Problem 120 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The value of the integral is \( \ln \left( \frac{4}{3} \right) \).

Step 1: Identify the Function

The integral to be evaluated is \( \int_{2}^{3} \frac{1}{z+1} \, dz \). The function we are dealing with is \( \frac{1}{z+1} \), which is a simple rational function.

Step 2: Determine the Antiderivative

The antiderivative of \( \frac{1}{z+1} \) is \( \ln|z+1| \). Thus, the antiderivative we will use is \( F(z) = \ln|z+1| \).

Step 3: Apply the Fundamental Theorem of Calculus

Use the Fundamental Theorem of Calculus which states that if \( F \) is an antiderivative of \( f \) on \([a, b]\), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). Here, \( a = 2 \) and \( b = 3 \).

Step 4: Evaluate the Antiderivative at the Upper Limit

Substitute \( z = 3 \) into the antiderivative: \( F(3) = \ln|3+1| = \ln 4 \).

Step 5: Evaluate the Antiderivative at the Lower Limit

Substitute \( z = 2 \) into the antiderivative: \( F(2) = \ln|2+1| = \ln 3 \).

Step 6: Calculate the Definite Integral

Now, subtract the value of the antiderivative at the lower limit from the value at the upper limit: \( \ln 4 - \ln 3 \).

Step 7: Simplify the Expression

Using logarithmic properties, \( \ln 4 - \ln 3 = \ln \left( \frac{4}{3} \right) \). This is the value of the definite integral.

Key Concepts

Antiderivative

In calculus, an antiderivative is a function that reverses the process of differentiation. It helps us find the original function from its derivative. For any function, its antiderivative is not unique and there can be many possibilities differing by a constant. However, in the context of definite integrals, we are usually interested in one specific antiderivative. For the problem of evaluating the integral \( \int_{2}^{3} \frac{1}{z+1} \, dz \), we have to identify the antiderivative of \( \frac{1}{z+1} \).

• This function is a simple rational function.
• The antiderivative is \( \ln|z+1| \).

Understanding how to find antiderivatives is crucial because it sets the foundation for solving integrals, allowing us to understand the behavior and area under curves. In this particular exercise, knowing the antiderivative allows us to evaluate the definite integral over the interval \([2, 3]\).

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between differentiation and integration. It consists of two main parts. The first part establishes the relationship between differentiation and antiderivatives. The second part, which is key for solving definite integrals, tells us that if \( F \) is an antiderivative of a function \( f \) on an interval \([a, b]\), then \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]. In our exercise with \( \int_{2}^{3} \frac{1}{z+1} \, dz \), this theorem means:

• Find an antiderivative of the function, which is \( \ln|z+1| \).
• Evaluate this antiderivative at the upper and lower limits of integration.
• Subtract these values to get the result.

This approach simplifies the process of definite integration, as it reduces the complex process of finding the area under a curve to a simple subtraction problem.

Rational Functions

Rational functions are quotients of polynomials and play a vital role in calculus. Many functions can be expressed as rational functions, making them essential for various calculations. In our given integral problem, the function \( \frac{1}{z+1} \) is a rational function because:

• It can be expressed as a polynomial divided by another polynomial.
• It is of the form \( \frac{P(z)}{Q(z)} \) where \( P(z) = 1 \) and \( Q(z) = z+1 \).

Why are rational functions significant? Because they often have simpler antiderivatives compared to more complex functions, allowing for straightforward computation of integrals. Real-world data is frequently modeled using rational functions, which are versatile and easily manipulated within calculus operations. Understanding how to handle these functions helps in tackling a wide array of mathematical problems, including the evaluation of definite integrals as seen in this exercise.