Problem 61 - Calculus Early Transcendentals

Question

Fill in the blanks with right or midpoint, an interval, and a value of \(n\). In some cases, more than one answer may work.\(\sum_{k=1}^{4} f(1.5+k) \cdot \)1 is a ______ Riemann sum for \(f\) on the interval [ ____ ,_____ ]with \(n=\) ______ .

Step-by-Step Solution

Answer

On which interval and with what value of \(n\) does it approximate the integral? Answer: The given sum represents a right Riemann sum for \(f\) on the interval [1.5, 5.5] with \(n=4\).

Step 1: Identify the type of Riemann sum

To identify the type of Riemann sum, we will compare the given sum to the standard formats for right and midpoint Riemann sums. A right Riemann sum has the general form: \(\sum_{k=1}^{n} f(x_k) \cdot \Delta x\) where \(x_k = a + k\Delta x\) A midpoint Riemann sum has the general form: \(\sum_{k=1}^{n} f(\overline{x}_k) \cdot \Delta x\) where \(\overline{x}_k = a + \frac{\Delta x}{2} + (k-1)\Delta x\) The given sum looks like: \(\sum_{k=1}^{4} f(1.5+k) \cdot\)1 Comparing this sum with the standard formats, we can see that it resembles more a right Riemann sum. So, it is a right Riemann sum.

Step 2: Determine the interval

To determine the interval for which the Riemann sum is calculated, we'll use the information from the general form of the right Riemann sum: \(\sum_{k=1}^{n} f(x_k) \cdot \Delta x\) where \(x_k = a + k\Delta x\) In our given sum, we have \(f(1.5 + k)\) and \(\Delta x = 1\). We can rewrite this as \(f(a + k\cdot 1)\). By comparing this with the general form \(f(a + k\Delta x)\), we obtain that \(a = 1.5\). Now, to find the interval, we need to determine the values of \(a\) and \(b\). Since \(n = 4\), we can obtain \(b\) as follows: \(a + n\Delta x = 1.5 + 4\cdot 1 = 1.5 + 4 = 5.5\) So, the interval is [1.5, 5.5].

Step 3: Determine the value of \(n\)

Since we have already identified the type of Riemann sum and the interval, we can quickly determine the value of \(n\). In our given sum, \(\sum_{k=1}^{4} f(1.5+k) \cdot\)1, the highest value of \(k\) is 4. Hence, \(n = 4\). Now, we can fill in the blanks for the given exercise: \(\sum_{k=1}^{4} f(1.5+k) \cdot\)1 is a right Riemann sum for \(f\) on the interval [1.5, 5.5] with \(n=\)4.

Key Concepts

Right Riemann Sum

The Right Riemann Sum is one of the methods used in calculus to approximate the area under a curve. In this approach, we divide the curve into equal sections or intervals. These intervals are matched with rectangles that touch the curve at the right endpoint of each interval.
It leverages the formula:

• \( \sum_{k=1}^{n} f(x_k) \cdot \Delta x \)

where \( x_k = a + k\Delta x \).Right Riemann sums take advantage of this technique by using the right edge of each interval to evaluate the function. This is especially useful when the curve is increasing, as it might give a better estimate of the total area.
In our exercise, when the term \(f(1.5 + k)\) appears with \(\Delta x = 1\), it signifies that each interval's right endpoint is evaluated, showcasing the right Riemann sum approach.

Midpoint Riemann Sum

While the Right Riemann Sum evaluates the function at the right edge of each interval, the Midpoint Riemann Sum takes a different approach. It evaluates the function at the midpoint of each interval, providing another means of estimating the area under a curve. The formula can be described as:

• \( \sum_{k=1}^{n} f(\overline{x}_k) \cdot \Delta x \)

where \( \overline{x}_k = a + \frac{\Delta x}{2} + (k-1)\Delta x \).This method often gives a better approximation compared to the Right Riemann Sum, especially when the function behaves more unpredictably. By evaluating in the middle of each interval, this sum can minimize the error that comes from the function's variations.
For example, if the function fluctuates, using the midpoint can "average out" these changes, potentially leading to a more accurate area calculation. Although not used in our current exercise, understanding this method broadens the options for tackling similar problems.

Interval Identification

In the context of Riemann sums, identifying the correct interval is crucial since it defines the boundaries of the area you are estimating. An interval is represented as [a, b], where "a" is the start and "b" is the end of the range over which you're summing.In the given exercise, we start by analyzing the function within known parameters. Using the formula \( a + n\Delta x = b \), one determines the endpoints. Given that \(a = 1.5\) and \(n = 4\), with \(\Delta x = 1\), the boundaries were calculated as

• \( a = 1.5 \)
• \( b = 5.5 \)

Therefore, the interval over which the sum proceeds is [1.5, 5.5]. Intervals guide the calculation process, ensuring the sum accurately reflects the function's behavior across the specified domain.

Delta x

In the context of Riemann sums, \( \Delta x \) represents the width of each interval. It is crucial because it determines how finely the area under the curve is sliced, directly affecting the accuracy of the approximation.Defined as \( \Delta x = \frac{b-a}{n} \), it facilitates the partitioning of the interval into \( n \) sub-intervals. For our task, since \( \Delta x = 1 \) was given, it implies each slice or interval is one unit wide.Understanding \( \Delta x \) helps in accurately setting up the sum needed for the approximation. A larger \( \Delta x \) suggests fewer, broader intervals, which might lead to less accuracy. Conversely, a smaller \( \Delta x \) results in more, narrower intervals, increasing the approximation precision.