Find ${\int_{0}^{3}{(3x -8)}}\,dx$.

Solution Since the integrand $f(x) = 3x - 8$ is continuous on the closed interval $[0, 3]$, the function $f$ is integrable over $[0, 3].$ Although we can use any partition of $[0,3]$ whose norm can be made as close to $0$ as we please, and we can choose any $u_{i}$ in each subinterval$,$ we use a regular partition and choose $u_{i}$ as the right endpoint of each subinterval. This will result in a simple expression for the Riemann sums.

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We partition $[0,3]$ into $n$ subintervals, each of length $\Delta x=\dfrac{3-0}{n}=\dfrac{3}{n}$. The endpoints of each subinterval of the partition, written in terms of $n$, are $$\begin{eqnarray*} x_{0}&=&0,\ x_{1}=\dfrac{3}{n},\ {x_{2}=2}\left({\dfrac{3}{n}}\right), \ldots , x_{i-1}=(i-1) \left(\dfrac{3}{n}\right),\\ x_{i} &=& i\left(\dfrac{3}{n}\right) , \ldots , x_{n}=n\left(\dfrac{3}{n}\right)=3 \end{eqnarray*}$$

The Riemann sums of $f(x)=3x - 8$ from $0$ to $3$, using $u_{i}= x_{i}= \dfrac{3i}{n}$ (the right endpoint) and $\Delta x=\dfrac{3}{n},$ are 357 $$\begin{eqnarray*} \sum\limits_{i=1}^{n}\,f(u_{i})\,\Delta x_{i} &=&\sum\limits_{i=1}^{n}\,f(x_{i})\,\Delta x \underset{\underset{{\color{#0066A7}{\hbox{\(\Delta x=\dfrac{3}{n}\)}}}}{\color{#0066A7}{\left\uparrow\right.}}} {=} \sum\limits_{i=1}^{n}(3x_{i}-8)\dfrac{3}{n} \underset{\underset{{\color{#0066A7}{\hbox{\(x_{i}=\dfrac{3i}{n}\)}}}}{\color{#0066A7}{\left\uparrow\right.}}} {=} \sum\limits_{i=1}^{n}\left[ 3\left( {\dfrac{{3i}}{{n}}}\right) -8\right] \dfrac{3}{n} \\ &=& \sum \limits_{i=1}^{n}\left( \dfrac{27i}{n^{2}}-\dfrac{24}{n}\right) =\dfrac{27}{n^{2}}\sum\limits_{i=1}^{n}{i}-\dfrac{24}{n} \sum\limits_{i=1}^{n}1\\[5pt] &=&\dfrac{27}{n^{2}}\cdot \dfrac{n(n+1)}{{2}}-\dfrac{24}{{n} }\cdot n=\dfrac{27}{2}+\dfrac{27}{2n}-24 =-\dfrac{21}{2}+\dfrac{27}{2n} \end{eqnarray*}$$

Now $${\int_{0}^{3}{(3x-8)\,dx=}}\lim\limits_{n\rightarrow \infty }\left( -\dfrac{21}{2}+\dfrac{27}{2n}\right) =-\dfrac{21}{2}$$