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Magazine Chapter 1: Problem 133
Suppose that \([a, b]\) can be partitioned. taking
\(a=a_{0}
Short Answer
Expert verified
The average value of \(f\) over \([a, b]\) is 1.
Step by step solution
01
Understanding the Problem
We are given a function \(f\) defined over a closed interval \([a,b]\) which is partitioned into subintervals \([a_{i-1}, a_{i}]\) such that the average value of \(f\) over each subinterval is 1. Our task is to show that the average value of \(f\) over the entire interval \([a, b]\) is also 1.
02
Calculate Individual Subinterval Averages
The average value of \(f\) over each subinterval \([a_{i-1}, a_i]\) is given by the formula: \[\frac{1}{a_{i} - a_{i-1}} \int_{a_{i-1}}^{a_{i}} f(x) \, dx = 1\]This formula implies that the integral of \(f\) over each subinterval is \[\int_{a_{i-1}}^{a_{i}} f(x) \, dx = a_{i} - a_{i-1}\].
03
Sum the Integrals Over All Subintervals
To find the total integral over the entire interval \([a, b]\), we sum the integrals over all subintervals: \[\int_{a}^{b} f(x) \, dx = \sum_{i=1}^{N} \int_{a_{i-1}}^{a_{i}} f(x) \, dx = \sum_{i=1}^{N} (a_{i} - a_{i-1})\]
04
Calculate the Average Value Over the Entire Interval
The average value of \(f\) over the entire interval \([a, b]\) is given by:\[\frac{1}{b-a} \int_{a}^{b} f(x) \, dx = \frac{1}{b-a} \sum_{i=1}^{N} (a_{i} - a_{i-1})\]Since the total length of the interval is \(b-a\), it follows that: \[\frac{b-a}{b-a} = 1\].
05
Conclude with the Result
Since the expression simplies to 1, the average value of the function \(f\) over the entirety of \([a, b]\) is equal to 1.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partition of an Interval
In mathematics, a partition of an interval is a crucial concept used to break down a larger interval \( [a, b] \) into smaller sections. This process is extremely helpful in various calculus applications, such as integration and analysis. A partition can be represented by a finite sequence \( a = a_0 < a_1 < ext{...} < a_N = b \). By dividing an entire interval into these subintervals, you allow for more precise computations and a better understanding of the function's behavior over the interval.
- Each subinterval \( [a_{i-1}, a_i] \) is defined by adjacent endpoints of the partition.
- These partitions help in averaging, which is needed to determine how each part of the interval contributes to the whole.
- The size and number of these subintervals can vary, based on the problem at hand, leading to different levels of precision.
Riemann Sum
Riemann sums are a foundational concept in calculus used to approximate the area under a curve. The Riemann sum is particularly effective when a function is divided into subintervals and summed up. For a function defined on a closed interval \( [a, b] \), its Riemann sum is an approximation of the area's integral:
- It involves summing the products of values of the function at specified points within subintervals and the length of these subintervals.
- Mathematically, this can be represented by \( \sum_{i=1}^{N} f(x_i^*) \cdot (a_i - a_{i-1}) \), where \( x_i^* \) is a point in the subinterval \( [a_{i-1}, a_i] \).
- Riemann Sums are crucial for approaching the concept of definite integrals.
Definite Integral
The definite integral is used to calculate the accumulation of quantities and is the exact value representing the area under a curve over an interval. It builds directly upon the concept of Riemann sums. Rather than approximating, the integral provides exactness when the number of partitions increases indefinitely:
- The expression for a definite integral from \( a \) to \( b \) is written as \( \int_{a}^{b} f(x) \, dx \).
- It encapsulates the net signed area, taking into account the function's positive and negative values over the interval.
- In this context, the definite integral plays a key role in confirming the consistency of all subinterval averages in the partition.
Subinterval Analysis
Analyzing subintervals is crucial for understanding the behavior of functions over a specified domain. By examining subintervals, one can infer information about the larger interval. This is particularly important when computing average values or integrals:
- Each subinterval \( [a_{i-1}, a_i] \) is assessed to determine the function's average value locally.
- The behavior in each piece contributes to conclusions about the broader interval, as the average over the entire interval is influenced by these components.
- Through subinterval analysis, variations within the interval are acknowledged and accounted for in computations.
