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 maxx(a,b)f(x)=1 and minx(a,b)f(x)={0 if 1(a,b),1 otherwise.




Let nN and let P=(Pi)iI be any partition such that Pj=(11n,1+1n) for some j. Clearly U(f;P)=2. Notice that L(f;P)=22n. Since n was chosen arbitrarily, we can get U(f;P)L(f;P) as small as we want.


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