Q 18. - Calculus

Question

The given 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.

$\sum _{k=0}^2\ln \left(2+\frac{k}{3}\right)\left(\frac{1}{3}\right)$

Step-by-Step Solution

Answer

$\begin{array}{rcl}f\left(x\right)& =& x(x-1)\\ N& =& 4\\ ∆x& =& \frac{1}{2}\\ x_k& =& 1+\frac{k}{3}\\ a& =& 1\\ b& =& 3\\ & & \end{array}$

Step 1. Given information

Sum is $\sum _{k=0}^2\ln \left(2+\frac{k}{3}\right)\left(\frac{1}{3}\right)$

Step 2. Explanation

Consider the sum $\sum _{k=0}^2\ln \left(2+\frac{k}{3}\right)\left(\frac{1}{3}\right)$

$\sum _{k=0}^2\ln \left(2+\frac{k}{3}\right)\left(\frac{1}{3}\right)=\ln \left(2+\frac{0}{3}\right)\left(\frac{1}{3}\right)+\ln \left(2+\frac{1}{3}\right)\left(\frac{1}{3}\right)+\ln \left(2+\frac{2}{3}\right)\left(\frac{1}{3}\right)$

width="98" style="max-width: none; vertical-align: -80px;" $\begin{array}{rcl}f\left(x\right)& =& x(x-1)\\ N& =& 4\\ ∆x& =& \frac{1}{2}\\ x_k& =& 1+\frac{k}{3}\\ & & \end{array}$

Therefore,

$\begin{array}{rcl}a& =& 1\\ b& =& 3\end{array}$