in exercises 9 - 12, use ram to estimate the area of the region enclosed between the graph of $f$ and the…

Explanation:

Step1: Define the Riemann - sum formula

The Right - Handed Approximating Riemann Sum (RAM) for a function $y = f(x)$ on the interval $[a,b]$ with $n$ subintervals is given by $R_n=\sum_{i = 1}^{n}f(x_i)\Delta x$, where $\Delta x=\frac{b - a}{n}$ and $x_i=a + i\Delta x$. Here, $a = 0$, $b = 3$, so $\Delta x=\frac{3-0}{n}=\frac{3}{n}$ and $x_i=0 + i\frac{3}{n}=\frac{3i}{n}$.

Step2: Substitute $x_i$ into the function

$f(x)=x^{2}-x + 3$, so $f(x_i)=(\frac{3i}{n})^{2}-\frac{3i}{n}+3=\frac{9i^{2}}{n^{2}}-\frac{3i}{n}+3$.

Step3: Calculate the Riemann - sum

$R_n=\sum_{i = 1}^{n}f(x_i)\Delta x=\sum_{i = 1}^{n}(\frac{9i^{2}}{n^{2}}-\frac{3i}{n}+3)\frac{3}{n}=\sum_{i = 1}^{n}(\frac{27i^{2}}{n^{3}}-\frac{9i}{n^{2}}+\frac{9}{n})$. Using the sum rules $\sum_{i = 1}^{n}c=cn$, $\sum_{i = 1}^{n}i=\frac{n(n + 1)}{2}$, and $\sum_{i = 1}^{n}i^{2}=\frac{n(n + 1)(2n+1)}{6}$, we have: [ \begin{align*} R_n&=\frac{27}{n^{3}}\sum_{i = 1}^{n}i^{2}-\frac{9}{n^{2}}\sum_{i = 1}^{n}i+\frac{9}{n}\sum_{i = 1}^{n}1\ &=\frac{27}{n^{3}}\cdot\frac{n(n + 1)(2n + 1)}{6}-\frac{9}{n^{2}}\cdot\frac{n(n + 1)}{2}+\frac{9}{n}\cdot n\ &=\frac{27}{6}\cdot\frac{(n + 1)(2n + 1)}{n^{2}}-\frac{9}{2}\cdot\frac{n + 1}{n}+9 \end{align*} ]

Step4: Find the limit as $n\to\infty$

[ \begin{align*} \lim_{n\to\infty}R_n&=\lim_{n\to\infty}\left(\frac{27}{6}\cdot\frac{2n^{2}+3n + 1}{n^{2}}-\frac{9}{2}\cdot\frac{n + 1}{n}+9\right)\ &=\lim_{n\to\infty}\left(\frac{27}{6}\left(2+\frac{3}{n}+\frac{1}{n^{2}}\right)-\frac{9}{2}\left(1+\frac{1}{n}\right)+9\right)\ &=\frac{27}{6}\times2-\frac{9}{2}\times1 + 9\ &=9-\frac{9}{2}+9\ &=\frac{18 - 9+18}{2}=\frac{27}{2}=13.5 \end{align*} ]

Answer:

$13.5$