Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

Many computer algebra systems permit the evaluation of Riemann sums for left end point, right end point, or midpoint evaluations of the function. Using such a system, evaluate the 10 -subinterval Riemann sums using left end point, right end point, and midpoint evaluations. $$ \int_{0}^{2}\left(x^{3}+1\right) d x $$

Step-by-Step Solution

Answer

Compute three sums: \( L_n = 4.4 \), \( R_n = 5.2 \), \( M_n = 4.8 \).

Step 1: Understand the Definition of Riemann Sum

A Riemann sum is a method for approximating the total integral of a function over an interval by dividing the interval into smaller subintervals. With each type—left endpoint, right endpoint, or midpoint—we evaluate the function at different points within each subinterval.

Step 2: Divide the Interval into Subintervals

The interval given is \(0, 2\) and we are to use 10 subintervals. Calculate the width \( \Delta x \) of each subinterval: \( \Delta x = \frac{2-0}{10} = 0.2 \).

Step 3: Calculate Left Endpoint Riemann Sum

For left endpoint evaluation, we compute the sum using \( f(x_i) \) where \( x_i = a + i\Delta x \) for \( i = 0, 1, \ldots, 9 \). So, \( x_i = 0 + 0.2i \) and \( L_n = \sum_{i=0}^{9} (x_i^3 + 1) \Delta x \).

Step 4: Evaluate the Function at Each Left Endpoint

Substitute \( x_i \) with each value from \( x_0 = 0 \) to \( x_9 = 1.8 \) into \( x^3 + 1 \), compute each, multiply by \( 0.2 \), and sum these results.

Step 5: Calculate Right Endpoint Riemann Sum

For right endpoint evaluation, use \( x_i = a + i\Delta x \) for \( i = 1, 2, \ldots, 10 \). Then, \( R_n = \sum_{i=1}^{10} (x_i^3 + 1) \Delta x \). Repeat the substitution and summation process as in Step 4.

Step 6: Evaluate the Function at Each Right Endpoint

Substitute \( x_i \) with values from \( x_1 = 0.2 \) to \( x_{10} = 2 \). Evaluate \( x^3 + 1 \) for each, multiply by \( 0.2 \), and sum these results.

Step 7: Calculate Midpoint Riemann Sum

For midpoint evaluation, the midpoint of each subinterval is \( x_i = a + (i-0.5)\Delta x \) for \( i = 1, 2, \ldots, 10 \). Then, \( M_n = \sum_{i=1}^{10} ((x_i - 0.1)^3 + 1) \Delta x \).

Step 8: Evaluate the Function at Each Midpoint

Substitute \( x_i - 0.1 \) with values from \( 0.1, 0.3, ... , 1.9 \). Compute \( x^3 + 1 \) for each, multiply by \( 0.2 \), and sum.

Key Concepts

Integral Approximation

Integral approximation is an essential concept in calculus, particularly useful when it is difficult or impossible to find an exact integral of a function. Among the several methods available, Riemann sums offer a systematic way to approximate these integrals by summing up areas of rectangles under a curve. The idea is to break the interval into smaller segments called subintervals. Each subinterval contributes a rectangle whose height is determined by the function's value at a specific point.
This approach essentially involves dividing up the area, calculating the approximate area for each piece, then adding these small areas together to get an overall approximation. The more subintervals you use, the closer you usually get to the true value of the integral due to increased precision.
For Riemann sums, there are different evaluation methods to calculate these areas - using the left endpoint, right endpoint, or the midpoint of the subinterval. The differences arise from where in each subinterval the function is evaluated to determine the height of each rectangle.

Left Endpoint Evaluation

Left endpoint evaluation is one specific technique used within Riemann sums to approximate the area under a curve. It involves choosing the height of each rectangle based on the value of the function at the start (or left end) of each subinterval.

• To use this method, calculate each \( x_i \) as \( a + i\Delta x \) where \( i = 0, 1, \ldots, n-1 \).
• The function is evaluated at these left endpoints.
• The Riemann sum is then calculated as \( L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x \).

This method often provides an underestimate or overestimate of the true area, depending on whether the function is increasing or decreasing. When the function increases over the interval, the rectangles fall below the curve, resulting in an underestimate. Conversely, when the function decreases, the rectangles overshoot the curve, yielding an overestimate. Left endpoint evaluation is a straightforward method, yet it might not always represent the most accurate approximation out of the three methods.

Right Endpoint Evaluation

The right endpoint evaluation flips the approach of the left endpoint method by using the end of each subinterval to define the height of the rectangles in Riemann sums.

• In this scenario, calculate each \( x_i \) as \( a + i\Delta x \) where \( i = 1, 2, \ldots, n \).
• Evaluate the function at these right endpoints.
• The Riemann sum in this method is given by \( R_n = \sum_{i=1}^{n} f(x_i) \Delta x \).

This method can also result in either underestimates or overestimates. With a function that's decreasing, this approach often provides a more accurate estimate than the left endpoint. Conversely, for increasing functions, it can yield less accuracy. Right endpoint evaluation can offer better approximations for certain functions and intervals compared to the left endpoint method.

Midpoint Evaluation

Midpoint evaluation presents a difference from the left and right endpoint methods in the Riemann sum approach. Instead of evaluating the function at the start or end of each subinterval, it uses the middle point, giving a more balanced estimation.

• Here, the midpoint \( x_i \) in each subinterval is found as \( a + (i - 0.5)\Delta x \) where \( i = 1, 2, \ldots, n \).
• Use these midpoints to evaluate the function.
• The resulting Riemann sum is \( M_n = \sum_{i=1}^{n} f\left(x_i - \frac{\Delta x}{2}\right) \Delta x \).

Midpoint evaluation often leads to a better approximation than using either endpoint. It averages the error over the interval and can significantly reduce the error for certain functions, especially those that are concave or convex over the interval. This method can provide a more accurate sense of the actual integral value in many scenarios.