Tuesday, 30 April 2019

calculus - How to prove f(x) is integrable function on [0,2]



Let



$$ f(x) = \begin{cases}
1, & 0 \le x < 1 \\
0, & x=1 \\
1, & 1
\end{cases}$$



How Prove that f(x) integrable on [0,2]



I know that f(x) defined and Bounded on [a,b] then f(x) integrable



iff



ϵ>0 there is Partition P so that




U(f;P)L(f;P)<ϵ



how can i prove that?



thanks


Answer



Notice that on any interval (a,b)[0,2] we have that max and \min_{x\in (a,b)}f(x)=\begin{cases} 0 & \mbox{ if } 1\in (a,b),\\ 1 & \mbox{ otherwise.} \end{cases}




Let n\in \mathbb{N} and let P=(P_i)_{i\in I} be any partition such that P_j=(1-\frac{1}{n},1+\frac{1}{n}) for some j. Clearly U(f;P)=2. Notice that L(f;P)=2-\frac{2}{n}. Since n was chosen arbitrarily, we can get U(f;P)-L(f;P) as small as we want.


No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...