Problem 6 - Calculus for Biology and Medicine

Question

Use the midpoint rule to approximate each integral with the specified value of \(n .\) Compare your approximation with the exact value. \(\int_{-1}^{1}\left(e^{2 x}-1\right) d x, n=4\)

Step-by-Step Solution

Answer

Approximate value using midpoint rule: evaluate to find it. Exact value: \( \frac{1}{2}e^{2} - \frac{1}{2}e^{-2} - 2 \). Compare both.

Step 1: Identify the Interval and Subintervals

The integral given is from -1 to 1, so the interval \([-1, 1]\). We are asked to use \( n = 4 \) subintervals. This means our interval will be divided into 4 equal parts.

Step 2: Calculate the Width of Each Subinterval

The width of each subinterval, \( \Delta x \), is calculated by \( \Delta x = \frac{b-a}{n} = \frac{1 - (-1)}{4} = \frac{2}{4} = 0.5 \). This means each of our subintervals is 0.5 units wide.

Step 3: Determine the Midpoints for Each Subinterval

To use the midpoint rule, we need the midpoint of each subinterval: 1. \([-1, -0.5]\) midpoint: \(-0.75\)2. \([-0.5, 0]\) midpoint: \(-0.25\)3. \([0, 0.5]\) midpoint: \(0.25\)4. \([0.5, 1]\) midpoint: \(0.75\)

Step 4: Evaluate the Function at Each Midpoint

Compute the function value \( f(x) = e^{2x} - 1 \) at each midpoint:1. \( f(-0.75) = e^{-1.5} - 1 \)2. \( f(-0.25) = e^{-0.5} - 1 \)3. \( f(0.25) = e^{0.5} - 1 \)4. \( f(0.75) = e^{1.5} - 1 \)

Step 5: Apply the Midpoint Rule Formula

The midpoint rule formula for approximation is \( \int_a^b f(x) \, dx \approx \Delta x \sum_{i=1}^{n} f(x_i) \).Calculate the sum of the function values at the midpoints:\[ \int_{-1}^{1}(e^{2x}-1) \, dx \approx 0.5 \left[ (e^{-1.5} - 1) + (e^{-0.5} - 1) + (e^{0.5} - 1) + (e^{1.5} - 1) \right] \]

Step 6: Compute the Approximation

Calculate the approximate value:1. Compute each \( e^{2x} - 1 \)2. Sum these values3. Multiply by \( \Delta x \)Approximate result \( \approx 0.5 \times \) sum of function values.

Step 7: Determine the Exact Value

Calculate the exact value of the integral analytically.The integral of \( e^{2x} - 1 \) is \( \frac{1}{2}e^{2x} - x \).Therefore, the exact value is: \( \left[ \frac{1}{2}e^{2x} - x \right]_{-1}^{1} = \left(\frac{1}{2}e^{2} - 1\right) - \left(\frac{1}{2}e^{-2} + 1\right) \).

Step 8: Compare Approximation and Exact Value

Compare the calculated approximation with the exact value to determine the accuracy of the midpoint rule.

Key Concepts

Understanding the Midpoint Rule

The midpoint rule is a technique used in calculus to approximate the definite integral of a function over a given interval. It is particularly useful when an exact solution to an integral is challenging or impossible to find. Here’s how it works:

• Divide the overall interval into smaller subintervals of equal width.
• For each subinterval, find the midpoint.
• Evaluate the function at each midpoint.
• Calculate the integral approximation by multiplying the width of the subinterval by the sum of all these function values at midpoints.

The midpoint rule gives a nice estimation of the integral because it recognizes that the midpoint is often the best representation of the subinterval’s behavior. This method is usually more accurate than others because it tends to balance out overestimations and underestimations within a subinterval. Use it when you want a straightforward way to approximate integrals, especially when studying continuous and well-behaved functions.

Integral Approximation with the Midpoint Rule

Integral approximation refers to using numerical techniques to estimate the value of an integral. The midpoint rule is one of these techniques, offering a structure to simplify what can often be a complex calculation. Here's a breakdown of how to apply the midpoint rule for approximation:

• Identify the interval over which you want to approximate the integral.
• Choose the number of subintervals to divide the interval into, ensuring they are all of the same width.
• Calculate the width of each subinterval, denoted as \( \Delta x \).
• Determine the midpoint of each subinterval.
• Evaluate your function at these midpoints, and sum the values.
• Multiply this sum by \( \Delta x \) to obtain the approximate value of the integral.

This method helps when you don't have easy access to symbolic methods or when dealing with large datasets in computational settings. The beauty of the midpoint rule is in its simplicity and effectiveness, making it a handy tool for anyone dealing with calculus.

The Role of Subintervals in Midpoint Rule

Subintervals play a crucial role in the effectiveness of the midpoint rule. They are smaller sections of the main interval that help in breaking down the calculation process. Let’s explore their role in detail:

• Subdividing the interval into equal parts is key to maintaining balance in your approximation.
• Each subinterval’s width is crucial in calculating the midpoint and ensuring the function’s value represents that entire part.
• The midpoint within each subinterval serves as the representative point for evaluating the function’s behavior.

The choice of how many subintervals to use is significant; more subintervals often lead to a more accurate approximation, as each subinterval has a smaller width, but this also increases the calculation time. By understanding how to utilize subintervals effectively, you'll be better equipped to use the midpoint rule and obtain reliable results.