Problem 41 - Calculus for Biology and Medicine

Question

$$ \text { In Problems } , \text { evaluate each definite integral. } $$ $$ \int_{2}^{3} \frac{1}{1-x} d x $$

Step-by-Step Solution

Answer

The evaluated integral is \(-\ln 2\).

Step 1: Identify the integral

We are given the definite integral \( \int_{2}^{3} \frac{1}{1-x} \, dx \). Our task is to evaluate this integral from \( x = 2 \) to \( x = 3 \).

Step 2: Simplify the integrand

Rewrite the integrand \( \frac{1}{1-x} \) as \( -\frac{1}{x-1} \) by factoring out and changing the sign.

Step 3: Integrate the simplified function

Find the indefinite integral of \( -\frac{1}{x-1} \), which is \(-\ln|x-1| + C \) where \( C \) is the constant of integration.

Step 4: Apply the limits of integration

Use the Fundamental Theorem of Calculus to evaluate the definite integral. Substitute the upper limit 3 and the lower limit 2 into the antiderivative: \[- (\ln|3-1| - \ln|2-1|) = -(\ln 2 - \ln 1) \]

Step 5: Simplify the expression

\( \ln 1 = 0 \), so the expression simplifies to:\[- (\ln 2 - 0) = - \ln 2 \]

Step 6: Final Step: Conclude with the evaluated integral

The evaluated integral \( \int_{2}^{3} \frac{1}{1-x} \, dx \) is \(- \ln 2 \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus beautifully connects the concept of differentiation and integration, two seemingly distinct operations in calculus. It states that if a function is continuous over a closed interval, like our example from 2 to 3, then the definite integral of the function over that interval can be found using the antiderivative. This theorem has two parts:

• First Part: This part provides an assurance that if a function is integrable, then its integral function is differentiable and its derivative returns back to the original function within the interval.
• Second Part: The part we use when calculating definite integrals. It tells us that to find the exact value of an integral, we need an antiderivative of the function, and we must evaluate it at the upper and lower limits. The difference gives us the exact value. In this exercise, we found the antiderivative of the function \(\frac{1}{1-x}\) and evaluated it from 2 to 3.

By using the Fundamental Theorem of Calculus, complex area problems are simplified. In this case, the theorem makes evaluating the integral from 2 to 3 straightforward, ultimately showing that the value is \- \ln 2\.

natural logarithm

Natural logarithms are logarithms to the base \(e\), which is an irrational number approximately equal to 2.71828. This type of logarithm is commonly represented by "ln." Unlike logarithms with other bases, natural logarithms have unique properties that simplify calculus operations.

• Property of Natural Logarithms: One key property is that the natural logarithm of 1 is 0, \(\ln 1 = 0\). This helps simplify expressions with logarithms, like when finding antiderivatives.
• Appearance in Calculus: In integration, natural logarithms often appear when dealing with rational functions. For our integral, the antiderivative of \(-\frac{1}{x-1}\) gives us \(-\ln|x-1|\).

These properties often make calculations easier. We saw that by identifying the antiderivative and applying natural logarithm properties, the definite integral was quickly solved.

antiderivative

An antiderivative is essentially a "reverse" operation to differentiation. While differentiation gives you the rate at which a function changes, finding an antiderivative gives you back the original function (up to a constant). An integral is just another name for collectively considering the antiderivatives, especially when considering definite integrals.

• Finding Antiderivatives: To solve our definite integral, we first transformed \(\frac{1}{1-x}\) into a simpler form \(-\frac{1}{x-1}\), making it easier to integrate.
• Using Antiderivatives in Definite Integrals: The second part of the Fundamental Theorem says to evaluate an antiderivative at specific limits. In our problem, after finding the antiderivative \(-\ln|x-1|\), we used limits 2 and 3 to calculate the final answer \(-\ln 2\).

Understanding and identifying antiderivatives is key to solving integration problems, allowing us to move smoothly from discovering functions to evaluating specific values at given points.