Problem 44 - Calculus for Biology and Medicine

Question

As definite integrals. Note that (1) \(P=\left[x_{0}, x_{1}, \ldots, x_{n}\right]\) is a partition of the indicated interval, (2) \(c_{k} \in\left[x_{k-1}, x_{k}\right]\), and (3) \(\Delta x_{k}=x_{k}-x_{k-1}\). \(\lim _{\|P\| \rightarrow 0} \sum_{k=1}^{n} \frac{1}{c_{k}+1} \Delta x_{k}\), where \(P\) is a partition of \([1,2]\)

Step-by-Step Solution

Answer

The limit of the sum is the integral \(\int_{1}^{2} \frac{1}{x+1} \, dx = \ln \left(\frac{3}{2}\right)\).

Step 1: Recognize the Riemann Sum

The limit of a sum that you see, \(\lim _{\|P\| \rightarrow 0} \, \sum_{k=1}^{n} \frac{1}{c_{k}+1} \, \Delta x_{k}\), represents the Riemann sum for a definite integral. As the partition size \(\|P\|\) approaches zero, this sum approaches an integral over the interval \([1, 2]\).

Step 2: Identify the Function and Interval

In the Riemann sum, the term \(\frac{1}{c_{k}+1}\) resembles the function \(f(x) = \frac{1}{x+1}\). The sum indicates that this function should be integrated over the interval \([1, 2]\).

Step 3: Convert Riemann Sum to Definite Integral

Recognize that \(\lim _{\|P\| \rightarrow 0} \, \sum_{k=1}^{n} f(c_{k}) \, \Delta x_{k}\) becomes \(\int_{1}^{2} \frac{1}{x+1} \, dx\) when expressed as a definite integral.

Step 4: Calculate the Definite Integral

To solve the integral \(\int_{1}^{2} \frac{1}{x+1} \, dx\), set \(u = x+1\), which means \(du = dx\). Transforming the limits: at \(x = 1\), \(u = 2\) and at \(x = 2\), \(u = 3\). The integral becomes \(\int_{2}^{3} \frac{1}{u} \, du\).

Step 5: Evaluate the Transformed Integral

The integral \(\int \frac{1}{u} \, du\) has an antiderivative of \(\ln|u|\). Evaluate from 2 to 3: \(\ln|3| - \ln|2| = \ln{3} - \ln{2}\).

Step 6: Simplify the Result

Use the properties of logarithms to simplify \(\ln{3} - \ln{2}\) to \(\ln \left(\frac{3}{2}\right)\). Thus, the original Riemann sum converges to \(\ln \left(\frac{3}{2}\right)\).

Key Concepts

Definite Integral

The concept of a definite integral is fundamental in calculus. It's a way to calculate the accumulation of quantities, often representing the area under a curve over a specific interval. Imagine slicing this area into a number of rectangles, which get smaller and more numerous as we refine our approach.
For instance, take the following equation from our example: \[\lim _{\|P\| \rightarrow 0} \, \sum_{k=1}^{n} \frac{1}{c_{k}+1} \, \Delta x_{k} \] This is a Riemann sum — a sum that approximates the area under a curve \(f(x)\) across an interval. As the number of slices increases and their width approaches zero, this sum converges to a definite integral. \[\int_{1}^{2} \frac{1}{x+1} \, dx \] Thus, the definite integral provides the exact area, achieving what the Riemann sum approximates.

Partition

A partition in the context of integrals and Riemann sums involves dividing an interval into smaller segments. It's like breaking a chocolate bar into smaller pieces to measure it more accurately. You will often see a partition represented as \(P = [x_0, x_1, \ldots, x_n]\).

• The partition points \(x_0, x_1, \dots, x_n\) divide the interval.
• The difference between these points, \(\Delta x_k = x_k - x_{k-1}\), represents the width of each subinterval.

When the norm of a partition \(\|P\|\) approaches zero, it means these subintervals become finer. This refinement is key to ensuring that the Riemann sum approaches the true value of the integral.

Antiderivative

Understanding antiderivatives is essential for solving definite integrals. An antiderivative of a function \(f(x)\) is another function whose derivative is \(f(x)\). Simply put, if you find a function \(F(x)\) such that \(F'(x) = f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\).
For the definite integral we are dealing with, \[\int \frac{1}{u} \, du\]The antiderivative is \(\ln|u|\), signifying the natural logarithm of \(u\). This means that solving the integral involves evaluating \(\ln|u|\) at the upper and lower limits of the interval, yielding:\[\ln|3| - \ln|2|\] Antiderivatives enable the transition from individual terms of an integral to a comprehensive solution.

Logarithm Properties

Logarithms come with a set of properties that are particularly handy when dealing with integrals. The exercise at hand involves simplifying terms into a single, concise expression using these properties. Consider the evaluation \[\ln|3| - \ln|2|\] Using logarithm properties, this can be combined into a single logarithm:

• The difference of two logs: \(\ln a - \ln b\) becomes \(\ln \left(\frac{a}{b}\right)\).

By applying this property, we simplify \[\ln|3| - \ln|2| = \ln \left(\frac{3}{2}\right)\]Logarithm properties simplify calculations immensely, turning complex expressions into manageable forms.