Problem 122 - Calculus for Biology and Medicine

Question

Evaluate the definite integrals. $$ \int_{2}^{3} \frac{2}{t-1} d t $$

Step-by-Step Solution

Answer

The definite integral is \( 2 \ln 2 \).

Step 1: Identify the Integral Form

The given integral is \( \int_{2}^{3} \frac{2}{t-1} \, dt \). This is a basic form that matches the integral of \( \frac{1}{t-a} \), which is \( \ln|t-a| \).

Step 2: Use the Basic Integration Rule

The integral \( \int \frac{1}{t-a} \, dt \) results in \( \ln|t-a| \). In our problem, we have \( \frac{2}{t-1} \), where \( a = 1 \). Thus, the integral of \( \frac{2}{t-1} \) is \( 2 \ln|t-1| \).

Step 3: Evaluate the Indefinite Integral

Write the indefinite integral form as the result of the integral: \( 2 \ln|t-1| + C \), where \( C \) is the constant of integration. For definite integrals, however, it will be omitted in the next step.

Step 4: Apply the Limits of Integration

Substitute the upper and lower limits into the indefinite integral result: \[ 2 \ln|3-1| - 2 \ln|2-1| \].

Step 5: Simplify the Expression

Calculate the natural logarithms: \( 2 \ln|2| = 2 \ln 2 \) and \( 2 \ln|1| = 0 \). Thus, the definite integral evaluates to \( 2 \ln 2 - 0 \). Hence, the result is \( 2 \ln 2 \).

Key Concepts

Integration Techniques

Integration techniques are methods used to solve integral equations. When evaluating definite integrals, different techniques can be applied depending on the form of the integrand (the function being integrated).

Some common integration techniques include:

• Substitution: This technique involves replacing the variable in the integrand with another variable that simplifies the integrand, making it easier to integrate.
• Integration by Parts: This method is based on the product rule for differentiation and is useful when the integrand is a product of functions.
• Using Known Integrals: Sometimes, integrals can be solved by recognizing the form of the integrand and using an already known integral result.

In the problem presented, knowing the standard integral form \(\int \frac{1}{t-a} \, dt = \ln|t-a|\) is crucial. Recognizing this form allows the problem to be solved quickly and efficiently using Basic Integration Rules.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a logarithm to the base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.718281828459.

It is an essential function in calculus, particularly in integration and differentiation. The natural logarithm has the following key properties:

• \(\ln e = 1\): This is due to the definition of logarithms, as \(e^1 = e\).
• \(\ln 1 = 0\): Because any number to the power of zero is 1, the log of 1 for any base, including \(e\), is zero.
• Derivative: The derivative of \(\ln x\) with respect to \(x\) is \(\frac{1}{x}\), making it particularly useful for integration, as seen in the integral \(\int \frac{1}{t-a} \, dt = \ln|t-a|\).

These features make the natural logarithm indispensable for solving integrals that have rational function forms.

Basic Integration Rules

Basic integration rules provide a foundation for solving a variety of integral problems. Here are some common rules that facilitate the integration process:

• Constant Rule: \(\int c \, dx = cx + C\), where \(c\) is a constant.
• Power Rule: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), for \(n eq -1\).
• Logarithmic Rule: For integrals of the form \(\int \frac{1}{x} \, dx\), the result is \(\ln|x| + C\).

Applying these rules in practice involves identifying the form of the integral.

In the exercise, the natural logarithm rule is used for the integrand \(\frac{2}{t-1}\) by acknowledging it parallels \(\frac{1}{t-a}\). The constant factor of 2 is then factored out, leading to \(2 \ln|t-1|\) before applying the definite integration limits of 2 and 3.