Problem 41 - Calculus Single Variable

Question

Suppose that \(f(x) \geq 0\) for \(x\) in \(I=[a, b]\). If for each subinterval \(\left[x_{j-1}, x_{j}\right]\) that arises from the uniform partition \(\mathcal{P}=\left\\{x_{0}, x_{1}, \ldots, x_{N}\right\\}\) of \(I,\) we use the left endpoint \(x_{j-1}\) instead of the right endpoint in formula \((5.1 .9),\) then we obtain the left endpoint approximation $$ \sum_{j=1}^{N} f\left(x_{j-1}\right) \cdot \Delta x $$ of the area \(A\) of the region below the graph of \(f\) and above \(I\) (see Figure 10 ). As can be seen from Figure \(10,\) if \(f\) increases on \(I,\) then the right endpoint approximation overestimates \(A,\) and the left endpoint approximation underestimates \(A .\) In each of Exercises \(39-44,\) calculate the average of the left and right endpoint approximations. (For purposes of comparison, the exact value of \(A\) is given. Notice that your answer is more accurate than both the left and right endpoint approximations.) $$ f(x)=(2 x+1) /(x+1) \quad I=[0,2], N=2, A=4-\ln (3) $$

Step-by-Step Solution

Answer

The average endpoint approximation is \(\frac{17}{3}\).

Step 1: Define the Partition and Subintervals

The interval, given by \(I = [0, 2]\), is divided equally into \(N = 2\) subintervals by a uniform partition. Therefore, each subinterval length is \(\Delta x = \frac{b-a}{N} = \frac{2-0}{2} = 1\). The partition set is \(\mathcal{P} = \{x_0, x_1, x_2\} = \{0, 1, 2\}\).

Step 2: Calculate the Left Endpoint Approximation

For the left endpoint approximation, we calculate \(f(x_{j-1})\) and multiply by \(\Delta x\). The sum for each subinterval is: \[ \text{Left Approximation} = \sum_{j=1}^{2} f(x_{j-1}) \cdot \Delta x = f(x_0) \cdot 1 + f(x_1) \cdot 1 \]. We compute each term: - \(f(x_0) = f(0) = \frac{2 \cdot 0 + 1}{0 + 1} = 1\) - \(f(x_1) = f(1) = \frac{2 \cdot 1 + 1}{1 + 1} = \frac{3}{2}\)Thus, the left endpoint approximation is: \[ 1 \cdot 1 + \frac{3}{2} \cdot 1 = 1 + \frac{3}{2} = \frac{5}{2} \].

Step 3: Calculate the Right Endpoint Approximation

For the right endpoint approximation, we calculate \(f(x_j)\) and multiply by \(\Delta x\). The sum is: \[ \text{Right Approximation} = \sum_{j=1}^{2} f(x_j) \cdot \Delta x = f(x_1) \cdot 1 + f(x_2) \cdot 1 \]. Compute each term: - \(f(x_1) = \frac{3}{2}\) from the previous calculation- \(f(x_2) = f(2) = \frac{2 \cdot 2 + 1}{2 + 1} = \frac{5}{3}\)Thus, the right endpoint approximation is: \[ \frac{3}{2} \cdot 1 + \frac{5}{3} \cdot 1 = \frac{3}{2} + \frac{5}{3} = \frac{9}{6} + \frac{10}{6} = \frac{19}{6} \].

Step 4: Calculate the Average of the Two Approximations

To find the average of the left and right endpoint approximations, add them and divide by 2:\[ \text{Average} = \frac{\frac{5}{2} + \frac{19}{6}}{2} = \frac{15}{6} + \frac{19}{6} = \frac{34}{6} = \frac{17}{3} \].

Step 5: Compare to the Exact Value

The exact area \(A\) is given as \(4 - \ln(3)\). Comparing the average \(\frac{17}{3}\) to the exact value shows why the average approximation is more accurate.

Key Concepts

Riemann Sums

Riemann sums are a fundamental concept in calculus used to approximate the total area under a curve, represented by a function, from a given interval. It is a way to estimate the integral of a function using a finite sum. When working with Riemann sums, the interval \[ I = [a, b] \]is divided into smaller subintervals. You can choose either the left endpoint, right endpoint, or even the midpoint of these subintervals to make calculations easier. The idea is simple: add up small rectangles' areas under the curve to get an approximation of the total area.

Each rectangle's width is denoted as \( \Delta x \), which is equal across all subintervals, and the height is the function value at the chosen point within each subinterval, such as \( f(x_{j-1}) \) or \( f(x_j) \). Thus, the Riemann sum for left endpoint approximation is given by: \[ \sum_{j=1}^{N} f(x_{j-1}) \cdot \Delta x \].Riemann sums can serve as a great introduction to the more precise concept of definite integrals. Calculating the left and right endpoint approximations helps understand how closely these sums approach the actual area under the curve. As you increase the number of subintervals \( N \), the Riemann sum becomes more accurate, eventually becoming the integral in the limit as \( N \to \infty \).

Definite Integrals

Definite integrals are a core concept in calculus used to compute the exact area under a function's graph over a specific interval \( I = [a, b] \). They give the precise measurement of the total accumulation of quantities, like area, between two bounds. The notation for a definite integral of a function \( f(x) \) over an interval is:\[\int_{a}^{b} f(x) \, dx.\]The result of this integral is a number that represents the net signed area between the function and the x-axis over \( I \).

While Riemann sums are valuable for approximations, definite integrals provide the actual value when limits are applied to those approximations. For instance, considering the exercise above, the given function and interval lead us to a definite area of \( 4 - \ln(3) \). Understanding definite integrals allows you to solve more complex problems in calculus, where you can determine not just areas, but masses, volumes, and other cumulative quantities which Riemann sums initially approximate.

Numerical Approximation

Numerical approximation is a technique used in calculus when it's challenging or impossible to compute an integral analytically. When functions are too complex or lack antiderivatives in closed form, numerical methods such as Riemann sums become indispensable. Methods like these allow us to obtain approximate results that are often sufficient for practical purposes.One common numerical approximation approach, as discussed in the exercise, is the averaging of left and right endpoint approximations. By calculating both and taking their mean, we achieve a result that is typically more accurate than using either endpoint alone. In our example, calculating the average of Riemann sums provides a better approximation of the actual area \( 4 - \ln(3) \).In summary, numerical approximation techniques in calculus allow for solving problems where exact solutions aren't feasible. It opens up possibilities for analyzing real-world data and functions that aren't easily handled with standard analytical methods. Understanding and applying these techniques enable a deeper comprehension of integral calculus and its applications in various scientific fields.