Riemann Integrable function solved Question

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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

\[ f\in R\left[0,3\right] and \int_{2}^{4}{f(x)}dx=3\]


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