Real Analysis (Riemann Integral) 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 
                             $$|R(f,P)-R|< \varepsilon$$

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.
