Problem 7 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Use Definition of Integral II to evaluate $$\int_{1}^{2} \frac{1}{t^{2}} d t$$ Partition [1,2] in \(n\) equal subintervals by $$t_{0}=1, \quad t_{1}=1+\frac{1}{n}, \quad \cdots, \quad t_{k-1}=1+\frac{k-1}{n}, \quad t_{k}=1+\frac{k}{n},\quad\cdots \quad t_{n}=1+\frac{n}{n}$$ Let $$\tau_{k}=\sqrt{t_{k-1} \times t_{k}}, \quad k=1,2, \cdots, n$$ a. Show that \(t_{k-1} \leq \tau_{k} \leq t_{k}\). b. Write Equation 9.16 , $$\int_{a}^{b} f(t) d t=\lim _{\|\Delta\| \rightarrow 0} \sum_{k=1}^{n} f\left(\tau_{k}\right) \times\left(t_{k}-t_{k-1}\right),$$ for \(\int_{1}^{2} \frac{1}{t^{2}} d t,\) the given partition and values of \(\tau_{k}\). c. Show that $$\int_{1}^{2} \frac{1}{t^{2}} d t=\lim _{\|\Delta\| \rightarrow 0} \sum_{k=1}^{n}\left(\frac{1}{t_{k-1}}-\frac{1}{t_{k}}\right)$$ d. Write the previous sum in long form and show that $$\int_{1}^{2} \frac{1}{t^{2}} d t=\lim _{\|\Delta\| \rightarrow 0}\left(1-\frac{1}{2}\right)=\frac{1}{2}$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{2} \).

Step 1: Establish Inequality for Midpoint

To show that \( t_{k-1} \leq \tau_{k} \leq t_{k} \), note that \( \tau_k = \sqrt{t_{k-1} \times t_{k}} = \sqrt{(1+\frac{k-1}{n})(1+\frac{k}{n})} \). Since \( \sqrt{ab} \) is the geometric mean, it lies between \( a \) and \( b \). Hence, \( t_{k-1} \leq \tau_k \leq t_k \).

Step 2: Rewrite Integral Definition

According to the integral definition, \[ \int_{1}^{2} \frac{1}{t^{2}} dt = \lim _{\|\Delta\| \rightarrow 0} \sum_{k=1}^{n} \frac{1}{(\tau_k)^2} \cdot \left(t_{k} - t_{k-1}\right) \] Use \( \tau_k = \sqrt{t_{k-1} t_k} \) to express the equation specifically for this partition.

Step 3: Simplify the Expression

To evaluate the limit, approximate \( \frac{1}{(\tau_k)^2} \approx \frac{1}{t_{k-1}^2} \) as \( n \rightarrow \infty \). Therefore, \[ \int_{1}^{2} \frac{1}{t^{2}} dt = \lim _{\|\Delta\| \rightarrow 0} \sum_{k=1}^{n} \left( \frac{1}{t_{k-1}} - \frac{1}{t_{k}} \right) \] as shown in equation.

Step 4: Expand and Evaluate the Sum

The sum expands to \[ \left( \frac{1}{1} - \frac{1}{1+\frac{1}{n}} \right) + \left( \frac{1}{1+\frac{1}{n}} - \frac{1}{1+\frac{2}{n}} \right) + \cdots + \left( \frac{1}{1+\frac{n-1}{n}} - \frac{1}{2} \right) \] This is a telescoping series where intermediate terms cancel, leaving \( 1 - \frac{1}{2} \).

Step 5: Conclude Using Limit

The final expression becomes \[ \int_{1}^{2} \frac{1}{t^{2}} dt = \lim _{\|\Delta\| \rightarrow 0} \left( 1 - \frac{1}{2} \right) = \frac{1}{2} \] This satisfies the original integral evaluation.

Key Concepts

Riemann Sum

A Riemann Sum is a method used to approximate the definite integral of a function. It involves dividing the area under the curve of a function into small, simpler shapes, such as rectangles, and summing their areas to approximate the total area. In the context of our exercise, the interval [1, 2] is divided into subintervals using points called partition points.

• Each subinterval is associated with a point within it, often a midpoint or other sample point like \( \tau_k \)).
• The product of the function value at this point and the width of the subinterval gives a rectangle's area, which approximates the area under the curve.

The more subintervals we have, the better our approximation. As the number of subintervals becomes infinitely large (i.e., as \( n \rightarrow \infty \)), the Riemann Sum converges to the exact value of the integral, which is shown as a limit.

Geometric Mean

Geometric Mean is a type of average, often used for sets of numbers where differences in scale might otherwise skew a simple average. In this exercise, the geometric mean of two partition points \( t_{k-1} \) and \( t_k \) gives us a sample point \( \tau_k = \sqrt{t_{k-1} \times t_k} \).

• This is strategically chosen as the sample point, since it's always assured that \( t_{k-1} \leq \tau_k \leq \ t_k \), allowing \tau_k \ to be representative of its respective subinterval.
• This ensures that the Riemann Sum calculation using these midpoints is both effective and convenient.

The geometric mean helps balance the values at the ends of each subinterval and provides a symmetrical sample point, especially when dealing with increasing or decreasing functions.

Telescoping Series

A Telescoping Series is a series where most terms cancel out in a pattern. In our exercise, the sum expands into a telescoping series:

• Each term in the sequence cancels with a subsequent or preceding term.
• This simplifies the evaluation of the series since we are left with only the first and last terms.

In the definite integral exercise, this series is used to simplify the evaluation of the limit. After expansion, many intermediary terms disappear, ultimately leaving us only the endpoints, namely \( 1 - \frac{1}{2} \). This is why telescoping series are particularly powerful tools when evaluating sums and limits.

Integral Calculus

Integral Calculus focuses on finding the area under curves, which is represented by the definite integral. It involves processes like antidifferentiation, but primarily addresses using integrals to compute accumulated quantities like areas.

• In the exercise, we start with the integral \( \int_{1}^{2} \frac{1}{t^2} dt \).
• By using the limit of a Riemann Sum approach, we translate the problem into manageable segments by breaking the integral into smaller calculable parts.

The ingenuity of integral calculus lies in its ability to convert these tiny sums into precise calculations of total areas, providing mathematicians and scientists a tool to solve complex real-world problems involving changes and areas.