Problem 5 - 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. $$ \int_{1}^{2} \frac{1}{x} d x, n=4 $$

Step-by-Step Solution

Answer

The midpoint rule approximation is 0.6912, and the exact value is 0.6931.

Step 1: Divide the Interval

First, determine the width of each subinterval, \( \Delta x \), by dividing the interval \([1, 2]\) into \(n = 4\) subintervals.\[ \Delta x = \frac{2 - 1}{4} = \frac{1}{4} \]

Step 2: Find Midpoints of Each Subinterval

Identify the midpoints of each of the 4 subintervals. The midpoints are calculated as the average of the endpoints of each subinterval:- First subinterval \([1, 1.25]\): midpoint = \(\frac{1 + 1.25}{2} = 1.125\)- Second subinterval \([1.25, 1.5]\): midpoint = \(\frac{1.25 + 1.5}{2} = 1.375\)- Third subinterval \([1.5, 1.75]\): midpoint = \(\frac{1.5 + 1.75}{2} = 1.625\)- Fourth subinterval \([1.75, 2]\): midpoint = \(\frac{1.75 + 2}{2} = 1.875\)

Step 3: Apply the Midpoint Rule

Calculate the approximate value of the integral using the midpoint rule:\[ \int_{1}^{2} \frac{1}{x} \, dx \approx \Delta x \sum_{i=1}^{4} f(m_i) \]Where \(f(m_i)\) is \(\frac{1}{x}\) evaluated at each midpoint.- \(f(1.125) \approx 0.8889\)- \(f(1.375) \approx 0.7273\)- \(f(1.625) \approx 0.6154\)- \(f(1.875) \approx 0.5333\)Therefore:\[ \approx \frac{1}{4} \left( 0.8889 + 0.7273 + 0.6154 + 0.5333 \right) \]\[ \approx \frac{1}{4} \times 2.7649 = 0.6912 \]

Step 4: Calculate the Exact Value of the Integral

Find the exact value of the integral \(\int_{1}^{2} \frac{1}{x} \, dx\) using the antiderivative:\[ \int \frac{1}{x} \, dx = \ln |x| + C \]Evaluate from 1 to 2:\[ \ln |2| - \ln |1| = \ln 2 \approx 0.6931 \]

Step 5: Compare the Midpoint Rule Approximation with the Exact Value

Compare the approximate value (0.6912) with the exact value (0.6931). The difference between these values is \(0.6931 - 0.6912 = 0.0019\), indicating that the midpoint rule gives a reasonably close approximation to the exact value.

Key Concepts

Numerical Integration

Numerical integration is an essential mathematical technique used when it is challenging or impossible to determine an integral analytically. While some functions have easily computable antiderivatives, many do not. This is where numerical methods, such as the midpoint rule, come in handy. The main objective of numerical integration is to find an approximate value for a definite integral. This is particularly useful in real-world applications where precise calculations are necessary, but exact solutions are hard to obtain. There are various methods to perform numerical integration, such as:

• Trapezoidal Rule
• Simpson's Rule
• Midpoint Rule

Each method has its pros and cons, and often the choice depends on the specific problem at hand. The midpoint rule is particularly useful for functions that are approximately linear over small intervals. It involves taking the average value of the function at the midpoints of subintervals and multiplying it by the width of the interval.

Approximate Integral

When we talk about an approximate integral, we refer to the numerical estimation of the definite integral of a function over a given interval. It's crucial for situations where finding an exact integral is difficult due to complex functions or when we simply need a quick estimation.In the midpoint rule, the integral is divided into smaller, manageable regions. The average of the function's values at the midpoints of these regions gives an approximation of the integral. To apply the midpoint rule, follow these steps:

• Divide the interval into equal subintervals.
• Calculate the midpoints of each subinterval.
• Evaluate the function at these midpoints.
• Sum the values and multiply by the subinterval width.

In our example, the integral from 1 to 2 of \( \frac{1}{x} \) was divided into four subintervals, calculated using midpoints. Each value at the midpoint was evaluated for \( \frac{1}{x} \), summed, and then multiplied by the subinterval width to get an approximation.

Exact Value Comparison

Comparing the approximate integral obtained from numerical methods with the exact value is crucial to understanding the accuracy of our numerical approximation. It helps us assess how close our approximation is to the true value and understand the potential error involved.For the integral \( \int_{1}^{2} \frac{1}{x} \, dx \), we derived the exact value using its antiderivative, which is \( \ln |x| + C \). Evaluating this from 1 to 2 gives:\[ \ln 2 - \ln 1 = \ln 2 \approx 0.6931 \]The approximation using the midpoint rule was \( \approx 0.6912 \). Comparing these results, the difference is: \[ 0.6931 - 0.6912 = 0.0019 \]Though small, this difference shows the inherent approximation error. However, it also demonstrates how effective the midpoint rule can be, providing a result that's extremely close to the exact integral. Keeping an eye on such differences helps in choosing suitable numerical methods and understanding the accuracy required for specific applications.