Applications: Approximation of Area

Problem 247: Consider the integral

(1) $\begin{equation*} \int_{5}^{9} \left( \frac{2}{x} + 1\right) \ dx \end{equation*}$

(a) Find the Riemann sum for this integral using right endpoints and $n=4$ .

(b) Find the Riemann sum for this same integral, using left endpoints and $n=4$ .

Solution: The Riemann sum is given by

(2) $\begin{equation*} R_n = \sum_{i=1}^{n} f(x_i) \Delta x, \end{equation*}$

where $n$ is the number of triangles we use and $Delta x$ is the size of the width of the rectangles.

(a) In this case, $n= 4$ . Since we are using rectangles with the same size, we have

(3) $\begin{equation*}\Delta x = \frac{b-a}{n} = \frac{9-5}{4}=\frac{4}{4} = 1. \end{equation*}$

Using (1), we have

(4) $\begin{equation*} R_4 = \sum_{i=1}^{4}f(x_i) \cdot 1. \end{equation*}$

The only thing we are missing now is the $x_i$ . Since we are using right endpoints, we will go from $a=5$ to $i$ by a step of $\Delta x$ . That is, $x_i = 5 + \Delta x \cdot i$ . Therefore,

(5) $\begin{align*} R_4 & = \sum_{i=1}^{4}f(x_i) \cdot 1 \\ & = f(x_1) + f(x_2) + f(x_3) + f(x_4) \\ & = \left(\frac{2}{x_1} +1\right) + \left(\frac{2}{x_2} +1\right) + \left(\frac{2}{x_3} +1\right) + \left(\frac{2}{x_4} +1\right) \\ & = \left(\frac{2}{5+1} +1\right) + \left(\frac{2}{6+1} +1\right) + \left(\frac{2}{7+1} +1\right) + \left(\frac{2}{8+1} +1\right) \\ & = \left(\frac{2}{6} +1\right) + \left(\frac{2}{7} +1\right) + \left(\frac{2}{8} +1\right) + \left(\frac{2}{9} +1\right) \\ & = \frac{1}{3} + \frac{2}{7} + \frac{1}{4} + \frac{2}{9} + 4 \\ & = 5.09126984126984.\end{align*}$

(b) In this case, the only thing that change is the $x_i$ since we are using left endpoints. We have $x_i = a + \Delta x(i-1)$ . That is, $x_i= 5 - 1 = 4.$

(6) $\begin{align*} R_4 & = \sum_{i=1}^{4}f(x_i) \cdot 1 \\ & = \left(\frac{2}{x_1} +1\right) + \left(\frac{2}{x_2} +1\right) + \left(\frac{2}{x_3} +1\right) + \left(\frac{2}{x_4} +1\right) \\ & = \left(\frac{2}{4+1} +1\right) + \left(\frac{2}{5+1} +1\right) + \left(\frac{2}{6+1} +1\right) + \left(\frac{2}{7+1} +1\right) \\ & = \left(\frac{2}{5} +1\right) + \left(\frac{2}{6} +1\right) + \left(\frac{2}{7} +1\right) + \left(\frac{2}{8} +1\right) \\ & = \frac{2}{5} + \frac{1}{2} + \frac{2}{7} + \frac{1}{4} + 4 \\ & = 5.26904761904762.\end{align*}$

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