Let a partition $\{t_0,\ldots,t_n\}$ of the interval $[a,b]$ and let $f$ an integrable function. (we may also assume that $f$ is differentiable on $[a,b]$)
We know that the Right Riemann sum is
$$R(f)=\sum_{i=1}^n f(t_{i})(t_i-t_{i-1})$$
What's the relation ( in order sense terms) between $R(f)$ and the integral $\displaystyle \int_a^b f(x)dx$?
Can I say that there exists a positive constant $C>0$ such that $\displaystyle \int_a^b f(x)dx\leq C\cdot R(f)$? I know that if $f$ is increasing function, this is correct, but in the general case, i may say that?
EDIT: Assume that $f(t)\geq 0$ on $[a,b]$
From Wikipedia I get the following formula
$$\left|\int_a^b f(t)dt-R(f)\right|\leq \dfrac{M_1(a-b)^2}{2n}$$
Then,
$$\int_a^b f(t)dt\leq \dfrac{M_1(a-b)^2}{2n}+R(f)\leq \max\left\{\dfrac{M_1(a-b)^2}{2n},1\right\}(1+R(f))$$
I'd like a constant multiplying $R(f)$.
Could this be improved?