Problem 73 - Calculus Early Transcendentals

Question

Bounds on \(e\) Use a left Riemann sum with at least \(n=2\) subintervals of equal length to approximate \(\ln 2=\int_{1}^{2} \frac{d t}{t}\) and show that \(\ln 2<1 .\) Use a right Riemann sum with \(n=7\) subintervals of equal length to approximate \(\ln 3=\int_{1}^{3} \frac{d t}{t}\) and show that \(\ln 3>1 .\)

Step-by-Step Solution

Answer

Based on the left and right Riemann sums, we can approximate the natural logarithms, \(\ln 2\) and \(\ln 3\). By using a left Riemann sum with 2 subintervals, we determined that \(\ln 2 < 1\), as the sum was approximately 0.833. On the other hand, using a right Riemann sum with 7 subintervals, we found that \(\ln 3 > 1\), as the sum was approximately 1.035.

Step 1: Approximate \(\ln 2\) using Left Riemann Sum

To approximate \(\ln 2 = \int_{1}^{2} \frac{dt}{t}\) using a left Riemann sum, we first need to divide the interval \([1,2]\) into at least 2 equal subintervals. Let's divide it into 2 subintervals: \([1, 1.5]\) and \([1.5, 2]\). Using the left Riemann sum, we'll use the left endpoint of each subinterval as the value for t, so for the first subinterval, we will use t=1. For the second, we will use t=1.5. Our left Riemann sum can be computed as follows: Left Riemann Sum = \(\Delta t \left(\frac{1}{t_1} + \frac{1}{t_2}\right)\) Where \(\Delta t = \frac{2-1}{2} = 0.5\), and \(t_1 = 1\), and \(t_2 = 1.5\). Now, let's plug in the values for \(\Delta t\), \(t_1\), and \(t_2\): Left Riemann Sum = \(0.5\left(\frac{1}{1} + \frac{1}{1.5}\right)\)

Step 2: Calculate the Left Riemann Sum and show \(\ln 2 < 1\)

Now that we have set up our left Riemann sum for approximating \(\ln 2\), let's calculate its value: Left Riemann Sum = \(0.5\left(1 + \frac{1}{1.5}\right) = 0.5\left(1 + \frac{2}{3}\right) = 0.5\left(\frac{5}{3}\right) = \frac{5}{6} = 0.8333...\) Since the left Riemann sum of \(\int_{1}^{2} \frac{dt}{t}\) is less than 1, we can conclude that \(\ln 2 < 1\).

Step 3: Approximate \(\ln 3\) using Right Riemann Sum

To approximate \(\ln 3 = \int_{1}^{3} \frac{dt}{t}\) using a right Riemann sum, we first need to divide the interval [1, 3] into 7 equal subintervals. The equal length of the subintervals are: \(\Delta t = \frac{3-1}{7} = \frac{2}{7}\). Using right Riemann sum, we'll use the right endpoints of the subintervals as the value for t. Let's denote those values as \(t_i\), where \(i=1,2,\ldots,7\). Right Riemann Sum = \(\Delta t \left(\frac{1}{t_1} + \cdots + \frac{1}{t_7}\right)\) Now, let's calculate the \(t_i\) values: - t1 = 1 + \(\frac{2}{7}\) = \(\frac{9}{7}\) - t2 = 1 + \(\frac{4}{7}\) = \(\frac{11}{7}\) - ... - t7 = 1 + \(\frac{14}{7}\) = 3

Step 4: Calculate the Right Riemann Sum and show \(\ln 3 > 1\)

Now that we calculated the \(t_i\) values for approximating \(\ln 3\), let's plug them into the right Riemann sum expression: Right Riemann Sum = \(\frac{2}{7}\left(\frac{1}{\frac{9}{7}} +\frac{1}{\frac{11}{7}} + \cdots + \frac{1}{3}\right) = \frac{2}{7}\left(\frac{7}{9}+\frac{7}{11}+\cdots+\frac{7}{21}\right)\) Now, let's calculate the value: Right Riemann Sum = \(\frac{2}{7}(0.7778+0.6364+\cdots+0.3333) \approx 1.035\) Since the right Riemann sum of \(\int_{1}^{3} \frac{dt}{t}\) is greater than 1, we can conclude that \(\ln 3 > 1\).

Key Concepts

Left Riemann Sum

When we want to approximate the integral of a function over an interval, a Left Riemann Sum can be a handy tool. This method uses the left endpoint of each subinterval to form rectangles under the curve of the function. For each rectangle, the height is determined by the function value at the left endpoint. Calculate the area by multiplying this height by the width of the subinterval.

For example, when estimating \(\ln 2\ = \int_{1}^{2} \frac{dt}{t}\), dividing the interval \[1,2\] into 2 equal parts gives us subintervals \[1, 1.5\] and \[1.5, 2\]. We use \(t=1\) for the first subinterval and \(t=1.5\) for the second. These values are then used to calculate the Left Riemann Sum:

• \(\Delta t = \frac{1}{2} = 0.5\)
• Sum is given by: \(0.5\left(\frac{1}{1} + \frac{1}{1.5}\right)\)

This yields an approximate value of 0.8333. Since this value is less than 1, we conclude that \(\ln 2 < 1\).

Right Riemann Sum

The Right Riemann Sum calculates an approximation of an integral using the right endpoint of each subinterval. This method similarly forms rectangles, but uses the right endpoint value to determine the height, thus providing a slightly different result than the left sum.

To approximate \(\ln 3\ = \int_{1}^{3} \frac{dt}{t}\), the interval \[1, 3\] is divided into 7 parts, resulting in a width \(\Delta t = \frac{2}{7}\). For each of these subintervals, use the right endpoint value \(t_i\) to calculate the heights:

• Values are \(t_1 = \frac{9}{7}, t_2 = \frac{11}{7}, \ldots, t_7 = 3\)
• Sum is given by: \(\frac{2}{7}\left(\frac{1}{t_1} +\frac{1}{t_2} + \cdots + \frac{1}{t_7}\right)\)

This approach yields a sum that approximately equals 1.035. Since this is greater than 1, it supports that \(\ln 3 > 1\).

Natural Logarithm

The natural logarithm is a logarithmic function with base \(e\), a fundamental mathematical constant approximately 2.71828. Denoted as \(\ln x\), it is widely used in mathematics to represent the time needed for an amount to grow to a certain point under continuous compounding.

In calculus, natural logarithms often appear in integration, especially when dealing with functions of the form \(rac{1}{t}\). Integrals like \(\ln 2 = \int_{1}^{2} \frac{dt}{t}\) provide practical situations where natural logarithms are used to represent growth or decay over intervals. Understanding these logarithms is crucial for solving real-world exponential problems in physics, finance, and other fields.

• It shows how quickly quantities increase
• Important for both theoretical and applied mathematics

Integral Approximation

Integral approximation is a technique used to estimate the value of an integral when it cannot be easily calculated or determined analytically. Methods like the Riemann Sum facilitate understanding of how functions behave over intervals when exact solutions are not immediately accessible.

These approximations are vital in various practical applications. By subdividing an interval and evaluating at specific points—whether through Left or Right Riemann Sums or other methods—we can make educated guesses about the area under a curve. Those sums reflect how continuous and piecewise functions behave over selected x-values.

Using approximations allows for:

• easier computation of complicated functions
• insights into functions with no straightforward antiderivatives
• becoming a precursor for understanding advanced numerical methods

They form a foundation upon which further numerical approximation techniques, such as Trapezoidal and Simpson's rule, are built.