Q 20. - Calculus

Question

The sum approximates the area between the graph of some function f and the x-axis from x = a to x = b. Do some “reverse engineering” to determine the

type of approximation (left sum, midpoint sum, etc.) and identify f(x), a, b, n, $∆x$, and $x_k$. Then sketch the approximation described.

$\sum _{k=1}^9\sqrt{\frac{\left(1+{\displaystyle \frac{k-1}{3}}\right)+\left(1+{\displaystyle \frac{k}{3}}\right)}{2}}\left(\frac{1}{3}\right)$

Step-by-Step Solution

Answer

$\begin{array}{rcl}f\left(x\right)& =& \sin x\\ N& =& 100\\ ∆x& =& 0.04\\ x_k& =& 0+0.04k\\ a& =& 0\\ b& =& 4\end{array}$

Step 1. Given information

The sum is $\sum _{k=1}^9\sqrt{\frac{\left(1+{\displaystyle \frac{k-1}{3}}\right)+\left(1+{\displaystyle \frac{k}{3}}\right)}{2}}\left(\frac{1}{3}\right)$

Step 2. Explanation

Consider the sum $\sum _{k=1}^9\sqrt{\frac{\left(1+{\displaystyle \frac{k-1}{3}}\right)+\left(1+{\displaystyle \frac{k}{3}}\right)}{2}}\left(\frac{1}{3}\right)$

$\sum _{k=1}^9\sqrt{\frac{\left(1+{\displaystyle \frac{k-1}{3}}\right)+\left(1+{\displaystyle \frac{k}{3}}\right)}{2}}\left(\frac{1}{3}\right)=\sqrt{\frac{\left(1+{\displaystyle \frac{1-1}{3}}\right)+\left(1+{\displaystyle \frac{1}{3}}\right)}{2}}\left(\frac{1}{3}\right)+\sqrt{\frac{\left(1+{\displaystyle \frac{2-1}{3}}\right)+\left(1+{\displaystyle \frac{2}{3}}\right)}{2}}\left(\frac{1}{3}\right)+\sqrt{\frac{\left(1+{\displaystyle \frac{3-1}{3}}\right)+\left(1+{\displaystyle \frac{3}{3}}\right)}{2}}\left(\frac{1}{3}\right)+...$

$\begin{array}{rcl}f\left(x\right)& =& \sin x\\ N& =& 100\\ ∆x& =& 0.04\\ x_k& =& 0+0.04k\\ a& =& 0\\ b& =& 4\end{array}$