Find a formula for the Riemann sum obtained by dividing the interval [a, b] into n equal subintervals and using the right-hand endpoint for each $\mathrm { c } _ { \mathrm { k } }$. Then take a limit of these sums as $\mathrm { n } \rightarrow \infty$ to calculate the area under the curve over [a, b].
-$f ( x ) = 5 x ^ { 2 } + 4$ over the interval $[ 0,3 ]$.
A) $12 + \frac { 270 n ^ { 3 } + 405 n ^ { 2 } + 135 n } { 6 n ^ { 4 } }$; Area $= 12$
B) $12 + \frac { 270 n ^ { 3 } + 405 n ^ { 2 } + 135 n } { 6 n ^ { 3 } } ;$ Area $= 45$
C) $12 + \frac { 270 n ^ { 3 } + 405 n ^ { 2 } + 135 n } { 6 n ^ { 4 } } ;$ Area $= 57$
D) $12 + \frac { 270 n ^ { 3 } + 405 n ^ { 2 } + 135 n } { 6 n ^ { 3 } } ;$ Area $= 57$

Question

Quizplus · Accepted Answer

Using the right-hand endpoint, we have $\mathrm { c } _ { \mathrm { k } }= a + k \Delta \mathrm { x }$, where $\Delta \mathrm { x } = \frac { b - a } { n }$ is the width of each subinterval. Thus, the Riemann sum is
$$\sum _ { k = 1 } ^ { n } f \left( \mathrm { c } _ { \mathrm { k } } \right) \Delta \mathrm { x } = \sum _ { k = 1 } ^ { n } f \left( a + k \Delta \mathrm { x } \right) \Delta \mathrm { x }.$$
For the given function $f(x) = 5x^2+4$ over the interval $[0,3]$, we have
\begin{align*}
\sum _ { k = 1 } ^ { n } f \left( a + k \Delta \mathrm { x } \right) \Delta \mathrm { x } &= \sum _ { k = 1 } ^ { n } f \left( 0 + k \frac{3-0}{n} \right) \frac{3-0}{n} \
&= \sum _ { k = 1 } ^ { n } f \left( \frac{3k}{n} \right) \frac{3}{n} \
&= \frac{3}{n} \sum _ { k = 1 } ^ { n } \left(5\cdot \left(\frac{3k}{n}\right)^2 + 4 \right) \
&= \frac{3}{n} \left[ 5\sum _ { k = 1 } ^ { n } \left(\frac{3k}{n}\right)^2 + 4\sum _ { k = 1 } ^ { n } 1 \right] \
&= \frac{3}{n} \left[ 5\sum _ { k = 1 } ^ { n } \frac{9k^2}{n^2} + 4n \right] \
&= \frac{3}{n} \left[ 5\cdot \frac{9}{n^2} \sum _ { k = 1 } ^ { n } k^2 + 4n \right] \
&= \frac{3}{n} \left[ 5\cdot \frac{9}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} + 4n \right] \
&= \frac{9}{2} + \frac{45n^2+30n}{2n^2} \
&= 12 + \frac { 270 n ^ { 3 } + 405 n ^ { 2 } + 135 n } { 6 n ^ { 3 } },
\end{align*}
where we have used the formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$.
Taking the limit as $\mathrm { n } \rightarrow \infty$ gives the area under the curve:
$$\lim\limits_{n \to \infty} \left( 12+ \frac { 270 n ^ { 3 } + 405 n ^ { 2 } + 135 n } { 6 n ^ { 3 } } \right) = 57.$$

Find a Formula for the Riemann Sum Obtained by Dividing ck\mathrm { c } _ { \mathrm { k } }ck​

Find a Formula for the Riemann Sum Obtained by Dividing $\mathrm { c } _ { \mathrm { k } }$