Problem 57 - 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. $$f(x)=\frac{-2}{x+1}$$

Step-by-Step Solution

Answer

The sum is \(-\frac{176}{105}\).

Step 1: Understand the Sum

We need to evaluate \(\sum_{i=1}^{4} f(x_{i}) \Delta x\), which means we'll calculate \(f(x_i)\) for each given \(x_i\) and multiply it by \(\Delta x\). Then, sum these products.

Step 2: Calculate \(f(x_1)\)

Substitute \(x_1 = 0\) into \(f(x) = \frac{-2}{x+1}\): \[f(x_1) = \frac{-2}{0+1} = -2\]

Step 3: Calculate \(f(x_2)\)

Substitute \(x_2 = 2\) into the function:\[f(x_2) = \frac{-2}{2+1} = \frac{-2}{3}\]

Step 4: Calculate \(f(x_3)\)

Substitute \(x_3 = 4\) into the function:\[f(x_3) = \frac{-2}{4+1} = \frac{-2}{5}\]

Step 5: Calculate \(f(x_4)\)

Substitute \(x_4 = 6\) into the function:\[f(x_4) = \frac{-2}{6+1} = \frac{-2}{7}\]

Step 6: Multiply Each Result by \(\Delta x\)

Each \(f(x_i)\) is multiplied by \(\Delta x = 0.5\): - \(f(x_1)\): \(-2 \times 0.5 = -1\)- \(f(x_2)\): \(\frac{-2}{3} \times 0.5 = \frac{-1}{3}\)- \(f(x_3)\): \(\frac{-2}{5} \times 0.5 = \frac{-1}{5}\)- \(f(x_4)\): \(\frac{-2}{7} \times 0.5 = \frac{-1}{7}\)

Step 7: Sum the Products

Sum all the multiplied products:\[-1 + \frac{-1}{3} + \frac{-1}{5} + \frac{-1}{7}\]

Step 8: Simplify the Sum

Calculate the final answer by combining the terms: \[-1 - \frac{1}{3} - \frac{1}{5} - \frac{1}{7} = -1 - \frac{35}{105} - \frac{21}{105} - \frac{15}{105}\] Combine them over a common denominator:\[-1 - \frac{71}{105} = -\frac{176}{105}\]

Key Concepts

Function Evaluation

When approaching a problem involving Riemann sums, understanding how to evaluate a function is crucial. In this exercise, we need to calculate the function value at specific points. These points are given by the values of \(x_1, x_2, x_3,\) and \(x_4\). Each of these values is substituted into the given function \(f(x) = \frac{-2}{x+1}\). Evaluating a function involves replacing the variable \(x\) with a specific number and simplifying the expression that results. This step is critical in any calculus problem, as it helps determine part of the value needed for a larger calculation or sum.
It's important that you handle each substitution carefully, following the order of operations to ensure accuracy. For example, substituting \(x_1 = 0\) gives \(f(x_1) = \frac{-2}{0+1} = -2\), where we first add one to zero, then divide \(-2\) by the result.

Substitution Method

The substitution method is a technique used to make complex problems more manageable. In our exercise, it involves inserting the specific \(x\) values into the function one by one. This lets us evaluate \(f(x)\) for each \(x_i\) in the sum.

• Substitute \(x_1 = 0\) to get \(f(x_1) = -2\).
• Substitute \(x_2 = 2\) to get \(f(x_2) = \frac{-2}{3}\).
• Substitute \(x_3 = 4\) to get \(f(x_3) = \frac{-2}{5}\).
• Substitute \(x_4 = 6\) to get \(f(x_4) = \frac{-2}{7}\).

Once the substitution is complete, you have all the values needed for the next steps in solving the problem. This method is effective because it breaks down a complex operation into simpler, concrete calculations.

Fraction Simplification

Fractions often appear in mathematical problems, and simplifying them is key to obtaining an accurate result. In our exercise, after evaluating the function for each \(x_i\), we need to multiply by the given \(\Delta x = 0.5\). Each product becomes a fraction that must be simplified:

• \(-2 \times 0.5 = -1\)
• \(\frac{-2}{3} \times 0.5 = \frac{-1}{3}\)
• \(\frac{-2}{5} \times 0.5 = \frac{-1}{5}\)
• \(\frac{-2}{7} \times 0.5 = \frac{-1}{7}\)

The simplification process helps in converting these into a manageable form that is easier to combine or compare. Always aim to reduce fractions to their simplest form by dividing the numerator and the denominator by their greatest common factor.

Definite Integrals

A definite integral represents the signed area under a curve, from one specific point to another along the x-axis. The result from evaluating a Riemann sum approximates this area. In our exercise, the Riemann sum involves a discrete sum of area segments defined by \(\Delta x = 0.5\) and calculated function values at intervals \(x_1, x_2, x_3,\) and \(x_4\).
By multiplying each evaluated function by \(\Delta x\), we find a series of small areas which add together to approximate the integral from \(x_1\) to \(x_4\). The sum of the products is simplified:\[-1 + \frac{-1}{3} + \frac{-1}{5} + \frac{-1}{7}\]Simplifying further helps estimate the total area: \[-1 - \frac{71}{105} = -\frac{176}{105}\].This result approaches what you would find if solving the definite integral using other methods. Definite integrals are fundamental in calculus for understanding accumulation of quantities.