Problem 120 - Calculus for Biology and Medicine

Question

Evaluate the definite integrals. $$ \int_{0}^{1} \frac{1}{z+1} d z $$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln 2 \).

Step 1: Determine the Antiderivative

The first step in evaluating the definite integral is to find the antiderivative of the function \( \frac{1}{z+1} \). The antiderivative of \( \frac{1}{z+1} \) is \( \ln|z+1| + C \), where \( C \) is the constant of integration. Since we are dealing with a definite integral, the constant \( C \) is not needed.

Step 2: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F \) is the antiderivative of \( f \) on an interval \([a, b]\), then the definite integral of \( f \) from \( a \) to \( b \) is \( F(b) - F(a) \). In our case, the antiderivative \( F(z) \) is \( \ln|z+1| \). Thus, evaluate \( F(z) \) at the upper and lower limits: \( F(1) = \ln|1+1| = \ln 2 \) and \( F(0) = \ln|0+1| = \ln 1 \).

Step 3: Calculate the Definite Integral

Now, we subtract the value of the antiderivative at the lower limit from the value at the upper limit: \( \ln 2 - \ln 1 \). Since \( \ln 1 = 0 \), the result simplifies to \( \ln 2 \).

Key Concepts

Antiderivative

An antiderivative of a function is essentially the reverse of taking a derivative. For any given function, the antiderivative is another function that, when differentiated, gives you the original function back.

• Finding an antiderivative is a key step in solving definite integrals because it allows us to measure the accumulation of changes across an interval.
• In the exercise, the given function is \( \frac{1}{z+1} \), and its antiderivative is \( \ln|z+1| + C \).
• The constant \( C \) appears in indefinite integrals but is irrelevant for definite integrals, which are concerned with the change between two points.

Understanding how to find the antiderivative is crucial when working with integrals, as it converts the problem from finding an area under a curve to a simpler subtraction problem. This concept is essential for applying the Fundamental Theorem of Calculus later on.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a powerful tool in the field of calculus, linking differentiation with integration, two of the central concepts.

• The theorem states that if \( F \) is an antiderivative of \( f \) over an interval \([a, b]\), then the definite integral of \( f \) from \( a \) to \( b \) is given by \( F(b) - F(a) \).
• This theorem simplifies the process of finding the area under a curve by turning it into evaluating the antiderivative at the boundary points.

In the problem, you evaluated \( F(1) = \ln 2 \) and \( F(0) = \ln 1 \). By subtracting these results, you find the total change, or the integral, over the interval from \( 0 \) to \( 1 \), which results in \( \ln 2 \).
This theorem provides the bridge between calculating areas, summing infinitely small lines, and more intricate applications of calculus.

Logarithmic Functions

Logarithmic functions are important in many fields including calculus, where they often appear as antiderivatives.

• The natural logarithm, denoted \( \ln \), is a logarithm with base \( e \), where \( e \approx 2.718 \).
• The function \( \ln(x) \) is the inverse of the exponential function \( e^x \).

In the problem, you identify the antiderivative of \( \frac{1}{z+1} \) as \( \ln|z+1| \). This highlights a special rule in calculus: the derivative of \( \ln(x) \) is \( \frac{1}{x} \), which is the reverse process used here. Understanding logarithmic functions are essential for tackling integrals involving relative changes and growth problems, making them a vital tool in your calculus toolkit.