Question

Estimating error Refer to Theorem 8.1 in the following exercises. Let \(f(x)=\sqrt{x^{3}+1}\) a. Find a Midpoint Rule approximation to \(\int_{1}^{6} \sqrt{x^{3}+1} d x\) using \(n=50\) subintervals. b. Calculate \(f^{\prime \prime}(x)\) c. Use the fact that \(f^{\text {- }}\) is decreasing and positive on [1,6] to show that \(\left|f^{*}(x)\right| \leq 15 /(8 \sqrt{2})\) on [1,6] d. Use Theorem 8.1 to find an upper bound on the absolute error in the estimate found in part (a).

Short answer

Question: Estimate the integral \(\int_{1}^{6} \sqrt{x^3+1} dx\) using the Midpoint Rule with n=50 subintervals and find the error bound. Answer: The Midpoint Rule approximation, \(M_{50}\), of the integral is calculated as: $$M_{50} = \frac{1}{10} \sum_{i=1}^{50} f(x_i),$$ where \(f(x) = \sqrt{x^3+1}\) and \(x_i = 1 + \frac{2i - 1}{2} \cdot \frac{1}{10}\). The second derivative of the function is: $$f''(x) = \frac{-3x^4+12x^3+6x^2}{4(x^3+1)^{3/2}}.$$ We showed that the magnitude of the second derivative is less than or equal to \(\frac{15}{8\sqrt{2}}\) on the interval [1, 6]. Using this information and Theorem 8.1, we find the error bound to be: $$\mathrm{Absolute \ Error} \le 0.014696.$$ Thus, the Midpoint Rule approximation provides an estimate of the integral with an error bound of less than or equal to 0.014696.

Step-by-step solution

(a) Calculate the Midpoint Rule approximation using n=50 subintervals

In order to compute the integral \(\int_{1}^{6} \sqrt{x^3+1} dx\), we will use the Midpoint Rule formula to approximate the integral: $$M_n = \Delta x \sum_{i=1}^{n} f(x_i),$$ where \(n\) is the number of subintervals, \(\Delta x = \frac{b-a}{n}\), and \(x_i = a + \frac{2i - 1}{2} \Delta x\). For our problem, a = 1, b = 6, and n = 50. Now let's calculate the Midpoint Rule approximation: 1. Calculate \(\Delta x\): $$\Delta x = \frac{6-1}{50} = \frac{1}{10}.$$ 2. Create a loop to iterate over the subintervals (i = 1 to 50) and calculate \(x_i\) in each iteration: $$x_i = 1 + \frac{2i - 1}{2} \cdot \frac{1}{10}.$$ 3. For each \(x_i\), evaluate the function \(f(x) = \sqrt{x^3+1}\). 4. Calculate the sum of all the evaluated function values: $$\sum_{i=1}^{50} f(x_i).$$ 5. Multiply the sum by \(\Delta x\): $$M_{50} = \frac{1}{10} \sum_{i=1}^{50} f(x_i).$$ Calculate the result to get the Midpoint Rule approximation.

(b) Calculate the second derivative of the given function

Before finding the second derivative, we need to find the first derivative. Let $$f(x) = \sqrt{x^3+1} = (x^3+1)^{1/2}.$$ Now, find the first derivative \(f'(x)\) using the chain rule: $$f'(x) = \frac{1}{2}(x^3+1)^{-1/2} \cdot 3x^2.$$ Simplify the expression: $$f'(x) = \frac{3x^2}{2\sqrt{x^3+1}}.$$ Next, we find the second derivative \(f''(x)\): $$f''(x) = \frac{\mathrm{d}}{{\mathrm{d}x}} \left( \frac{3x^2}{2\sqrt{x^3+1}}\right).$$ Using the quotient rule, we get: $$f''(x) = \frac{(6x)(2\sqrt{x^3+1})-(3x^2)(\frac{3x^2}{2\sqrt{x^3+1}})}{(2\sqrt{x^3+1})^2}.$$ Simplify the expression to get the final result: $$f''(x) = \frac{-3x^4+12x^3+6x^2}{4(x^3+1)^{3/2}}.$$

(c) Show that the magnitude of the second derivative is less than or equal to \(15 / (8\sqrt{2})\) on [1, 6]

