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

Explanation:

Step1: Find 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 $x_k=a + k\Delta x=0 + k\cdot\frac{1}{n}=\frac{k}{n}$ for $k = 1,2,\cdots,n$.

Step3: Evaluate the function at the right - hand endpoints

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

Step4: Form the Riemann sum

$S_n=\sum_{k = 1}^{n}f(x_k)\Delta x=\sum_{k = 1}^{n}(\frac{k}{n}+\frac{11k^{2}}{n^{2}})\frac{1}{n}=\sum_{k = 1}^{n}(\frac{k}{n^{2}}+\frac{11k^{2}}{n^{3}})$. 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: $S_n=\frac{1}{n^{2}}\sum_{k = 1}^{n}k+\frac{11}{n^{3}}\sum_{k = 1}^{n}k^{2}=\frac{1}{n^{2}}\cdot\frac{n(n + 1)}{2}+\frac{11}{n^{3}}\cdot\frac{n(n + 1)(2n + 1)}{6}$.

Step5: Simplify the Riemann sum

$S_n=\frac{n^2 + n}{2n^{2}}+\frac{11(2n^{3}+3n^{2}+n)}{6n^{3}}=\frac{1}{2}+\frac{1}{2n}+\frac{11}{3}+\frac{11}{2n}+\frac{11}{6n^{2}}$.

Step6: Find the limit as $n\to\infty$

$\lim_{n\to\infty}S_n=\lim_{n\to\infty}(\frac{1}{2}+\frac{1}{2n}+\frac{11}{3}+\frac{11}{2n}+\frac{11}{6n^{2}})$. As $n\to\infty$, $\frac{1}{n}\to0$ and $\frac{1}{n^{2}}\to0$. So $\lim_{n\to\infty}S_n=\frac{1}{2}+\frac{11}{3}=\frac{3 + 22}{6}=\frac{25}{6}$.

Answer:

$\frac{25}{6}$