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Chapter 11: Problem 65

Write the terms of \(\sum_{i=1}^{4} f\left(x_{i}\right) \Delta x,\) with \(x_{1}=0, x_{2}=2, x_{3}=4, x_{4}=6,\) and \(\Delta x=0.5,\) for =ach function. Evaluate the sum. $$f(x)=\frac{-2}{x+1}$$

Short Answer

Expert verified
The sum is approximately -1.676.

Step by step solution

01

- Identify the given values

The values given are: - \( x_1 = 0 \) - \( x_2 = 2 \) - \( x_3 = 4 \) - \( x_4 = 6 \) - \( \Delta x = 0.5 \)
02

- Write the function values

Calculate each term \( f(x_i) \): - \( f(x_1) = f(0) = \frac{-2}{0+1} = -2 \) - \( f(x_2) = f(2) = \frac{-2}{2+1} = \frac{-2}{3} \) - \( f(x_3) = f(4) = \frac{-2}{4+1} = \frac{-2}{5} \) - \( f(x_4) = f(6) = \frac{-2}{6+1} = \frac{-2}{7} \)
03

- Multiply each function value by \( \Delta x \)

Each term should be multiplied by \( \Delta x \): - \( f(0) \Delta x = -2 \times 0.5 = -1 \) - \( f(2) \Delta x = \frac{-2}{3} \times 0.5 = \frac{-1}{3} \) - \[ f(4) \Delta x = \frac{-2}{5} \times 0.5 = \frac{-1}{5} \] - \( f(6) \Delta x = \frac{-2}{7} \times 0.5 = \frac{-1}{7} \)
04

- Sum the terms

The sum of the terms is: \[ (-1) + \left( \frac{-1}{3} \right) + \left( \frac{-1}{5} \right) + \left( \frac{-1}{7} \right) \] Simplify the sum: \[ -1 - \frac{1}{3} - \frac{1}{5} - \frac{1}{7} \] Group and simplify: \[-1 - \frac{1}{3} - \frac{1}{5} - \frac{1}{7}\approx -1-0.333 -0.2 -0.143 = -1.676 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integrals
Definite integrals are a fundamental concept in calculus. They represent the total accumulation of quantities, which can be thought of as the area under the curve of a function. The notation for a definite integral is \(\int_{a}^{b} f(x) \, dx\), where \([a, b]\) defines the interval over which the function \(~f(x)~\) is integrated.
If you imagine the graph of a function, the definite integral sums up the infinitely small areas between the graph and the x-axis, from point \(a to b\). This is useful in applications like finding distances, areas, volumes, and more.
Function Evaluation
Function evaluation simply means finding the value of a function at a specific point. For instance, if we have a function \(f(x)\) and we need to find \(f(a)\), we substitute \(a\) into the function.
In the example given, \(f(x)=\frac{-2}{x+1}\), evaluating this function at specific points involves:
  • Substituting \(x_1=0\) into the function: \(f(0)= \frac{-2}{0+1}= -2\)
  • Substituting \(x_2=2\) into the function: \(f(2)=\frac{-2}{2+1}=\frac{-2}{3}\)
  • Substituting \(x_3=4\) into the function: \(f(4)=\frac{-2}{4+1}=\frac{-2}{5}\)
  • Substituting \(x_4=6\) into the function: \(f(6)=\frac{-2}{6+1}=\frac{-2}{7}\)
These steps are straightforward but important as they lay the groundwork for further calculations like those in a Riemann Sum.
Summation Notation
Summation notation, represented by the Greek letter sigma (\(\sum\)), is a concise way to write the sum of a sequence of terms. It is essential for expressing series and can significantly simplify the representation of sums.
In the problem given, \(\sum_{i=1}^{4} f(x_{i}) \Delta x\) means summing up the terms \(f(x_i) \Delta x\) for \(i\) from 1 to 4. Here, each term \(f(x_{i})\) is the function evaluated at \(x_{i}\), and then multiplied by \(\Delta x\). The steps were:
  • Calculate the function values: \[f(0)=-2, f(2)=\frac{-2}{3}, f(4)=\frac{-2}{5}, f(6)=\frac{-2}{7}\]
  • Multiply each by \(\Delta x\) (here, 0.5): \[f(0) \cdot 0.5=-1, f(2) \cdot 0.5=\frac{-1}{3}, f(4) \cdot 0.5=\frac{-1}{5}, f(6) \cdot 0.5=\frac{-1}{7}\]
  • Sum the resulting terms: \(-1+\frac{-1}{3}+\frac{-1}{5}+\frac{-1}{7}=-1-0.333-0.2-0.143\approx -1.676\)
This method breaks down a complicated integration problem into manageable steps.

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