Problem 120 - Calculus Volume 1

Question

In the following exercises, compute the average value using the left Riemann sums \(L_{N}\) for \(N=1,10,100\) . How does the accuracy compare with the given exact value? [T] \(y=x^{2}-4\) over the interval \([0,2] ;\) the exact solution is \(-\frac{8}{3}\)

Step-by-Step Solution

Answer

\(L_1 = -8\), \(L_{10} = -5.8\), \(L_{100} \approx -5.34\); accuracy improves with more subdivisions, but is less accurate than the exact value \(-\frac{8}{3}\).

Step 1: Define the Problem

We are tasked with finding the average value of the function \( y = x^2 - 4 \) over the interval \([0, 2]\) using the left Riemann sums \(L_N\) for \(N=1, 10, 100\).

Step 2: Setup for Left Riemann Sum

The left Riemann sum is calculated as \( L_N = \sum_{i=0}^{N-1} f(x_i) \Delta x \), where \( \Delta x = \frac{b-a}{N} = \frac{2-0}{N} = \frac{2}{N} \). The points \( x_i \) are left endpoints given by \( x_i = a + i \Delta x = 0 + i \frac{2}{N} \).

Step 3: Calculate \(L_1\)

For \(N=1\), \( \Delta x = 2 \). The left endpoint is \( x_0 = 0 \), so \( L_1 = f(0) \cdot 2 = (0^2 - 4) \cdot 2 = -8 \).

Step 4: Calculate \(L_{10}\)

For \(N=10\), \( \Delta x = 0.2 \). Calculate left endpoints: \( x_i = 0, 0.2, 0.4, ..., 1.8 \). So, \( L_{10} = (f(0) + f(0.2) + ... + f(1.8)) \cdot 0.2 = \sum_{i=0}^{9}(x_i^2 - 4) \cdot 0.2 \). After computation, we find \( L_{10} = -5.8 \).

Step 5: Calculate \(L_{100}\)

For \(N=100\), \( \Delta x = 0.02 \). Calculate left endpoints: \( x_i = 0, 0.02, 0.04, ..., 1.98 \). So, \( L_{100} = \sum_{i=0}^{99}(x_i^2 - 4) \cdot 0.02 \). After computation, we find \( L_{100} \approx -5.34 \).

Step 6: Compare with Exact Solution

The exact average value over the interval from 0 to 2 is given to be \( -\frac{8}{3} \approx -2.667 \). Compare each Riemann sum result to this value. \( L_1 = -8 \), \( L_{10} = -5.8 \), and \( L_{100} \approx -5.34 \) slightly improve but are less accurate compared to the exact value.

Key Concepts

Left Riemann Sum

The Left Riemann Sum is a method to approximate the area under a curve, which is useful for estimating the value of definite integrals. It divides the area into rectangles and sums up their areas to get an approximation of the total area. In this exercise, we focus on the function \( y = x^2 - 4 \) over the interval \([0, 2]\).

For each rectangle in the left Riemann sum, the height is determined by the function value at the left endpoint of each subinterval, hence the name "left" Riemann sum. The formula for the left Riemann sum \( L_N \) is:

• \( L_N = \sum_{i=0}^{N-1} f(x_i) \Delta x \)
• Where \( \Delta x = \frac{b-a}{N} = \frac{2}{N} \)
• \( x_i = a + i \Delta x = 0 + i \frac{2}{N} \)

An interesting thing about solving with different \( N \) values, like 1, 10, and 100, is witnessing how the approximation improves. More rectangles mean you get closer to the actual area under the curve.

Average Value of Function

The average value of a function over an interval provides a way to find a single value representing the entire set of infinitely many values that the function takes from the start to the end of the interval. The general formula for calculating the average value of a continuous function \( f(x) \) over the interval \([a, b]\) is:

• \( \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \)

In this exercise, the exact average value of the function \( y = x^2 - 4 \) over \([0, 2]\) is calculated using integration and given as \(-\frac{8}{3} \), or approximately \(-2.667\).

The goal of using Riemann sums, like the left Riemann sum in this exercise, is to estimate this average value. As we increase \( N \) the computed approximation should approach the true average value.

Computational Accuracy

Computational accuracy deals with how close our approximate results are to the actual or exact result. In this exercise, we compare the results of the left Riemann sums with different \( N \) values to the exact solution, which is \(-\frac{8}{3}\).

We observed:

• For \( L_1 \), the accuracy was low, with the result being \(-8\).
• For \( L_{10} \), the result improved to \(-5.8\).
• For \( L_{100} \), it further improved to \(-5.34\).

This process highlights that increasing the number of subdivisions \( N \) in Riemann Sums enhances our approximation. However, it also illustrates the inherent limit in computation — you need increasingly more subdivisions to achieve high accuracy, which may become computationally expensive.

Understanding these limitations is crucial when dealing with larger or more complex integrals, where finding a balance between computational efficiency and accuracy is key.