Problem 58 - A Graphical Approach to College Algebra

Question

Evaluate 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 each function. $$f(x)=\frac{5}{2 x-1}$$

Step-by-Step Solution

Answer

The sum is \( \frac{-250}{231} \).

Step 1: Substitute into the Function

First, we need to evaluate the function \( f(x) = \frac{5}{2x-1} \) for each \( x_i \). Start with \( x_1 = 0 \):\[ f(x_1) = \frac{5}{2 \times 0 - 1} = \frac{5}{-1} = -5. \]

Step 2: Evaluate for Next x-value

Now evaluate the function for \( x_2 = 2 \):\[ f(x_2) = \frac{5}{2 \times 2 - 1} = \frac{5}{3}. \]

Step 3: Continue Evaluating the Function

Next, evaluate the function for \( x_3 = 4 \):\[ f(x_3) = \frac{5}{2 \times 4 - 1} = \frac{5}{7}. \]

Step 4: Evaluate for Last x-value

Finally, substitute \( x_4 = 6 \) into the function:\[ f(x_4) = \frac{5}{2 \times 6 - 1} = \frac{5}{11}. \]

Step 5: Sum the Function Values

Now, we sum all function values and multiply by \( \Delta x \):\[ \left( f(x_1) + f(x_2) + f(x_3) + f(x_4) \right) \Delta x = \left( -5 + \frac{5}{3} + \frac{5}{7} + \frac{5}{11} \right) \times 0.5. \]

Step 6: Calculate the Expression

Change each term \( -5 \) to \( \frac{-15}{3} \) to combine with \( \frac{5}{3} \), and combine all fractions:\[ \left( \frac{-15}{3} + \frac{5}{3} + \frac{5}{7} + \frac{5}{11} \right) = \frac{-10}{3} + \frac{5}{7} + \frac{5}{11}. \]

Step 7: Common Denominator and Calculate the Sum

Use a common denominator to calculate the sum:\[ \frac{-770}{231} + \frac{165}{231} + \frac{105}{231} = \frac{-770 + 165 + 105}{231} = \frac{-500}{231}. \]

Step 8: Multiply by Δx

Finally, multiply the result by \( \Delta x = 0.5 \):\[ \frac{-500}{231} \times 0.5 = \frac{-250}{231}. \]

Key Concepts

Definite Integrals

A definite integral is a concept from calculus that represents the area under the curve of a function between two points on the x-axis. Practically, it tells the sum or accumulation of quantities over an interval. For example, in physics, it can represent the total distance traveled or total accumulation of change.

• The limits of integration, often denoted as \(a\) and \(b\), define the start and end points, respectively, of this interval.
• In definite integrals, the function itself is continuous on the interval \([a, b]\). This means no breaks, only a smooth journey from start to finish.
• The result is a single number that summarises the entire range of the function on that interval.

In the problem at hand, we use Riemann sum as an approximation of the integral calculation. By adding up slices of rectangles with heights determined by function values and widths by \(\Delta x\), we approximate the area under the curve.

Numerical Methods

Numerical methods are techniques used to approximate solutions for mathematical problems which may not easily produce exact results. These are especially useful when tackling complex equations or functions where analytical solutions are hard to find.

• Riemann sums provide one such approach to evaluating integrals numerically. In this method, the partition of the interval into subintervals and calculating function values at specific points gives us manageable pieces to approximate the total area.
• By choosing more intervals or subintervals and varying their width (\(\Delta x\)), numerical methods can potentially increase accuracy.
• It is important to remember that these methods offer approximations that approach the "true" results as the process is refined.

Referring back to the original exercise, Riemann sums allow us to accumulate the approximated areas under the curve by choosing specific points (midpoints, left or right endpoints) for evaluation.

Function Evaluation

Function evaluation is the process of determining the output of a function given specific inputs. In our exercise, it entails substituting each \(x_i\) value into a function formula \(f(x)\) and computing the result.

• The function given here is \(f(x) = \frac{5}{2x-1}\). We evaluate it for each specified \(x_i\) value: \(x_1 = 0\), \(x_2 = 2\), \(x_3 = 4\), and \(x_4 = 6\).
• This evaluation results in various function outputs that represent the height of rectangles in a Riemann sum approximation.
• Each calculated value forms a part of the sum that, when multiplied by the interval width \(\Delta x\), contributes to the total integral approximation.

This process is a vital step to execute the numerical method for approximating the integral, ensuring that each contribution to the sum is accurately accounted for.