for the function given below, find a formula for the riemann sum obtained by dividing the interval 0,4 into…

Explanation:

Step1: Find the width of sub - intervals

The interval is $[0,4]$, and we divide it into $n$ equal sub - intervals. The width of each sub - interval $\Delta x=\frac{4 - 0}{n}=\frac{4}{n}$.

Step2: Find the right - hand endpoints $c_k$

The right - hand endpoints of the sub - intervals are $c_k=a + k\Delta x$, where $a = 0$ and $\Delta x=\frac{4}{n}$. So $c_k=0 + k\cdot\frac{4}{n}=\frac{4k}{n}$ for $k = 1,2,\cdots,n$.

Step3: Find the Riemann sum $S_n$

The function is $f(x)=3x^{2}$. Then $f(c_k)=3(\frac{4k}{n})^{2}=3\cdot\frac{16k^{2}}{n^{2}}=\frac{48k^{2}}{n^{2}}$. The Riemann sum $S_n=\sum_{k = 1}^{n}f(c_k)\Delta x$. Substituting $f(c_k)$ and $\Delta x$ we get $S_n=\sum_{k = 1}^{n}\frac{48k^{2}}{n^{2}}\cdot\frac{4}{n}=\sum_{k = 1}^{n}\frac{192k^{2}}{n^{3}}$. Using the formula $\sum_{k=1}^{n}k^{2}=\frac{n(n + 1)(2n+1)}{6}$, we have $S_n=\frac{192}{n^{3}}\cdot\frac{n(n + 1)(2n + 1)}{6}$.

Step4: Simplify the Riemann sum

$S_n=\frac{192}{6}\cdot\frac{n(n + 1)(2n + 1)}{n^{3}}=32\cdot\frac{2n^{3}+3n^{2}+n}{n^{3}}=32(2+\frac{3}{n}+\frac{1}{n^{2}})$.

Step5: Find the limit as $n\to\infty$

$\lim_{n\rightarrow\infty}S_n=\lim_{n\rightarrow\infty}32(2+\frac{3}{n}+\frac{1}{n^{2}})$. As $n\to\infty$, $\frac{3}{n}\to0$ and $\frac{1}{n^{2}}\to0$. So $\lim_{n\rightarrow\infty}S_n=32\times2 = 64$.

Answer:

64