Darboux Integral

Question

Let $f$ be defined on $[0,1]$ by $f(x):=1$ if $x \not= 1$ and $f(1):=0$, Show that the Darboux Integral exist and find its value.

I know I want my partition to be $P_\epsilon := (0, 1-\epsilon/2, 1+\epsilon/2,2)$

I'm having trouble defining my Upper sum and Lower sum is this correct?

Define the partition $P_\epsilon := (0, 1-\epsilon/2, 1+\epsilon/2,2)$ then we get the lower sum $L(f, P_\epsilon)= 1(1-\epsilon/2)+ 1(1+\epsilon/2 -(1-\epsilon/2)) + 2(1-\epsilon/2) = 1-\epsilon/2 + \epsilon + 2(1-\epsilon/2)=1-\epsilon/2 +\epsilon +2-\epsilon=3-\epsilon/2$

therefore the lower integral satisfies $L(f)\ge3$

Similarly we get the upper sum $U(f,P_\epsilon)= 1(1-\epsilon/2)+ 2(1+\epsilon/2 -(1-\epsilon/2)) + 2(1-\epsilon/2)=3+ \epsilon/2$

Therefore the upper integral satisfies $U(f)\le3$

Thus $L(f)=U(f)=3$ and the Darboux integral of f is $\int_0^2 f=3$

Answer 1

There are some mistakes in your question as the integral has a value of $2$ and not $3$

Here's an easier and more clear way of showing that $f$ is integrable on $[0,2]$:

Let $P=\{t_o,...,t_{j-1},t_j,...,t_n\}$ be a partition of $[0,2]: t_{j-1}<1<t_j$

$\implies$

• $m_i=M_i=1,$ when $i \ne j$
• $m_i=0, M_i=1,$ when $i=j$

The lower sum is $L(f,P)=\sum_{i=1}^{j-1}{m_i(t_i-t_{i-1})}+m_j(t_j-t_{j-1})+\sum_{i=j+1}^{n}{m_i(t_i-t_{i-1})}$

The upper sum is $U(f,P)=\sum_{i=1}^{j-1}{M_i(t_i-t_{i-1})}+M_j(t_j-t_{j-1})+\sum_{i=j+1}^{n}{M_i(t_i-t_{i-1})}$

$\implies U(f,P)-L(f,P)=t_j-t_{j-1}$

Let $\epsilon > 0$

So for $P$ such that $t_{j-1}<1<t_j, t_j-t_{j-1}< \epsilon$ we have that $U(f,P)-L(f,P)< \epsilon$

This means that $f$ is integrable on $[0,2]$

Also, it is obvious that $L(f,P) \le 2 \le U(f,P)$ $\forall P$ partition of $[0,2]$

So $\int_{0}^{2} {f}=2$