Problem 36 - Calculus Single Variable

Question

In each of Exercises \(27-38\), calculate the right endpoint approximation of the area of the region that lies below the graph of the given function \(f\) and above the given interval \(I\) of the \(x\) -axis. Use the uniform partition of given order \(N\). $$ f(x)=\sec (x) \quad I=[-\pi / 3, \pi / 3], N=4 $$

Step-by-Step Solution

Answer

The right endpoint approximation is approximately 1.81.

Step 1: Determine Partition Width

First, calculate the width of each subinterval, denoted as \( \Delta x \). The interval \( I = [-\frac{\pi}{3}, \frac{\pi}{3}] \) has a total length of \( \frac{\pi}{3} - (-\frac{\pi}{3}) = \frac{2\pi}{3} \). With \( N = 4 \) subintervals, the width \( \Delta x \) is given by:\[ \Delta x = \frac{\frac{2\pi}{3}}{4} = \frac{\pi}{6} \]

Step 2: Identify Right Endpoints

Next, calculate the right endpoints of each subinterval. Starting from \(-\frac{\pi}{3}\) and moving to the right with step size \( \Delta x = \frac{\pi}{6} \), the right endpoints \( x_i \) are:- \( x_1 = -\frac{\pi}{3} + \frac{\pi}{6} = -\frac{\pi}{6} \)- \( x_2 = -\frac{\pi}{3} + 2 \times \frac{\pi}{6} = 0 \)- \( x_3 = -\frac{\pi}{3} + 3 \times \frac{\pi}{6} = \frac{\pi}{6} \)- \( x_4 = -\frac{\pi}{3} + 4 \times \frac{\pi}{6} = \frac{\pi}{3} \)

Step 3: Evaluate the Function at Right Endpoints

Evaluate \( f(x) = \sec(x) \) at each right endpoint:- \( f(-\frac{\pi}{6}) = \sec(-\frac{\pi}{6}) = \frac{2}{\sqrt{3}} \)- \( f(0) = \sec(0) = 1 \)- \( f(\frac{\pi}{6}) = \sec(\frac{\pi}{6}) = \frac{2}{\sqrt{3}} \)- \( f(\frac{\pi}{3}) = \sec(\frac{\pi}{3}) = 2 \)

Step 4: Calculate the Right Endpoint Approximation

Calculate the approximation using the formula:\[ R_N = \Delta x \left( f(x_1) + f(x_2) + f(x_3) + f(x_4) \right) \]Substituting the values, we have:\[ R_4 = \frac{\pi}{6} \left( \frac{2}{\sqrt{3}} + 1 + \frac{2}{\sqrt{3}} + 2 \right) \]Calculating the sum inside the parentheses:\[ \frac{2}{\sqrt{3}} + 1 + \frac{2}{\sqrt{3}} + 2 = \frac{4}{\sqrt{3}} + 3 \]Finally, calculate:\[ R_4 = \frac{\pi}{6} \left( \frac{4}{\sqrt{3}} + 3 \right) \approx \frac{\pi}{6} \left( \frac{4}{1.732} + 3 \right) \approx 1.81 \]

Key Concepts

Partition Width

The partition width, denoted as \( \Delta x \), is a crucial concept when dividing an interval into smaller parts, known as subintervals. When calculating the right endpoint approximation, the function within a given interval is divided into equal parts. This ensures that each subinterval has the same width. The interval \( I = \left[ -\frac{\pi}{3}, \frac{\pi}{3} \right] \) is segmented based on the number \( N \) of subintervals required. To find the partition width, we compute the total interval length and divide it by \( N \), which is 4 in this scenario.

• Total interval length: \( \frac{2\pi}{3} \)
• Partition width: \( \Delta x = \frac{2\pi/3}{4} = \frac{\pi}{6} \)

This consistent subinterval length simplifies the task of locating right endpoints, as each step between points is precisely the partition width \( \Delta x \).

Right Endpoints

In the context of right endpoint approximation, identifying the right endpoints is essential for evaluating the function across the specified interval. These endpoints are the finishing points of each subinterval. To find them, start at the beginning of the interval and add the partition width \( \Delta x \) successively.

• Starting at \(-\frac{\pi}{3}\), increment by \( \frac{\pi}{6} \) to find each endpoint:
• \( x_1 = -\frac{\pi}{3} + \frac{\pi}{6} = -\frac{\pi}{6} \)
• \( x_2 = 0 \)
• \( x_3 = \frac{\pi}{6} \)
• \( x_4 = \frac{\pi}{3} \)

Right endpoints play a pivotal role in approximating the integral of the function by serving as the points at which the function \( f(x) \) is evaluated.

Secant Function

The secant function, denoted \( f(x) = \sec(x) \), is a trigonometric function related to the cosine function. Specifically, it is the reciprocal of cosine, expressed mathematically as \( \sec(x) = \frac{1}{\cos(x)} \). Understanding how this function behaves on our interval is crucial for approximation calculations.
When evaluating the secant function at the right endpoints:

• \( f(-\frac{\pi}{6}) = \sec(-\frac{\pi}{6}) = \frac{2}{\sqrt{3}} \)
• \( f(0) = \sec(0) = 1 \)
• \( f(\frac{\pi}{6}) = \frac{2}{\sqrt{3}} \)
• \( f(\frac{\pi}{3}) = 2 \)

The secant function values at these endpoints are crucial for computing the sum that forms part of the endpoint approximation formula.

Subinterval

Subintervals are the smaller divisions of the main interval \( I \) and play a fundamental role in the approximation process. By dividing the main interval into equal parts, subintervals help in analyzing and estimating the area under a curve with greater precision.

• In our example, the interval \( I = \left[ -\frac{\pi}{3}, \frac{\pi}{3} \right] \) is divided into \( N = 4 \) subintervals.
• Each subinterval has a width \( \Delta x = \frac{\pi}{6} \).

Each subinterval contains one right endpoint where the function is evaluated, and the sum of these evaluations, weighted by \( \Delta x \), gives us the approximation of the region’s area under the curve. This systematic approach helps to break down complex problems into manageable parts, providing deeper insights into integrals and area estimation.