for the function given below, find a formula for the riemann sum obtained by dividing the interval a,b into…

Explanation:

Step1: Calculate the width of sub - intervals

The interval is $[a,b]=[0,1]$. The width of each of the $n$ equal sub - intervals, $\Delta x=\frac{b - a}{n}=\frac{1-0}{n}=\frac{1}{n}$.

Step2: Find the right - hand endpoints

The right - hand endpoints $x_k=a + k\Delta x$, with $a = 0$ and $\Delta x=\frac{1}{n}$, so $x_k=\frac{k}{n}$ for $k = 1,2,\cdots,n$.

Step3: Evaluate the function at the right - hand endpoints

$f(x)=x + 11x^{2}$, so $f(x_k)=\frac{k}{n}+11(\frac{k}{n})^{2}=\frac{k}{n}+\frac{11k^{2}}{n^{2}}$.

Step4: Form the Riemann sum

The Riemann sum $S_n=\sum_{k = 1}^{n}f(x_k)\Delta x$. Substitute $f(x_k)$ and $\Delta x$ into the sum: [ \begin{align*} S_n&=\sum_{k = 1}^{n}\left(\frac{k}{n}+\frac{11k^{2}}{n^{2}}\right)\frac{1}{n}\ &=\sum_{k = 1}^{n}\left(\frac{k}{n^{2}}+\frac{11k^{2}}{n^{3}}\right)\ &=\frac{1}{n^{2}}\sum_{k = 1}^{n}k+\frac{11}{n^{3}}\sum_{k = 1}^{n}k^{2} \end{align*} ] Using the sum formulas $\sum_{k=1}^{n}k=\frac{n(n + 1)}{2}$ and $\sum_{k = 1}^{n}k^{2}=\frac{n(n + 1)(2n+1)}{6}$, we have: [ \begin{align*} S_n&=\frac{1}{n^{2}}\cdot\frac{n(n + 1)}{2}+\frac{11}{n^{3}}\cdot\frac{n(n + 1)(2n + 1)}{6}\ &=\frac{n+1}{2n}+\frac{11(n + 1)(2n + 1)}{6n^{2}} \end{align*} ]

Answer:

$\frac{n + 1}{2n}+\frac{11(n + 1)(2n + 1)}{6n^{2}}$