Problem 20 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Suppose that \(a_{k}=f(k)\) for \(k=1,2,3, \ldots\), where \(f(x)\) is positive, decreasing, and continuous on \([1, \infty)\). Put the following expressions in order, from smallest to largest. Explain your reasoning with a picture or two. \(\sum_{k=2}^{n-1} a_{k}, \quad \sum_{k=3}^{n} a_{k}, \quad \int_{2}^{n} f(x) d x\)

Step-by-Step Solution

Answer

The expressions in ascending order are \(\sum_{k=2}^{n-1} a_{k}\), \(\sum_{k=3}^{n} a_{k}\), \(\int_{2}^{n} f(x) dx\).

Step 1: Understand the Integral and Summation Expressions

Given a decreasing continuous function \(f\), the element \(a_{k}\) can be interpreted as the value of the function at its integer points. The integral \(\int_{2}^{n} f(x) dx\) represents an area under the curve \(f(x)\) from \(x = 2\) to \(x = n\), which is approximated by the summation values. The bigger granularity of summation, the closer its value is to the integral value. By comparing the position of \(k\) in two summations, we can say \(\sum_{k=2}^{n-1} a_{k}\) skips the last element of \(\sum_{k=3}^{n} a_{k}\), so \(\sum_{k=2}^{n-1} a_{k}\) is smaller than \(\sum_{k=3}^{n} a_{k}\).

Step 2: Compare Integral with Summation Expressions

As the summation \(\sum_{k=2}^{n-1} a_{k}\) excludes the last element to \(\sum_{k=3}^{n} a_{k}\) and starts from \(k = 2\), while the integral \(\int_{2}^{n} f(x) dx\) covers the same interval, but in a more detailed manner (since it involves the entire curve), therefore, the integrated area would exceed the cumulative sum of the function's values over the same range. Hence, \(\int_{2}^{n} f(x) dx\) is larger than \(\sum_{k=2}^{n-1} a_{k}\).

Step 3: Order the expressions

Taking Step 1 and Step 2 into consideration, we can order these quantities from smallest to largest as follows: \(\sum_{k=2}^{n-1} a_{k} < \sum_{k=3}^{n} a_{k} < \int_{2}^{n} f(x) dx\).

Key Concepts

Definite Integral

A definite integral can be thought of as the total accumulation of a quantity across an interval. It's akin to summing up infinitesimal pieces, which represents areas under the curve on a graph. In the context of our exercise, the definite integral \(\int_{2}^{n} f(x) dx\) signifies the area under the function \(f(x)\) starting at \(x = 2\) and ending at \(x = n\).

Because \(f(x)\) is a continuous and decreasing function, the area under the curve from 2 to \(n\) is a clear depiction of all the values \(f\) takes on within the bounds. This understanding allows students to visualize how the definite integral relates to the sum of the values of \(f\) at discrete points and why it usually provides a larger value than a summation.

Summation of Series

• Summation involves adding up a series of numbers, which are the values of \(f(x)\) at specific integer points.
• In our case, summations \(\sum_{k=2}^{n-1} a_{k}\) and \(\sum_{k=3}^{n} a_{k}\) represent the sum of function values from \(k = 2\) to \(k = n-1\) and from \(k = 3\) to \(k = n\), respectively.

When you compare these summations to the definite integral, you notice that the sums represent step-like approximations of the area under the curve. This idea of summation as a series of steps helps illustrate why a definite integral provides a more complete measure — it fills in the gaps between the steps defined by the summations.

Continuous Functions

Continuous functions, like the one described in our exercise, have no breaks, jumps, or gaps in their domain. The feature of continuity is essential for definite integrals, as it ensures that every point in the interval contributes to the total area under the curve. The function's continuity makes the concept of integration very natural since every infinitesimal change in \(x\) leads to a corresponding change in \(f(x)\), allowing us to smoothly add up all the values.

Decreasing Functions

A decreasing function is one where, for any two points on the graph, if the second point lies to the right of the first, its function value is less than or equal to the value at the first point. In this exercise, we're dealing with such a function that continuously decreases over its domain. This property implies that as \(x\) increases, \(f(x)\) becomes smaller.

Such a characteristic is vital to understand when comparing definite integrals and summation. It explains why \(\sum_{k=2}^{n-1} a_{k}\) would be less than \(\sum_{k=3}^{n} a_{k}\) – because as the index \(k\) increases, the terms being added in the summation become smaller. It demonstrates the importance of the position of the first and last term in the summation when it comes to ordering expressions from smallest to largest.