Problem 57 - A Graphical Approach to Precalculus with Limits

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. $$f(x)=2 x^{2}$$

Step-by-Step Solution

Answer

The evaluated sum is 56.

Step 1: Evaluate the Function at the First Point

Find \(f(x_1)\) by substituting \(x_1 = 0\) into the function \(f(x) = 2x^2\). Calculate: \[f(x_1) = 2(0)^2 = 0\]

Step 2: Evaluate the Function at the Second Point

Find \(f(x_2)\) by substituting \(x_2 = 2\) into the function \(f(x) = 2x^2\). Calculate: \[f(x_2) = 2(2)^2 = 8\]

Step 3: Evaluate the Function at the Third Point

Find \(f(x_3)\) by substituting \(x_3 = 4\) into the function \(f(x) = 2x^2\). Calculate: \[f(x_3) = 2(4)^2 = 32\]

Step 4: Evaluate the Function at the Fourth Point

Find \(f(x_4)\) by substituting \(x_4 = 6\) into the function \(f(x) = 2x^2\). Calculate: \[f(x_4) = 2(6)^2 = 72\]

Step 5: Calculate Each Term with Δx

Multiply each function evaluation by \(\Delta x\) to compute the terms of the sum.- First term: \[f(x_1) \Delta x = 0 \times 0.5 = 0\]- Second term: \[f(x_2) \Delta x = 8 \times 0.5 = 4\]- Third term: \[f(x_3) \Delta x = 32 \times 0.5 = 16\]- Fourth term: \[f(x_4) \Delta x = 72 \times 0.5 = 36\]

Step 6: Sum the Terms

Add all the terms to find the total sum:\[0 + 4 + 16 + 36 = 56\]

Key Concepts

Function Evaluation

Function evaluation involves substituting a given number into a function to find its corresponding value. In this exercise, the process involves using the function \(f(x) = 2x^2\) at various points \(x_1 = 0, x_2 = 2, x_3 = 4, x_4 = 6\).
Evaluating the function means plugging these values into the function individually and performing the arithmetic.
Here’s how it works in simple steps:

• For \(x_1 = 0\), substitute 0 into the function: \(f(x_1) = 2(0)^2 = 0\).
• For \(x_2 = 2\), substitute 2 into the function: \(f(x_2) = 2(2)^2 = 8\).
• For \(x_3 = 4\), substitute 4: \(f(x_3) = 2(4)^2 = 32\).
• For \(x_4 = 6\), substitute 6: \(f(x_4) = 2(6)^2 = 72\).

This step is crucial because it forms the foundation for the calculation in a Riemann Sum, offering snapshots of the function’s behavior at certain points.

Discrete Approximation

The term discrete approximation refers to the idea of using specific values to estimate a continuous quantity. In mathematics, continuous data is broken down into a finite number of intervals or points, like the ones chosen in the exercise (\(x_1, x_2, x_3, x_4\)).
In this exercise, we pretend that the function forms straight horizontal lines over small subintervals.The Riemann Sum helps in building a sum of these horizontal slices to approximate the area under a curve. To do this:

• The value of the function at each chosen point is taken.
• Each function evaluation is then multiplied by a fixed small width, \(\Delta x = 0.5\), representing the width of each interval.

This method enables approximate integration by replacing a potentially complex curve with steps, similar to staircases, thereby making calculations feasible with known values.

Summation Notation

Summation notation is a concise way of expressing the addition of a sequence of numbers. It's represented by the Greek letter sigma (\(\sum\)). In Riemann Sums, the summation symbol indicates that multiple terms need to be added together.For this exercise, the notation \(\sum_{i=1}^{4} f(x_i) \Delta x\) represents four evaluations of the function at specific points, each multiplied by the common width \(\Delta x = 0.5\), being added together.
Here's how it translates step-by-step:

• Every term \(f(x_i) \Delta x\) corresponds to a specific piece of the area approximation under the curve.
• To find the complete approximate area, compute and add all terms: \(f(x_1) \Delta x + f(x_2) \Delta x + f(x_3) \Delta x + f(x_4) \Delta x\).

In this context, summation notation simplifies the calculation of the total estimate, conveniently aggregating the discrete contributions into one compact formula.