We can analyze the second derivative to determine its maximum value in the interval [1, 6]. Note that, as f(x) is a decreasing positive function, we can limit our interval further to [1, 4]. Let \(g(x) = -3x^4+12x^3+6x^2\). The first derivative of \(g(x)\) is: $$g'(x) = -12x^3+36x^2+12x.$$ By setting \(g'(x)\) to zero, we can find the critical points of \(g(x)\): $$-12x^3+36x^2+12x = 0 \Rightarrow 12x(x^2-3x-1) = 0.$$ The critical point x=0 is out of our interval, so we can ignore it. The other critical points can be found by applying the quadratic formula to \(x^2-3x-1\), which yields the roots x≈0.79 and x≈1.6. For these x-values, we find \(|f''(x)|\). In addition, we also check the interval endpoints x=1 and x=4: $$|f''(1)| = \frac{15}{8\sqrt{2}}$$ $$|f''(4)| = \frac{15}{16\sqrt{2}}$$ Since we validate that for x=1 and x=4, the highest possible value for the magnitude of \(f''(x)\) is \(\frac{15}{8\sqrt{2}}\), we can state that: $$|f''(x)| \le \frac{15}{8\sqrt{2}} \text{ on } [1, 6].$$

(d) Find the error bound using Theorem 8.1

Theorem 8.1 states that if \(M \ge |f''(x)|\) for all x in the interval [a, b], then the error bound for the Midpoint Rule approximation is given by: $$\mathrm{Absolute \ Error} \le \frac{K(b-a)^3}{24n^2} \text{ where } K \ge |f''(x)|$$ In our case, we have \(a=1\), \(b=6\), and \(n=50\). From part (c), we know that the maximum possible value of the magnitude of the second derivative on the interval [1, 6] is \(\frac{15}{8\sqrt{2}}\). Therefore, we can use this value as \(K\) to find the error bound: $$\mathrm{Absolute \ Error} \le \frac{\frac{15}{8\sqrt{2}}(6-1)^3}{24 \cdot 50^2}.$$ Calculate the value to get the error bound: $$\mathrm{Absolute \ Error} \le 0.014696.$$

Key concepts

Numerical Integration

Numerical integration is a crucial technique in mathematical analysis and applied mathematics, where it serves as a tool for approximating the definite integral of functions that can be hard or even impossible to integrate analytically. The definite integral can be thought of as the area under the curve of a function on a given interval. However, not all functions have nice, easy-to-calculate antiderivatives, which is where numerical methods like the Midpoint Rule come into play.

Numerical integration methods fall into two broad categories: Newton-Cotes formulas, such as the Trapezoidal Rule and Simpson's Rule, and Gaussian quadrature, which involves selecting optimal points and weights for the integration. The Midpoint Rule is one of the simplest Newton-Cotes formulas and often provides a good balance between simplicity and accuracy. It works by dividing the interval of integration into equal subintervals and then taking the height of the function at the midpoint of each subinterval to create a series of rectangles. The sum of the areas of these rectangles serves as the approximation of the integral. It's particularly effective for functions that are reasonably smooth over the interval.

In practical applications, this method is widely used across various fields such as physical sciences, engineering, and economics, where quick approximations are needed for complex integrals. What's impressive about numerical integration is its capability to yield sufficiently accurate results with relatively simple calculations, making it an indispensable tool for both students and professionals alike.

Error Estimation in Numerical Methods

Error estimation in numerical methods is a vital aspect of computational mathematics, providing an insight into how close an approximate solution is to the true value. When we use numerical methods like the Midpoint Rule for numerical integration, there is inevitably some level of error introduced. The primary goal of error estimation is to assess the quality of these approximations and to ensure they meet the required precision for given applications.

Error estimation can involve a priori estimates, which are predictions made before the computation, and a posteriori estimates, which are based on actual results of the computation. A variety of techniques are employed to estimate errors in numerical integration, but a common approach includes analyzing the behavior of the function's derivatives within the integration interval. Theorem 8.1, for instance, involves using the second derivative of the function to bound the absolute error.

Understanding error bounds is crucial as they indicate the maximum error we can expect and thus, the reliability of the numerical solution. For students dealing with numerical methods, grasping the concepts behind error bounds and how to calculate them is key to verifying the accuracy of their computational results. In an educational setting, it's imperative to emphasize these concepts, ensuring that learners not only perform the calculations but also comprehend the implications and limitations of the estimates they are producing.

Second Derivative Test

The second derivative test is a standard procedural test in calculus used to find the concavity of a function and identify possible points of inflection. This test centers around the second derivative of a function. When we're given a continuous function, and we find its second derivative, we can get valuable information about the curvature of the original function.

If the second derivative is positive, the function is concave up, resembling the shape of a 'U', and if it's negative, the function is concave down, similar to an upside-down 'U'. This information is particularly useful when one wants to infer about the maxima and minima of functions, as it can indicate whether a critical point is a local maximum, a local minimum, or a point of inflection.

In the context of numerical integration, the second derivative is significant because it appears in the error bound formulas, as seen in Theorem 8.1. Analyzing the second derivative is part of the error estimation process because areas where the function has high curvature could lead to greater approximation errors when using methods like the Midpoint Rule. This connection between the second derivative test and the error bounds in numerical methods is crucial for students to understand, as it underscores the importance of the function's behavior on the overall error in numerical calculations.