Introduction : Any function that satisfies conditions of Riemann integral conditions is a Riemann integrable function. An example is solved.
Question : Show that following function
\[f\left(x\right)=1 , if 2≤x<32\][f\left(x\right)=2, if 3≤x≤4\]
is Riemann integral on [2,4].
Solution :
Given function
\[f\left(x\right)=1 , if 2≤x<32, f\left(x\right)=2, if 3≤x≤4\]
Let any descretization of the interval [2,4]
\[\left\{x_0,x_1,\ x_2,..,,,\ x_n\right\}\]
Such that
\[\left(x_{i+1}-x_i\right)=\delta>0\]
for all i=0,1,2,….n-1
Now f will be Riemann integrable in [2,4] if
\[\left|\sum_{i=0}^{n-1}{f\left(x_i\right)\left(x_{i+1}-x_i\right)-3}\right|<\epsilon\] \[\left|\delta\sum_{i=0}^{n-1}{f\left(x_i\right)-3}\right|\le\delta\left|\sum_{i=0}^{n-1}f\left(x_i\right)\right|+3\le2n\delta+3<\epsilon\] So \[2n\delta+3<\epsilon\]
Thus if we choose
\[\delta<\frac{\left(\epsilon-3\right)}{2n}\]
Then f(x) will be Riemann integrable in [2, 4]. that is
\[ f\in R\left[0,3\right] and \int_{2}^{4}{f(x)}dx=3\]