Right Riemann sum Error bound proof

Question

How do we prove the right Riemann sum error bound? In wikipedia (https://en.wikipedia.org/wiki/Riemann_sum#Right_Riemann_sum) they have mentioned the following bound, but no proof. $$\left | \int_a^bf(x)dx - A_{\mbox{right}}\right | \leq \frac{M_1(b-a)^2}{2n}$$

Answer 1

This result is stated for a monotone function and uniformly spaced points. Assume first that $f$ is non-decreasing ( the proof for $f$ non-increasing is similar).

With a uniform partition $x_j = a + j\frac{b-a}{n}$, we can apply the mean value theorem to obtain

$$f(x_j) = f(x) + f'(\xi_j(x))(x_j - x)$$

where $\xi_j(x)$ is between $x$ and $x_j$.

Integrating and using $f' \geqslant 0$, we get the local error bound

$$f(x_j)(x_j - x_{j-1}) - \int_{x_{j-1}}^{x_j} f(x) \, dx = \int_{x_{j-1}}^{x_j} f'(\xi_j(x))(x_j - x) \, dx \leqslant M \frac{(x_j - x_{j-1})^2}{2} = M \frac{(b-a)^2}{2n^2}$$

and the global error bound is

$$\sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - \int_{a}^{b} f(x) \, dx = \sum_{j=1}^nf(x_j)(x_j - x_{j-1}) - \sum_{j=1}^n\int_{x_{j-1}}^{x_j} f(x) \, dx \\ \leqslant \sum_{j=1}^n M \frac{(b-a)^2}{2n^2} = M \frac{(b-a)^2}{2n} $$

Answer 2

This could be an exercise when learning the definite integrals. Here is a demo.

For convenience, suppose $f$ is differentiable on $[a,b]$. Let $x_j = a + j \varDelta x$, where $\varDelta x = (b-a)/n$, for $j =1,2,\dots, n$. Then $$ A _{\text{right}} = \sum_1^n f(x_j) \varDelta x. $$ Thus \begin{align*} \newcommand\Abs[1]{\left\vert #1 \right \vert} &\quad \Abs {-\int_a^b f(x) \,\mathrm dx + A_{\text{right}} } \\ &= \Abs {-\int_a^b f(x) \,\mathrm dx + \sum_1^n f(x_j) (x_{j} - x_{j-1})} \\ &= \Abs {-\sum_1^n \int_{x_{j-1}}^{x_j} f(x) \,\mathrm dx + \sum_1^n \int_{x_{j-1}}^{x_j}f(x_j) \,\mathrm dx } \\ &= \Abs {\sum_1^n \int_{x_{j-1}}^{x_j} (f(x_j) - f(x) )\,\mathrm dx } \\ &= \Abs { \sum_1^n \int_{x_{j-1}}^{x_j} f'(\eta_j) (x_j - x)\mathrm dx } [\text{Lagrange's MVT}], \end{align*} where $\eta_j \in (x_{j-1}, x_j)$ for each $j$.

Now use $-M_1 \leqslant f' \leqslant M_1$, and notice that $(x_j - x) \geqslant 0$, we have $$ -M_1 (x_j - x) \leqslant f'(\eta_j)(x_j -x) \leqslant M_1 (x_j -x), $$ hence $$ -M_1 \int_{x_{j-1}}^{x_j} (x_j - x) \mathrm dx \leqslant \int_{x_{j-1}}^{x_j} f'(\eta_j) (x_j - x)\mathrm dx = \int_{x_{j-1}}^{x_j} (f(x_j) - f(x))\mathrm dx \leqslant M_1 \int_{x_{j-1}}^{x_j} (x_j - x) \mathrm dx, $$ i.e. $$ \Abs { \int_{x_{j-1}}^{x_j} (f(x_j) - f(x))\mathrm dx } \leqslant M_1 \int_{x_{j-1}}^{x_j} (x_j -x)\mathrm dx = M_1 \frac {(b-a)^2}{2n^2}. $$ Therefore we have $$ \Abs {-\int_a^b f(x) \,\mathrm dx + A_{\text{right}} } \leqslant \sum_1^n \Abs { \int_{x_{j-1}}^{x_j} (f(x_j) - f(x))\mathrm dx } \leqslant n \cdot M_1 \frac {(b-a)^2}{2n^2} = \frac {M_1(b-a)^2}{2n}. $$