{\mathbf{(1)}}+\underbrace{\frac{\va repsilon}{32n}\sum_{k=1}^{n}\left(\sup_{x\in I_k}f(x)-\inf_{x\in I_k}f(x)\right)}_{\mathbf{(2)}}$
Note though that by construction $\displaystyle \mathbf{(1)}=2n\cdot \frac{\varepsilon}{4n}=\frac{\varepsilon}{2}$ and ,since $\displaystyle \displaystyle \inf_{x\in I_k}f(x)=0$,
$\displaystyle \displaystyle \mathbf{(2)}=\frac{\varepsilon}{32n}\sum_{k=1}^{n} \sup_{x\in I_k}f(x)\leqslant \frac{\varepsilon}{32n}\sum_{k=1}^{2n}8=\frac{\var epsilon}{2}$
And thus $\displaystyle U(P,f)-L(P,f)=\mathbf{(1)}+\mathbf{(2)}<\frac{\varepsilon }{2}+\frac{\varepsilon}{2}=\varepsilon$ and since $\displaystyle \varepsilon$ (after taking the min of $\displaystyle 1$ and $\displaystyle \varepsilon$ or any $\displaystyle \varepsilon$) was arbitrary it follows that[ math]f[/tex] is integrable. Thus,
$\displaystyle \displaystyle \int_0^2 f(x)\text{ }dx=\overline{\int_0^2}f(x)dx=\sup_{P\in\mathcal{P }}U(P,f)=\sup_{p\in\mathcal{P}}U(P,x^3)=\int_0^2 x^3\text{ }dx=4$
where we've used the obvious fact that $\displaystyle U(P,f)=U(P,x^3)$ for any $\displaystyle P\in\mathcal{P}$ and since $\displaystyle x^3\in\mathcal{R}([0,2])$ that $\displaystyle \displaystyle \sup_{P\in\mathcal{P}}U(P,x^3)=\int_0^2 x^3\text{ }dx$.