riemann integral

**yarlagaddatrinath** · March 24th 2008, 02:41 AM

If f is continous on [a,b],then f belongs to R[a,b],R-RIEMANN INTEGRAL

**Jhevon** · March 24th 2008, 08:34 AM

Originally Posted by yarlagaddatrinath

If f is continous on [a,b],then f belongs to R[a,b],R-RIEMANN INTEGRAL

this question is phrased weird. what do you mean "belongs to R[a,b]"? do you mean it is Riemann Integrable?

**ThePerfectHacker** · March 24th 2008, 09:34 AM

Originally Posted by yarlagaddatrinath

If f is continous on [a,b],then f belongs to R[a,b],R-RIEMANN INTEGRAL

A continous function on [a,b] is always Riemann integrable. Since $f$ is continous it is bounded by extreme value theorem. Furthermore, continous functions on compact (in particular closed) sets are uniformly continous. To prove Riemann integrability we need to show $S_P(f) - s_P(f) < \epsilon$ where $P$ is a partition of $[a,b]$ . Given any $\epsilon > 0$ there is a $\delta > 0$ such that $|x-y|<\delta \mbox{ and }x,y\in [a,b]\implies |f(x)-f(y)|<\epsilon$ . Pick a partition $P$ such that $(x_i - x_{i-1}) < \delta$ . Then $f$ assumes its maximum and minimum on each interval and so $S_P(f) - s_P(f) < \epsilon(b-a)$ , and this quantity can be made arbitrary small.

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