Question

 Consider any two positive real numbers and call it 𝑎 and 𝑏. Then consider the function defined as  

𝑓(𝑥) = {

𝑎, 0 ≤ 𝑥 < 1 𝑏,          𝑥 = 1 Find the 𝑈(𝑃,𝑓) and 𝐿(𝑃,𝑓) for the partition 𝑃 = {𝑥0,𝑥1,…𝑥𝑛} of [0,1]. Also check whether the function is Riemann integrable over [0,1] or not. 

Answer 1

$Let \space P=\{0, \frac{1}{n},\frac{2}{n},...,\frac{r-1}{n}, \frac{r}{n}...,\frac{n}{n}=1\}$ be any partition of [0,1]

The r^th sub interval of [0,1] $= I_r=[\frac{r-1}{n},\frac{r}{n}] ; r= 1,2,...n$

If $\delta_r$ be the length of I_r then, $\delta = \frac{r}{n}- \frac{r-1}{n}= \frac{1}{n}$

Let M_r and m_r be respectively the supremum and the infimum of f in I_r . Since f is increasing on [0,1] we have

$M_r = supf(x) =\frac{r^3}{n^3}\\ m_r = inff(x) =\frac{(r-1)^3}{n^3}\\ U(f,p)= \sum^n_{r=1}M_rdr=\sum_{r=1}^n \frac{r^3}{n^3}\frac{1}{n}=\frac{1}{4}(1+\frac{1}{n})^2\\ L(f,p)= \sum^n_{r=1}m_rdr=\sum_{r=1}^n \frac{(r-1)^3}{n^3}\frac{1}{n}=\frac{1}{4}(1-\frac{1}{n})^2$

$\int_a^{\bar{b}}f(x)dx = Lim_{n\to \infin} U(f,p)= Lim_{n\to \infin} \frac{1}{4}(1+\frac{1}{n})^2=\frac{1}{4}\\ \int_{\bar{a}}^bf(x)dx = Lim_{n\to \infin} L(f,p)= Lim_{n\to \infin} \frac{1}{4}(1-\frac{1}{n})^2=\frac{1}{4}\\ \int_a^{\bar{b}}f(x)dx=\int_{\bar{a}}^bf(x)dx$

Hence the function is Riemann integrable over [0,1]

$\int_0^1 f(x)dx =\frac{1}{4}$