Answered: Find the upper and lower sums for the…

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter7: Integration

Section7.CR: Chapter 7 Review

Problem 54CR

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Question

Find the upper and lower sums for the region bounded by the graph of the function and the x-axis on the given interval. Leave your answer in terms of n, the number of subintervals. Function f(x) = 4x + 1 Interval [2, 3]

Expert Solution

Step 1

Consider the following function:

$f (x) = 4 x + 1$ on interval $[2, 3]$

Partition of the interval $[2, 3]$ into $n$ subintervals:

$\begin{array}{rcl} ∆ x & = & \frac{b - a}{n} \\ = & \frac{3 - 2}{n} \\ = & \frac{1}{n} \end{array}$

Step 2

Left endpoints:

$\begin{array}{rcl} m_{i} & = & 2 + (i - 1) \frac{1}{n} \\ = & 2 + \frac{i - 1}{n} \end{array}$

Lower sum:

$\begin{array}{rcl} s (n) & = & \sum_{i = 1}^{n} f (m_{i}) ∆ x \\ = & \sum_{i = 1}^{n} [4 (2 + \frac{i - 1}{n}) + 1] (\frac{1}{n}) \\ = & \sum_{i = 1}^{n} [8 + \frac{4 (i - 1)}{n} + 1] (\frac{1}{n}) \\ = & \sum_{i = 1}^{n} 9 (\frac{1}{n}) + \sum_{i = 1}^{n} \frac{4 (i - 1)}{n} (\frac{1}{n}) \\ = & \frac{9}{n} \sum_{i = 1}^{n} 1 + \frac{4}{n^{2}} \sum_{i = 1}^{n} (i - 1) \end{array}$

$\begin{array}{rcl} s (n) & = & \frac{9}{n} \sum_{i = 1}^{n} 1 + \frac{4}{n^{2}} [\sum_{i = 1}^{n} i - \sum_{i = 1}^{n} 1] \\ = & \frac{9}{n} \cdot n + \frac{4}{n^{2}} \cdot \frac{n (n + 1)}{2} - \frac{4}{n^{2}} \cdot n \\ = & 9 + \frac{2 (n + 1)}{n} - \frac{4}{n} \\ = & 9 + \frac{2 n - 2}{n} \\ = & 11 - \frac{2}{n} \end{array}$

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