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Magazine Chapter 11: Problem 2
Suppose \(f\) is a continuous positive decreasing function for \(x \geqslant 1\) and \(a_{n}=f(n)\). By drawing a picture, rank the following three quantities in increasing order: $$ \int_{1}^{6} f(x) d x \quad \sum_{i=1}^{5} a_{i} \quad \sum_{i=2}^{6} a_{i} $$
Short Answer
Expert verified
\( \int_{1}^{6} f(x) \, dx < \sum_{i=1}^{5} a_{i} < \sum_{i=2}^{6} a_{i} \).
Step by step solution
01
Understand the Problem
We have a function \( f(x) \) which is continuous, positive, and decreasing from \( x \geq 1 \). We are given three quantities to rank: \( \int_{1}^{6} f(x) \, dx \), \( \sum_{i=1}^{5} a_{i} \), and \( \sum_{i=2}^{6} a_{i} \). Here, \( a_{n} = f(n) \). Our goal is to rank these quantities in increasing order.
02
Visualize the Function and Sums
Imagine \( f(x) \), decreasing as \( x \) increases past 1. Visualize it on a graph, noting that the integral \( \int_{1}^{6} f(x) \, dx \) corresponds to the area under \( f(x) \) from 1 to 6. The sum \( \sum_{i=1}^{5} a_{i} \) is the sum of the values \( f(1) + f(2) + \ldots + f(5) \), and \( \sum_{i=2}^{6} a_{i} \) is \( f(2) + f(3) + \ldots + f(6) \).
03
Compare \( \sum_{i=1}^{5} a_{i} \) and \( \int_{1}^{6} f(x) \, dx \)
Since \( f(x) \) is decreasing, \( \sum_{i=1}^{5} a_{i} = f(1) + f(2) + \ldots + f(5) \) represents rectangles that are higher than the integral's area under the curve over the same range, missing the part from 5 to 6. Therefore, \( \sum_{i=1}^{5} a_{i} > \int_{1}^{6} f(x) \, dx \).
04
Compare \( \sum_{i=2}^{6} a_{i} \) with \( \sum_{i=1}^{5} a_{i} \)
Notice that \( \sum_{i=2}^{6} a_{i} = f(2) + f(3) + \ldots + f(6) \). Here, each term \( a_n \) is slightly larger because it shifts forward, covering the value \( f(6) \) not included in \( \sum_{i=1}^{5} a_{i} \). Thus, \( \sum_{i=2}^{6} a_{i} > \sum_{i=1}^{5} a_{i} \).
05
Rank the Quantities
Given the conclusions from the above steps, we now have \( \int_{1}^{6} f(x) \, dx < \sum_{i=1}^{5} a_{i} < \sum_{i=2}^{6} a_{i} \). This ordering is based on both geometric intuition and the properties of decreasing functions, where sums across intervals generate larger values that include additional terms.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
In calculus, a definite integral helps us find the area under a curve over a specific interval. This is essentially the sum of infinite tiny rectangles under the curve. For our function, represented as \( f(x) \), the definite integral from 1 to 6, written as \( \int_{1}^{6} f(x) \, dx \), gives us the total area under the function from \( x=1 \) to \( x=6 \).
This is a fundamental concept used to calculate things like distance, area, and even volume when the function and its boundaries are known. In evaluating definite integrals, especially with a decreasing function, the result reflects how the function steadily reduces its value as \( x \) increases.
This is a fundamental concept used to calculate things like distance, area, and even volume when the function and its boundaries are known. In evaluating definite integrals, especially with a decreasing function, the result reflects how the function steadily reduces its value as \( x \) increases.
- The notation \( \int_{a}^{b} f(x) \, dx \) specifies the lower limit \( a \) and the upper limit \( b \) for the interval.
- This integral evaluates how much area is captured under the curve, offering a graphical interpretation of accumulation or total change.
Riemann Sums
Riemann sums are techniques to approximate the definite integral of a function. Think of it as summing up areas of rectangles to estimate the area under a curve. Each rectangle's height is determined by the function value at specific points.
There are generally three types of Riemann sums: left, right, and midpoint, dependent on which part of the interval is used for determining the height.
There are generally three types of Riemann sums: left, right, and midpoint, dependent on which part of the interval is used for determining the height.
- The left Riemann sum uses left endpoints, the right one uses the right endpoints, and the midpoint uses the midpoints for the rectangle heights.
- This technique is specifically advantageous for getting a rough estimate of the integral, especially helpful when a function is complex.
Decreasing Functions
A decreasing function is one where the value of the function decreases as its input increases. In simpler terms, as you move along the horizontal axis (\( x \)) to the right, the function moves downwards.
This characteristic plays a pivotal role when evaluating sums or integrals, particularly because the terms in a sum will have progressively lower values.
With decreasing functions like \( f(x) \), certain properties become evident:
This characteristic plays a pivotal role when evaluating sums or integrals, particularly because the terms in a sum will have progressively lower values.
With decreasing functions like \( f(x) \), certain properties become evident:
- For a decreasing function, \( f(a) > f(b) \) if \( a < b \).
- It implies that the function's graph slopes downward, and its derivative \( f'(x) \) is always negative.
Function Approximation
Function approximation involves estimating a complex function with simpler, often polynomial functions. It helps in making calculations and predictions manageable.
Riemann sums and definite integrals can be seen as forms of function approximation; however, these methods use rectangles instead of polynomials. Function approximation is particularly useful when working with continuous but decreasing functions like \( f(x) \) because:
Riemann sums and definite integrals can be seen as forms of function approximation; however, these methods use rectangles instead of polynomials. Function approximation is particularly useful when working with continuous but decreasing functions like \( f(x) \) because:
- They allow complex areas under curves to be estimated using simpler geometric shapes.
- They assist in visualizing how the summed value compares to the "exact" value (integral), especially when the function's values decline steadily.
