Question

In the following exercises, does the right-endpoint approximation overestimate or underestimate the exact area? Calculate the right endpoint estimate R50 and solve for the exact area. $$ y=-2^{-x} \text { over }[0,1] $$

Short answer

R50 underestimates the area as the function is decreasing. Exact area: \( \frac{1}{2\ln(2)} - \frac{1}{\ln(2)} \).

Step-by-step solution

Understand the Function and Interval

The function given is \( y = -2^{-x} \) over the interval \([0, 1]\). We need to estimate the area under this curve using the right-endpoint approximation and then solve for the exact area.

Setup Right-Endpoint Approximation R50

To approximate the area using the right-endpoint method, divide the interval \([0, 1]\) into 50 subintervals. Each subinterval has a width \( \Delta x = \frac{1-0}{50} = \frac{1}{50} \). The right-endpoints for each subinterval \(i\) are \( x_i = \frac{i}{50} \) for \( i = 1, 2, \, \ldots, \, 50 \).

Calculate Function Values at Right Endpoints

Compute the function value at each right endpoint. The function value for each subinterval is \( y_i = -2^{-x_i} = -2^{-(\frac{i}{50})} \).

Calculate the Sum of Rectangles (R50)

The approximate area \( R_{50} \) is the sum of areas of rectangles with heights \( y_i \) and width \( \Delta x = \frac{1}{50} \). Thus, \( R_{50} = \sum_{i=1}^{50} \left(-2^{-(\frac{i}{50})} \right) \times \frac{1}{50} \).

Evaluate the Exact Area Using Integration

The exact area under the curve can be found by integrating the function from 0 to 1. Thus, \( \text{Area} = \int_{0}^{1} -2^{-x} \, dx \). The antiderivative of \(-2^{-x}\) is \(-\frac{2^{-x}}{\ln(2)}\). Evaluate: \[ \text{Area} = \left[ -\frac{2^{-x}}{\ln(2)} \right]_{0}^{1} = \left(-\frac{1}{2\ln(2)} - (-\frac{1}{\ln(2)}) \right) = \frac{1}{2\ln(2)} - \frac{1}{\ln(2)} \].

Determine Overestimation or Underestimation

Since \( y = -2^{-x} \) is a strictly decreasing function, the right-endpoint approximation \( R_{50} \) underestimates the area compared to the exact integral, as each rectangle lies entirely beneath the curve.

Key concepts

Right-endpoint approximation

The right-endpoint approximation is a numerical method used in calculus to estimate the area under a curve over a specific interval. It is part of the Riemann sum concepts, which break down the calculation of areas into smaller, more manageable parts by dividing the interval into subintervals.

• **Dividing the Interval:** The given interval is divided into equal parts called subintervals. In the case of the exercise, the interval was [0, 1] and was divided into 50 subintervals, each with a width \( \Delta x = \frac{1}{50} \).
• **Choosing Endpoints:** For the right-endpoint approximation, the rightmost point of each subinterval is used to evaluate the function. This forms the height of rectangles used to approximate the area.
• **Calculating the Sum:** The height of each rectangle is the function's value at the right endpoint of each subinterval, and the width is \( \Delta x \). The approximate area \( R_{50} \) is then calculated as the sum of the areas of these rectangles, represented as \( R_{50} = \sum_{i=1}^{50} \, \left(-2^{-(\frac{i}{50})}\right) \times \frac{1}{50} \).

This specific method underestimates the area under a decreasing function because each rectangle's top edge is lower than the curve itself. Understanding this concept helps in analyzing the precision of approximation methods used in integral calculus.

Integration

Integration is a fundamental concept in calculus that deals with finding the accumulation of quantities, like area under curves. In this context, integration gives the exact area under the curve of a function over a specific interval.

• **Defining Integration:** It is the process of finding the integral of a function, which represents the total accumulation, such as area.
• **Antiderivatives:** To find the integral of a function, one needs to discover its antiderivative, which is a function whose derivative is the original. For the function \( y = -2^{-x} \), the antiderivative is \( -\frac{2^{-x}}{\ln(2)} \).
• **Evaluating Definite Integrals:** To find the exact area between two points on a graph, evaluate the antiderivative at these points and compute the difference: \( \int_{0}^{1} -2^{-x} \, dx \) evaluates to \( \frac{1}{2\ln(2)} - \frac{1}{\ln(2)} \).

Integration provides a precise measure, unlike approximations, and is crucial in accurately determining quantities like area, volume, and total change.

Decreasing functions

A function is considered decreasing if, as the input (x-value) increases, the output (function value) decreases. This concept plays a significant role in understanding how approximation methods perform regarding under or overestimation.

• **Behavior Characteristics:** For a decreasing function like \( y = -2^{-x} \), the derivative is negative. This indicates a downward slope as x increases, leading to lower function values over any interval.
• **Precision of Approximations:** When using right-endpoint approximations on decreasing functions, the approach tends to underestimate the area. This happens because the heights of the rectangles used for calculation are consistently lower than the true curve height, resulting in an area estimate that is smaller than the integral's actual value.
• **Graphical Understanding:** Visualizing the function helps illustrate that for each subinterval, the function value at the right endpoint is lower than elsewhere in the interval, emphasizing the underestimation.

Recognizing the nature of decreasing functions helps in effectively choosing approximation methods and predicting their impact on estimations.