Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 3-6, calculate the Riemann sum \(\sum_{i=1}^{n} f\left(\bar{x}_{i}\right) \Delta x_{i}\) for the given data. \(f(x)=x-1 ; P: 3<3.75<4.25<5.5<6<7\); \(\bar{x}_{1}=3, \bar{x}_{2}=4, \bar{x}_{3}=4.75, \bar{x}_{4}=6, \bar{x}_{5}=6.5\)

Step-by-Step Solution

Answer

The Riemann sum is 15.6875.

Step 1: Understand the Problem

We are given a function \( f(x) = x - 1 \) and a partition \( P: 3<3.75<4.25<5.5<6<7 \). The task is to calculate a Riemann sum with sample points \( \bar{x}_{i} \) for each subinterval. The sample points provided are: \( \bar{x}_{1}=3 \), \( \bar{x}_{2}=4 \), \( \bar{x}_{3}=4.75 \), \( \bar{x}_{4}=6 \), and \( \bar{x}_{5}=6.5 \).

Step 2: Determine Subintervals and Their Widths

Using the partition \( P \), we identify the subintervals: \( [3, 3.75] \), \( [3.75, 4.25] \), \( [4.25, 5.5] \), \( [5.5, 6] \), and \( [6, 7] \). Calculate the width of each subinterval: \( \Delta x_1 = 3.75 - 3 = 0.75 \), \( \Delta x_2 = 4.25 - 3.75 = 0.5 \), \( \Delta x_3 = 5.5 - 4.25 = 1.25 \), \( \Delta x_4 = 6 - 5.5 = 0.5 \), and \( \Delta x_5 = 7 - 6 = 1 \).

Step 3: Calculate Function Values at Sample Points

Evaluate the function \( f(x) = x - 1 \) at each sample point: \( f(\bar{x}_1) = 3 - 1 = 2 \), \( f(\bar{x}_2) = 4 - 1 = 3 \), \( f(\bar{x}_3) = 4.75 - 1 = 3.75 \), \( f(\bar{x}_4) = 6 - 1 = 5 \), and \( f(\bar{x}_5) = 6.5 - 1 = 5.5 \).

Step 4: Compute Each Term of the Riemann Sum

Calculate each term \( f(\bar{x}_i) \Delta x_i \): - For the first interval: \( 2 \times 0.75 = 1.5 \).- For the second interval: \( 3 \times 0.5 = 1.5 \).- For the third interval: \( 3.75 \times 1.25 = 4.6875 \).- For the fourth interval: \( 5 \times 0.5 = 2.5 \).- For the fifth interval: \( 5.5 \times 1 = 5.5 \).

Step 5: Sum Up the Terms

Sum all the calculated terms: \( 1.5 + 1.5 + 4.6875 + 2.5 + 5.5 = 15.6875 \). The Riemann sum is 15.6875.

Key Concepts

Calculus

Calculus is the branch of mathematics that deals with the study of change and motion. It's divided into two main areas: differentiation, which is about calculating instantaneous rates of change, and integration, which concerns the accumulation of quantities. A Riemann sum falls under integration. This technique allows us to estimate the area under a curve using rectangles. The broader goal is understanding definite integrals, which are exact areas under curves. Calculus makes use of limits to get very precise results, where Riemann sums serve as an introductory method to intuitively grasp the concept of integration.
Calculus is essential for solving problems involving varying motion and quantities in physics, engineering, and various sciences. By learning Riemann sums, students get a foundation they can build upon as they dive deeper into integral calculus and apply these concepts in real-world scenarios.

Subintervals

Subintervals are smaller divisions within a larger interval that help us to compute Riemann sums. Imagine slicing a cheesecake into pieces; each slice is a subinterval. When you calculate a Riemann sum, you look at each subinterval individually to approximate the area under a curve, like estimating the area of one slice of cake.
In our exercise, the interval from 3 to 7 was split into subintervals:

• [3, 3.75]
• [3.75, 4.25]
• [4.25, 5.5]
• [5.5, 6]
• [6, 7]

Each subinterval has a width, which helps us determine the height of each rectangular approximation in the Riemann sum. The finer the partition, the more subintervals you have, and the more accurate your Riemann sum becomes.

Function Evaluation

Function evaluation involves plugging specific inputs into a function to get outputs. In the Riemann sum problem, you must evaluate the function at certain sample points, which are specific values chosen within each subinterval. These values help determine the height of the rectangles used in a Riemann sum.
For instance, with our function \( f(x) = x - 1 \):

• At \( x = 3 \), \( f(x) = 2 \)
• At \( x = 4 \), \( f(x) = 3 \)
• At \( x = 4.75 \), \( f(x) = 3.75 \)
• At \( x = 6 \), \( f(x) = 5 \)
• At \( x = 6.5 \), \( f(x) = 5.5 \)

By determining the function's value at these points, you can find the height of each rectangle in the sum and thus estimate the total area more effectively.

Partition

Partitioning involves dividing an overall interval into subintervals, essentially creating small sections that can be managed individually. This approach simplifies complex calculations, such as integrating a function over a range. Each section is a subset of the whole, allowing for step-by-step calculations.
In our problem, the interval from 3 to 7 was partitioned, resulting in smaller, more manageable subintervals. The quality of a partition directly influences the precision of a Riemann sum. More subintervals typically lead to a more accurate approximation of the integral.
When partitioning an interval:

• Choose points within the interval's range.
• Ensure that these points are ordered sequentially to maintain accuracy.
• Each point represents a break in the interval, thus forming subinterval boundaries.

This method forms the backbone of many integral calculus applications, providing a straightforward way of approximating complex functions.