Problem 38 - Calculus for Biology and Medicine

Question

$$ \text { In Problems } , \text { evaluate each definite integral. } $$ $$ \int_{3}^{5} \frac{x}{x-1} d x $$

Step-by-Step Solution

Answer

The value of the integral is \( 2 + \ln 2 \).

Step 1: Identify the Integral Type

The given integral is \( \int_{3}^{5} \frac{x}{x-1} \, dx \), which is a rational function. Our task is to evaluate this definite integral from 3 to 5.

Step 2: Simplify the Integrand

Let's simplify the integrand \( \frac{x}{x-1} \). We can express it as \( 1 + \frac{1}{x-1} \) by performing polynomial long division. Thus, \( \int_{3}^{5} \frac{x}{x-1} \, dx = \int_{3}^{5} \left( 1 + \frac{1}{x-1} \right) \ dx \).

Step 3: Split the Integral

Now separate the integral into two parts: \( \int_{3}^{5} 1 \, dx + \int_{3}^{5} \frac{1}{x-1} \, dx \). This step allows us to handle each part independently.

Step 4: Evaluate the First Integral

The first integral \( \int_{3}^{5} 1 \, dx \) is straightforward. It evaluates to \( [x]_{3}^{5} = 5 - 3 = 2 \).

Step 5: Evaluate the Second Integral

For the second integral \( \int_{3}^{5} \frac{1}{x-1} \, dx \), we use the standard result for \( \int \frac{1}{x-a} \, dx = \ln |x-a| \). Thus, it evaluates to \([\ln|x-1|]_{3}^{5} = \ln|5-1| - \ln|3-1| = \ln 4 - \ln 2 \).

Step 6: Simplify the Result

Combine the results from Steps 4 and 5: \( 2 + (\ln 4 - \ln 2) \). Using the property of logarithms \( \ln a - \ln b = \ln \frac{a}{b} \), we simplify this to \( 2 + \ln 2 \).

Step 7: Present the Final Answer

The value of the definite integral \( \int_{3}^{5} \frac{x}{x-1} \, dx \) is \( 2 + \ln 2 \).

Key Concepts

Definite Integral

The definite integral is a central concept in calculus that computes the net area under a curve within a specified interval on the x-axis. Indicated by the notation \( \int_{a}^{b} f(x) \, dx \), it represents the accumulation of quantities between the limits \( a \) and \( b \). In essence, it provides a way to calculate the sum of infinitely many infinitesimally small quantities, which could represent anything from distance to probability.

For the integral \( \int_{3}^{5} \frac{x}{x-1} \, dx \), evaluation involves determining the signed area between the curve \( \frac{x}{x-1} \) and the x-axis from \( x=3 \) to \( x=5 \). This process includes both understanding and transforming the integrand through algebraic manipulation to facilitate easier integration.

Understanding definite integrals involves:

• Identifying the function forming the curve.
• Determining the bounds or limits of the integration.
• Calculating the net area using integral properties.

Mastery of definite integrals enables students to solve problems involving areas between curves, volumes of solids of revolution, and other applications in physics and engineering.

Rational Functions

Rational functions are expressions that take the form of fractions, where both the numerator and the denominator are polynomials. A simple example is \( \frac{x}{x-1} \). Such functions often appear in calculus problems because they can describe various real-world scenarios, including rates and populations.

Because of their structure, rational functions can be tricky to integrate directly. It's usually necessary to simplify them into more manageable parts. This process might involve:

• Performing polynomial long division if the degree of the numerator is greater than or equal to that of the denominator.
• Using partial fraction decomposition if applicable.

In our exercise, simplifying the function \( \frac{x}{x-1} \) into \( 1 + \frac{1}{x-1} \) allows for an easier separation into integrable components, ultimately making the definite integral process more straightforward.

Polynomial Long Division

Polynomial long division is a technique used to simplify rational expressions, much like long division in arithmetic but with variables. This method is particularly useful when dealing with rational functions where the degree of the numerator is equal to or greater than the degree of the denominator.

In our example problem, polynomial long division allows us to express \( \frac{x}{x-1} \) as \( 1 + \frac{1}{x-1} \). Here's a brief look at how it works:

• Divide the leading term of the numerator by the leading term of the denominator.
• Multiply the entire divisor by this quotient and subtract the result from the original numerator.
• Repeat the process with the new polynomial until the remainder is of lower degree than the divisor.

This breakdown transforms a potentially complex integration problem into simpler terms, making it easier to carry out the integration of individual components. Understanding and applying polynomial long division is an essential skill in tackling complicated rational functions within calculus.