Problem 13 - Calculus Single Variable

Question

A function \(f(x)\) and an interval \(I=[a, b]\) are given. Also given is the approximation \(\mathcal{M}_{10}\) of \(A=\int_{a}^{b} f(x) d x\) that is obtained by using the Midpoint Rule with a uniform partition of order \(10 .\) a. Use inequality \((5.8 .3)\) to find a lower bound \(\alpha\) for \(A\). b. Use inequality \((5.8 .3)\) to find an upper bound \(\beta\) for \(A\). c. Calculate \(A\) and ascertain that it lies in the interval \([\alpha, \beta]\) \(f(x)=(2+x)^{(-1 / 3)} \quad I=[-1,3], \quad \mathcal{M}_{10}=2.88409 \ldots\)

Step-by-Step Solution

Answer

\(\alpha \approx 2.85449\), \(\beta \approx 2.91369\). \(A \approx 2.88803\) lies within \([\alpha, \beta]\).

Step 1: Understanding the Midpoint Rule

The Midpoint Rule for approximating the integral \( A = \int_{a}^{b} f(x) \, dx \) is given by \( \mathcal{M}_n = \frac{b-a}{n} \sum_{i=1}^{n} f\left(\frac{x_{i-1} + x_i}{2}\right) \), where the interval \([a, b]\) is divided into \(n\) subintervals. For this exercise, \(n=10\),\( a=-1\), and \(b=3\).

Step 2: Identifying the Bounds

The inequality \((5.8.3)\) used here implies that the error bounds for the Midpoint Rule are given by \( \frac{(b-a)^3}{24n^2} \cdot \max_{x \in [a, b]} |f''(x)| \). We need to calculate the second derivative of \(f(x) = (2+x)^{(-1/3)}\) to use in this error bound.

Step 3: Calculating the Second Derivative

First, find \(f'(x)\): \[ f'(x) = -\frac{1}{3}(2+x)^{-4/3} \]. Then, calculate \(f''(x)\): \[ f''(x) = \frac{4}{9}(2+x)^{-7/3} \].

Step 4: Finding Maximum of \(|f''(x)|\)

Observe that \(f''(x)\) is positive and decreasing in \([a, b]\), implying its maximum occurs at \(x = -1\). Calculate this maximum value: \[ |f''(-1)| = \frac{4}{9}\cdot(1)^{-7/3} = \frac{4}{9} \].

Step 5: Calculating the Error Bound

Using the earlier formula, compute the error bound as: \[ E_{mid} \leq \frac{(3 - (-1))^3}{24 \times 10^2} \times \frac{4}{9} = \frac{64}{2160} = \frac{8}{270} \approx 0.0296 \].

Step 6: Determine \(\alpha\) and \(\beta\)

Use \(\mathcal{M}_{10} = 2.88409\) and the error bound to find:\[ \alpha = \mathcal{M}_{10} - E_{mid} = 2.88409 - 0.0296 = 2.85449 \].\[ \beta = \mathcal{M}_{10} + E_{mid} = 2.88409 + 0.0296 = 2.91369 \].

Step 7: Calculating \(A\) Using Integral

Calculate the actual integral \(A\) by: \[ \int_{-1}^{3} (2+x)^{-1/3} \, dx = \left[ \frac{3}{2}(2+x)^{2/3} \right]_{-1}^{3} = \frac{3}{2}\left(5^{2/3} - 1\right) \approx 2.88803 \].

Step 8: Verify \(A\) Lies in \([\alpha, \beta]\)

Check if actual value \(A \approx 2.88803\) falls within the bounds:\(2.85449 \leq 2.88803 \leq 2.91369\). The calculation verifies \(A\) lies within the predicted bounds.

Key Concepts

Midpoint Rule

The Midpoint Rule is a method for approximating definite integrals. It provides a simplified way to estimate the area under a curve over a given interval. Instead of considering the exact curve, this rule averages the function values at the midpoints of each subinterval. Consider an interval \([a, b]\) which is divided into \(n\) equal parts. For each subinterval, you calculate the function's value at its midpoint and multiply by the subinterval length, \(\frac{b-a}{n}\). Then sum up all these products. This approach gives an estimate of the integral value.

• In this exercise, the function's interval is divided into 10 subintervals (\(n = 10\)).
• The midpoint of each subinterval is used within the function to find an estimated area.
• This value is then multiplied by the subinterval's length and summed up for all intervals.

The result is an approximation of the integral, known here as \(\mathcal{M}_{10} = 2.88409\). The Midpoint Rule is simple and effective for functions that are not too complex over the interval.

Error Bounds

Error bounds assess how close an approximation, such as that from the Midpoint Rule, is to the true value. They provide a range wherein the actual integral might lie. To calculate the error bound, determining the derivative of the function involved is crucial. The error bound formula for the Midpoint Rule involves:

• The interval's length \( (b-a)^3\).
• Dividing by \(24n^2\), where \(n\) is the number of subintervals.
• Multiplying that by the maximum value of the absolute second derivative of \(f(x)\) over the interval.

In this exercise, the error was calculated as approximately \(0.0296\). This means the true value differs from the approximation \(\mathcal{M}_{10}\) by this amount or less. Thus, setting bounds on our approximation leads to a better understanding of its accuracy.

Second Derivative

The second derivative of a function gives insight into its curvature and is fundamental in calculating error bounds for numerical integration methods like the Midpoint Rule. For the function \(f(x) = (2+x)^{-1/3}\), the first derivative \(f'(x)\) shows the rate of change of \(f(x)\). Further differentiating leads to the second derivative \(f''(x),\), which in this case is \[f''(x) = \frac{4}{9}(2+x)^{-7/3}\].

• The second derivative is crucial for assessing the function's concavity.
• By examining \(|f''(x)|\), you determine how steeply angled or curved the graph is.
• In this scenario, \(|f''(x)|\) is maximal at \(x = -1\) with a value of \(\frac{4}{9}\).

Finding this maximum is key since it's used in the error bound formula. Hence, calculating and understanding the second derivative significantly impacts the precision of integral approximations.

Definite Integral

The concept of a definite integral represents the exact area under a curve between two points on a graph. Calculating a definite integral gives the total accumulation of the function's values over a specified interval \([a, b]\). It reflects the net value, taking into account any areas below the x-axis as negative. In this exercise, the task is to find the definite integral of \((2+x)^{-1/3}\) over \( [-1, 3]\). The essential steps to compute this are:

• First, find the antiderivative of the function.
• Then, evaluate it at the bounds: the upper bound minus the lower bound.

The calculated integral yields \(A \approx 2.88803\). This result is crucial for verifying whether the approximation using the Midpoint Rule falls within the error bounds estimated. Definite integrals aren't just about exact values but offer a method to confirm the reliability of approximation techniques.