Problem 31 - Calculus for Biology and Medicine

Question

Use the theoretical error bound to determine how large n should be. [Hint: In each case, find the second derivative of the integrand you can use a graphing calculator to find an upper bound on \(\left.\left|f^{\prime \prime}(x)\right| .\right]\) How large should \(n\) be so that the midpoint rule approximation of $$ \int_{0}^{2} e^{-x^{2} / 2} d x $$ is accurate to within \(10^{-4}\) ?

Step-by-Step Solution

Answer

To ensure accuracy within \(10^{-4}\), use \(n \geq 45\).

Step 1: Understanding the Midpoint Rule

The Midpoint Rule for approximating the definite integral \( \int_{a}^{b} f(x) \, dx \) is given by:\[M_n = \Delta x \sum_{i=1}^{n} f \left( x_i^* \right)\]where \( \Delta x = \frac{b-a}{n} \) and \( x_i^* \) is the midpoint of the \(i\)-th subinterval. The error bound for the Midpoint Rule is:\[E_M \leq \frac{(b-a)^3}{24n^2} \max_{a \leq x \leq b} \left| f''(x) \right|\]We need to determine how large \( n \) should be such that \( E_M \leq 10^{-4} \).

Step 2: Derive the Integrand's Second Derivative

We begin by finding the second derivative of the integrand \( f(x) = e^{-x^2/2} \). The first derivative is:\[f'(x) = -x \, e^{-x^2/2}\]Applying the product rule again, the second derivative is:\[f''(x) = (x^2 - 1) \, e^{-x^2/2}\]

Step 3: Find the Maximum of the Second Derivative

To find the maximum of \( |f''(x)| \) on the interval \([0, 2]\), we need to evaluate \( |x^2 - 1| \, e^{-x^2/2} \). By plotting this function or using a calculator, we find that the maximum occurs at \( x = 1 \) with \( f''(1) \approx -e^{-1/2} \). Hence,\[\max_{0 \leq x \leq 2} \left| f''(x) \right| \approx e^{-1/2} \approx 0.6065\]

Step 4: Set Up the Error Inequality

Now we use the error inequality:\[\frac{(2-0)^3}{24n^2} \cdot 0.6065 \leq 10^{-4}\]Simplifying:\[\frac{8 \cdot 0.6065}{24n^2} \leq 10^{-4}\]\[\frac{4.852}{24n^2} \leq 10^{-4}\]

Step 5: Solve for n

Solving the inequality for \( n \):\[\frac{0.20217}{n^2} \leq 10^{-4}\]\[n^2 \geq \frac{0.20217}{10^{-4}}\]\[n^2 \geq 2021.7\]Calculating \( n \), we find:\[n \geq \sqrt{2021.7} \approx 44.95\]Therefore, \( n \) should be at least 45.

Key Concepts

Understanding the Error Bound

The Error Bound in the Midpoint Rule is crucial for determining how accurate the approximation of an integral is. When using numerical methods like the Midpoint Rule, we approximate the area under a curve by dividing it into smaller slices. To ensure this approximation is within a certain accuracy, we use the error bound formula:

\[ E_M \leq \frac{(b-a)^3}{24n^2} \cdot \max_{a \leq x \leq b} \left| f''(x) \right| \]

This shows us how the error (\( E_M \)) depends on both the interval size and the number of slices \( n \). The smaller each slice (or subinterval), the more accurate our approximation becomes. Here’s how error bound components affect the Midpoint Rule:

• **Interval \( (b-a) \)**: The larger the interval, the potentially larger the error can get if \( n \) is not increased.
• **Number of Subintervals \( n \)**: As \( n \) increases, the error decreases. More slices mean a finer approximation.
• **Maximum of the Second Derivative **: Found by evaluating \( f'' \), which tells us how much the function's rate of change itself changes. This directly influences the error bound.

For our exercise, we needed \( n \) large enough to make \( E_M \leq 10^{-4} \), ensuring the approximation's error was within acceptable limits.

The Role of Second Derivative

The second derivative \( f''(x) \) of a function gives us insight into the curvature of the graph—how it bends. It's vital in numerical integration because it reflects how much the slope of the function is changing, which is crucial when estimating integrals.

By determining \( f''(x) \) for the function \( f(x) = e^{-x^2/2} \), we gain insight into the variability of the function across the interval. Calculating \( f''(x) \) involves applying calculus rules such as the product and chain rules.

In this specific problem, the second derivative emerges as \( f''(x) = (x^2 - 1) \, e^{-x^2/2} \). The next step is to evaluate \( |f''(x)| \) to find where it peaks in the interval \([0, 2]\). This point tells us how much the function's slope changes, influencing the Midpoint Rule's error bound.

When graphed, or calculated using tools, the maximum of \( |f''(x)| \) is found to be approximately 0.6065. This maximum value is plugged into the error bound formula to assess how many subintervals are necessary for our desired accuracy.

Numerical Integration Essentials

Numerical integration methods, like the Midpoint Rule, are techniques used to estimate the definite integral of a function when an exact solution is challenging or impossible. These methods are vital in applied sciences for approximating areas under curves.

The Midpoint Rule is particularly useful because of its balance between accuracy and computational simplicity. It works by:

• Dividing the interval into \( n \) subintervals.
• Using the midpoint of each subinterval to evaluate the function.
• Multiplying these function values by the subinterval width \( \Delta x = \frac{b-a}{n} \) and summing them up.

However, numerical methods like these come with potential errors. The goal is to reduce the error to an acceptable level, which might require increasing \( n \). For functions with rapidly changing slopes, more subintervals are necessary to achieve a precise approximation. Hence, understanding the function's curvature through its second derivative becomes critical in deciding how many slices are optimal.