Problem 117 - Calculus Volume 1

Question

In the following exercises, approximate the average value using Riemann sums \(L_{100}\) and \(R_{100} .\) How does your answer compare with the exact given answer? [T] \(y=e^{x / 2}\) over the interval \([0,1] ;\) the exact solution is 2\((\sqrt{e}-1)\)

Step-by-Step Solution

Answer

Compute \(L_{100}\) and \(R_{100}\), average them, and compare to \(2(\sqrt{e}-1)\).

Step 1: Understand the Riemann Sums

To approximate the average value using Riemann sums, we need both the left endpoint sum (\(L_{100}\)) and the right endpoint sum (\(R_{100}\)). This will help approximate the integral of the function \(y = e^{x/2}\) over the interval \([0, 1]\).

Step 2: Establish the Subintervals

Divide the interval \([0,1]\) into 100 subintervals, each with a width \(\Delta x = \frac{1-0}{100} = 0.01\).

Step 3: Calculate Left Riemann Sum \(L_{100}\)

For \(L_{100}\), evaluate the function at the left endpoint of each subinterval. It is computed as: \[ L_{100} = \sum_{i=0}^{99} e^{\left(\frac{i}{2}\right) \cdot 0.01} \cdot 0.01 \]

Step 4: Calculate Right Riemann Sum \(R_{100}\)

For \(R_{100}\), evaluate the function at the right endpoint of each subinterval. It is computed as: \[ R_{100} = \sum_{i=1}^{100} e^{\left(\frac{i}{2}\right) \cdot 0.01} \cdot 0.01 \]

Step 5: Compare Riemann Sums with Exact Solution

Calculate both sums and compare their average to the exact answer of \(2(\sqrt{e} - 1)\). The average Riemann sum is \[ \frac{L_{100} + R_{100}}{2} \] Compare this value with the exact solution given.

Key Concepts

Average Value of a Function

When we talk about the average value of a function over an interval, we are essentially finding the mean height that the function reaches across that interval. To do this mathematically, we use the definite integral. For a function, let's say \( f(x) \), over an interval \([a, b]\), the average value is given by the formula:\[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]This formula provides a way to understand the average behavior of \(f(x)\) over the specified interval. In the original exercise, \( f(x) = e^{x/2} \) and the interval is \([0,1]\). The exact solution to this problem is \( 2(\sqrt{e} - 1) \), which is derived using this integration process.
A Riemann sum approximates this integral by dividing the area under the curve into small rectangles and summing up the area of these rectangles instead.

Left and Right Endpoint Approximations

The process of approximating the integral using Riemann sums is a common method in calculus. This involves using either the left-hand or the right-hand endpoint of each subinterval to determine the height of the rectangle used in the approximation.

• Left Endpoint Approximation \(L_n\): In this method, we evaluate the function at the left endpoint of each subinterval. It is useful because it sometimes provides an underestimation of the integral when the function is increasing.
  
• Right Endpoint Approximation \(R_n\): Here, we evaluate the function at the right endpoint. When the function is increasing, this often overestimates the integral.
  

The average of \(L_{100}\) and \(R_{100}\) in this particular exercise will give us a middle-ground estimate of the average value of the function on \([0,1]\), thus approximating the integral.

Subintervals in Calculus

Subdividing an interval into smaller segments or 'subintervals' is crucial for calculating Riemann sums efficiently. The more subintervals you use, the more accurate your approximation will be. In this exercise, the interval \([0,1]\) is divided into 100 subintervals, each having a width of \( \Delta x = 0.01 \).

• A small \(\Delta x\) implies a finer division, leading to more precise estimates of the integral.
  
• Subintervals are important not only in Riemann sums but also in many numerical estimation techniques in calculus.
  
• In the context of this exercise, dividing the interval into 100 subintervals provides a balance between computational cost and accuracy.

By carefully choosing the number of subintervals and appropriately calculating the left and right endpoint sums, we can achieve a very close approximation of the integral, and thus, the average value of the function.