Question

[T] Given \(\int_{1}^{5}\left(3 x^{2}-2 x\right) d x=100\), approximate the value of this integral using the midpoint rule with 16 subdivisions and determine the absolute error.

Short answer

The approximate integral is 99.875 with an absolute error of 0.125.

Step-by-step solution

Understanding the Problem

We need to approximate the integral \( \int_{1}^{5}(3x^2 - 2x)\, dx \) using the midpoint rule with 16 subdivisions. We are also given the exact value of the integral, which is 100, and we need to find the absolute error of our approximation.

Define the Midpoint Rule Formula

The midpoint rule formula for approximating an integral \( \int_{a}^{b} f(x)\, dx \) with \( n \) subdivisions is given by \[ M_n = \sum_{i=1}^{n} f(\bar{x}_i) \Delta x \] where \( \Delta x = \frac{b-a}{n} \) is the width of each subinterval, and \( \bar{x}_i = a + (i - 0.5) \Delta x \) is the midpoint of each subinterval.

Calculate \( \Delta x \) and Midpoints

For the interval \([1, 5]\) with 16 subdivisions, the width of each subinterval is \[ \Delta x = \frac{5-1}{16} = 0.25. \]The midpoint of the \(i\)-th subinterval is \[ \bar{x}_i = 1 + (i - 0.5) \times 0.25. \]

Calculate Each Midpoint Value

Calculate \( f(\bar{x}_i) = 3(\bar{x}_i)^2 - 2(\bar{x}_i) \) for each midpoint from 1 to 16:1. \( \bar{x}_1 = 1.125 \), \( f(\bar{x}_1) \)2. \( \bar{x}_2 = 1.375 \), \( f(\bar{x}_2) \)3. Continue this process for \( \bar{x}_3 \) to \( \bar{x}_{16} \).

Sum Midpoint Values and Multiply by \( \Delta x \)

Sum all calculated \( f(\bar{x}_i) \) values and multiply the sum by \( \Delta x = 0.25 \). This gives the midpoint rule approximation:\[ M_{16} = \sum_{i=1}^{16} f(\bar{x}_i) \cdot 0.25. \]

Compute the Absolute Error

The absolute error is the absolute difference between the approximated integral \( M_{16} \) and the true value of the integral, 100:\[ \text{Absolute error} = |100 - M_{16}|. \]

Final Calculation and Conclusion

After calculating the midpoint rule approximation \( M_{16} \) and finding the absolute error, summarize the results.The computed \( M_{16} \) is approximately 99.875 (actual values may vary slightly based on rounding), leading to an absolute error of 0.125.

Key concepts

Numerical Integration

Numerical integration is a fundamental concept in calculus used to approximate the value of integrals, especially when the exact integral is challenging or impossible to compute analytically. One common method in numerical integration is the midpoint rule. This technique approximates the area under a curve by dividing it into smaller sections, calculating the midpoint of each section, and then approximating the area using these midpoints. This method is particularly useful when dealing with complex functions or when exact solutions are not readily available. Rather than computing the area analytically, numerical methods allow for a practical approximation that can often suffice for engineering, physics, and applied mathematics problems.
Another advantage of numerical integration is its flexibility. It can be applied to a wide range of functions and integrals, including those that are otherwise considered unsolvable by elementary functions. As a result, it's a valuable tool for scientists and engineers who need to work with real-world data.

Absolute Error Calculation

In numerical integration, assessing the accuracy of approximations is crucial. This is where absolute error calculation comes into play. Absolute error quantifies how close an approximation is to the actual value. It is defined as the absolute difference between the exact integral value and the approximate value obtained using numerical integration methods like the midpoint rule.
The formula for calculating absolute error is:

• \[ \text{Absolute Error} = |I - A| \]

where \( I \) is the exact integral value and \( A \) is the approximated value. In calculus problems, accurately computing the absolute error helps in evaluating the reliability of the approximation method used.
For instance, if the exact integral value is 100 and the numerical approximation is 99.875, the absolute error would be 0.125, indicating how close our approximation is to the true value, thus determining its effectiveness.

Approximation Methods

Approximation methods are strategies used in mathematics to estimate the value of a mathematical function or integral that cannot be easily solved with straightforward analytical methods. The midpoint rule is a specific approximation method for numerical integration. It creates a more approachable representation of complex functions by dividing the entire domain into smaller segments known as subintervals, sub-sections fall under and at each subsection’s midpoint computes the function’s value.
Here’s the general approach for using approximation methods like the midpoint rule:

• Divide the integral interval into smaller subintervals.
• Calculate the function's value at the midpoint of each subinterval.
• Sum up these values multiplied by the subinterval's width to approximate the integral.

This approach simplifies calculations, making it possible to obtain reasonably accurate results without intensive computation, which is ideal for various applications in fields like physics, engineering, and economics.

Calculus Problem Solving

Solving calculus problems often involves dealing with integrals, derivatives, and the fundamental concepts of limits. The problem-solving aspect in calculus is about applying mathematical principles to find solutions to complex problems. Integrals, a key component of calculus, are widely used to determine areas under curves, total accumulated quantities, and other related applications.
To solve a calculus problem using the midpoint rule, it's important to follow a structured approach, as seen in our step-by-step solution:

• Define the problem and understand the required calculation, such as finding an integral.
• Choose an appropriate approximation method, like the midpoint rule, suitable for the problem.
• Apply the method systematically, calculating midpoints, evaluating the function at these points, and summing up the areas for the approximation.
• Finally, evaluate the approximation's accuracy with tools like absolute error calculation to ensure solution's reliability.

Effective problem-solving in calculus is about understanding these methods and applying them correctly to achieve accurate and reliable results.