Problem 15 - University Calculus: Early Transcendentals

Question

Use a finite sum to estimate the average value of \(f\) on the given interval by partitioning the interval into four sub intervals of equal length and evaluating \(f\) at the sub interval midpoints. $$f(x)=x^{3} \quad \text { on } \quad[0,2]$$ (Graph cant copy)

Step-by-Step Solution

Answer

The estimated average value of the function on the interval [0, 2] is 1.9375.

Step 1: Determine Subintervals

Divide the interval [0, 2] into four subintervals of equal length. Since the total length of the interval is 2, each subinterval will have a length of \( \frac{2}{4} = 0.5 \). Thus, the subintervals are [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, 2].

Step 2: Find Midpoints of Subintervals

Identify the midpoint of each subinterval. The midpoints will be used to evaluate the function: - Midpoint of [0, 0.5] is 0.25. - Midpoint of [0.5, 1] is 0.75. - Midpoint of [1, 1.5] is 1.25. - Midpoint of [1.5, 2] is 1.75.

Step 3: Evaluate Function at Midpoints

Calculate the value of \( f(x) = x^3 \) at each midpoint:- \( f(0.25) = (0.25)^3 = 0.015625 \).- \( f(0.75) = (0.75)^3 = 0.421875 \).- \( f(1.25) = (1.25)^3 = 1.953125 \).- \( f(1.75) = (1.75)^3 = 5.359375 \).

Step 4: Calculate Sum of Function Values

Sum up the values of \( f(x) \) at the midpoints. This can be used to find the approximate average value: \( 0.015625 + 0.421875 + 1.953125 + 5.359375 = 7.75 \).

Step 5: Estimate the Average Value

To estimate the average value, divide the sum from the previous step by the number of subintervals (4): \( \text{Average Value} = \frac{7.75}{4} = 1.9375 \).

Key Concepts

Finite Sum

The concept of a finite sum is crucial when estimating the average value of a function over an interval. In this particular method, the integral of a continuous function is approximated by summing the values of the function at specific points within subintervals.
This sum provides a discrete approximation of the area under the curve of the function over the interval.
When estimating the average value using finite sums, the interval

• is divided into equal subintervals.
• The function is evaluated at chosen points, often at midpoints, within those subintervals.

The sum of these evaluations provides an estimate of the total integral for the function across the entire interval. This estimated sum is then divided by the length of the original interval to calculate the average value, giving you a good approximation without performing full integration.

Subinterval Midpoints

Choosing midpoints within each subinterval to evaluate the function is a common technique used in numerical integration.
This choice helps in providing a balanced representation of the function's behavior across the subinterval.

• Each subinterval is analyzed by finding its midpoint, calculated as the average of the subinterval's boundary points.
• These midpoints are where the function is evaluated to provide a good sample of data that represents the entire interval.

In the exercise, the interval [0, 2] is divided into four subintervals. Each midpoint, such as 0.25 for [0, 0.5], is crucial since it represents a central point in each subinterval.
These midpoints are used to determine and evaluate the function value, ultimately contributing to the finite sum.

Function Evaluation

The act of function evaluation involves computing the value of the function at specific points. In the context of this exercise, the function evaluated is given by: \[ f(x) = x^3 \]By substituting the midpoints of each subinterval into the function, you observe how the function behaves at those points.

• At midpoint 0.25, the function evaluates to \((0.25)^3 = 0.015625\).
• At midpoint 0.75, it evaluates to \((0.75)^3 = 0.421875\).
• At midpoint 1.25, it evaluates to \((1.25)^3 = 1.953125\).
• At midpoint 1.75, it evaluates to \((1.75)^3 = 5.359375\).

These evaluations help establish an approximation of the underlying function's average value. Without having to fully integrate the function, these evaluations provide an insightful look into the integral's value.

Integration Approximation

Integration approximation is a practical technique used to estimate the definite integral of a function over an interval. The idea is to replace the continuous process of integration with a discrete approach that provides near-accurate results.
In this method, the sum of the products of subinterval length and function evaluation at designated points—midpoints, in this case—approximates the integral.

• This sum, 7.75 in the exercise, provides the estimated value across the full interval, based on sampled points.
• By dividing this sum by the number of subintervals, 4 here, an approximate average is gained: \( \frac{7.75}{4} = 1.9375 \).

This approach allows students to understand how integrating can be simplified using finite values and replacements. It captures essential value characteristics of the continuous function through sample representations.