Q.Find an approximate value of $\displaystyle\int_1^{1.5} (2-x)\,dx$ by applying the mid-point rule with the partition $\{1.1,\ 1.2,\ 1.3,\ 1.4,\ 1.5\}$.

Given a bounded function $f(x)$ on a closed interval $[a,b]$, partition $[a,b]$ into $n$ subintervals $[x_0,x_1],[x_1,x_2],\ldots,[x_{n-1},x_n]$ with $a=x_0<x_1<\cdots<x_n=b$. In each subinterval $[x_{i-1},x_i]$ pick any point $\xi_i$ (with $x_{i-1}\le \xi_i\le x_i$) and form the Riemann sum

As $n\to\infty$ in such a way that the width of the widest subinterval $\max(x_i-x_{i-1})\to 0$, this sum tends to a single finite number $A$ (when $f$ is continuous, or piecewise continuous). That limit $A$ is the definite integral (or Riemann integral) of $f$ on $[a,b]$, written $\displaystyle\int_a^b f(x)\,dx$. When $a=b$, $\int_a^a f(x)\,dx=0$.

Three standard choices of $\xi_i$, all giving the same limit for a continuous $f$:

• Left-end rule: $\xi_i=x_{i-1}$, so $\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^n f(x_{i-1})(x_i-x_{i-1})$.
• Right-end rule: $\xi_i=x_i$, so $\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^n f(x_i)(x_i-x_{i-1})$.
• Mid-point rule: $\xi_i=\tfrac12(x_{i-1}+x_i)$, so $\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^n f\!\left(\tfrac{x_{i-1}+x_i}{2}\right)(x_i-x_{i-1})$.

On a finite partition (before the limit is taken), each rule only gives an approximate value of the integral — this is exactly how Exercise 9.1's five-point partitions are used.

Closed-form limit formula. Dividing $[a,b]$ into $n$ equal subintervals of width $h=\dfrac{b-a}{n}$, so $x_i=a+ih$, and using the right-end rule gives the standard working formula

$\int_a^b f(x)\,dx=\lim_{n\to\infty}\dfrac{b-a}{n}\sum_{r=1}^n f\!\left(a+\dfrac{(b-a)r}{n}\right).$ …