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In Exercises 15-18, use a finite sum to estimate the average value of \(f\) on the given interval by partitioning the interval into four subintervals of equal length and evaluating \(f\) at the subinterval midpoints. $$ f(x)=x^{3} \quad \text { on } \quad[0,2] $$

Short Answer

Expert verified
The estimated average value of \(f(x)\) on \([0, 2]\) is 3.875.

Step by step solution

01

Understand the Problem

We need to find the average value of the function \(f(x) = x^3\) on the interval \([0, 2]\) by partitioning the interval into four equal subintervals and evaluating the function at the midpoints of these subintervals.
02

Determine the Length of Each Subinterval

The interval \([0, 2]\) has a total length of 2. Dividing it into four equal subintervals gives each subinterval a length of \(\Delta x = \frac{2 - 0}{4} = 0.5\).
03

Identify Subintervals and Their Midpoints

The subintervals are \([0, 0.5]\), \([0.5, 1]\), \([1, 1.5]\), and \([1.5, 2]\). The midpoints are calculated as follows: \(0.25\), \(0.75\), \(1.25\), and \(1.75\).
04

Evaluate f at Each Midpoint

Evaluate \(f(x) = x^3\) at each midpoint: - For \(x = 0.25\), \(f(0.25) = (0.25)^3 = 0.015625\).- For \(x = 0.75\), \(f(0.75) = (0.75)^3 = 0.421875\).- For \(x = 1.25\), \(f(1.25) = (1.25)^3 = 1.953125\).- For \(x = 1.75\), \(f(1.75) = (1.75)^3 = 5.359375\).
05

Calculate the Finite Sum

Add the values of \(f(x)\) at the midpoints and multiply by \(\Delta x = 0.5\):\[\frac{1}{4} (0.015625 + 0.421875 + 1.953125 + 5.359375) = 0.5 \times (0.015625 + 0.421875 + 1.953125 + 5.359375)\]
06

Compute the Average Value

Calculate \[0.5 \times (0.015625 + 0.421875 + 1.953125 + 5.359375) = 0.5 \times 7.75 = 3.875\]So, the average value of the function over the interval is \(3.875\).

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Key Concepts

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

Finite Sum Approximation
When estimating the average value of a function over an interval, finite sum approximation is a crucial method. This approach helps us approximate an integral by summing a finite number of discrete terms. In simpler terms, instead of calculating the area under a curve precisely (which can be complex), we approximate it by adding up smaller, manageable pieces.

Consider the function \( f(x) = x^3 \) on the interval \([0, 2]\). We partition this interval into smaller equal parts to ease the computation. Think of it like dividing a cake into equal slices to better share with guests. Each slice represents a part of the area under the curve, and by calculating the function's value at certain points (like the midpoint), we can sum these up to find an approximate area.

This approximation becomes particularly handy in applications where we deal with real-life data that does not conform to precise mathematical functions. Finite sum approximation simplifies complex problems, making them accessible and manageable with limited computation resources.
Partitioning Intervals
Partitioning intervals is a technique that breaks down a given interval into smaller, equal parts. In the context of finding an average value of a function, it involves dividing the entire interval over which the function is defined into smaller subintervals. This helps in better managing data and simplifying calculations.

For the function \( f(x) = x^3 \) on the interval \([0, 2]\), we divide this range into four equal subintervals. Here, each subinterval length is computed as \( \Delta x = \frac{2 - 0}{4} = 0.5 \). The idea is that by shrinking the interval size, each part becomes easier to evaluate and analyze.

Partitioning this way facilitates evaluating the function at more strategic points, like the "midpoints". This not only provides a more accurate approximation but also helps check the function's behavior in smaller, more precise segments, which is critical in understanding the overall trend of the function throughout the interval.
Midpoint Evaluation
Midpoint evaluation is a precise method used to approximate the function's behavior over an interval, especially when it comes to finite sum approximation. By computing the value of the function at the midpoint of each subinterval, we get a better estimate of the average value for that segment.

In the example of the function \( f(x) = x^3 \) on the interval \([0, 2]\), the midpoints for each subinterval \([0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]\) are \(0.25, 0.75, 1.25,\) and \(1.75\) respectively. Evaluating the function at these midpoints gives the values \(0.015625, 0.421875, 1.953125,\) and \(5.359375\).

The strategy of taking midpoints helps in canceling out negative and positive deviations within each subinterval, thus giving an overall more balanced and accurate approximation of the function's true behavior. This technique is particularly useful because it accounts for local variations within the interval, providing insights into the function's characteristics that would be missed by just evaluating endpoints or random points.

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