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Let $f(x)=c$ for all $x$ in $[a,b]$ and some real number $c$. Show by definition below that $f$ is Riemann integrable on $[a,b]$, and $\int f(x) dx = c(b-a)$.

Definition: A function $f$ is Riemann integrable on $[a,b]$ if there is a real number $R$ such that for any $\varepsilon > 0$, there exists $\delta > 0$ such that for any a partition $P$ of $[a,b]$ satisfying $\|P\|< \delta$, and for any Riemann sum $R(f,P)$ of relative to $P$, we have $$|R(f,P)-R|< \varepsilon$$

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Can you show any progress you have made? –  Patrick McLaren Oct 2 '12 at 1:49
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You can get proof in any intro book buddy –  Ram Oct 2 '12 at 2:04
    
i search it already but it's proof not showed by this definition. Anyone can help me? :) –  Chiko Wa Oct 2 '12 at 13:08
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1 Answer

Let $R = c (b-a)$ and let $\varepsilon > 0$.

You want to show that there is $\delta > 0$ such that for all tagged partitions $P$ with $\|P\| < \delta$ you have $$ \left | \sum_{k=1}^n f(x_i) (t_{i+1}-t_i) - R \right | < \varepsilon$$

where $x_i \in [t_i , t_{i+1}] \subset [a,b]$ form a tagged partition of $[a,b]$. We have $f(x_i) = c$ hence $$ \left | \sum_{k=1}^n f(x_i) (t_{i+1}-t_i) - R \right | = \left |c \sum_{k=1}^n (t_{i+1}-t_i) - R \right | = \left | c (b-a) - R \right | = 0 < \varepsilon$$

hence $f$ is Riemann integrable with $$ \int_a^b f(x) dx = c (b - a)$$

Hope this helps.

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