Problem 121 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The integral evaluates to \( \ln\left(\frac{3}{2}\right) \).

Step 1: Identify the Integral to be Evaluated

The problem presents the definite integral \( \int_{-2}^{-1} \frac{1}{1-u} \, du \). The integrand is \( \frac{1}{1-u} \), and the limits of integration are from \(-2\) to \(-1\).

Step 2: Simplify the Integrand

The integrand \( \frac{1}{1-u} \) can be tricky because of the expression \( 1-u \) in the denominator. We will integrate it by recognizing its form. Let's rewrite \( \frac{1}{1-u} \) as \( -(\frac{1}{u-1}) \). This adjustment will simplify integration using known rules for logarithmic derivatives.

Step 3: Determine the Antiderivative

The antiderivative of \( \frac{1}{u-1} \) is \( \ln|u-1| \). Thus, the antiderivative of \( -(\frac{1}{u-1}) \) is \( -\ln|u-1| \).

Step 4: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, evaluate the antiderivative from \(-2\) to \(-1\):\[ -\ln|-1-1| - (-\ln|-2-1|) = -\ln|-2| + \ln|-3|. \]

Step 5: Simplify the Result

Simplify the expression:\[ -\ln(2) + \ln(3) = \ln\left(\frac{3}{2}\right). \]

Step 6: Conclusion

The value of the definite integral \( \int_{-2}^{-1} \frac{1}{1-u} \, du \) is \( \ln\left(\frac{3}{2}\right) \).

Key Concepts

Integrand Simplification

Before diving into finding an integral, it's beneficial to examine and potentially simplify the integrand itself. In the problem at hand, the integral is given by \( \int_{-2}^{-1} \frac{1}{1-u} \, du \). The integrand, \( \frac{1}{1-u} \), presents a potential difficulty because of the subtraction inside the denominator.
This expression can be rewritten by recognizing that

• \( 1-u \) can be expressed as \(-(u-1)\) since switching the signs reverses the subtraction.
• Therefore, we can simplify \( \frac{1}{1-u} \) to \(-\left(\frac{1}{u-1}\right)\).

This simplification might not seem significant at first glance, but it changes the form into one that's easier to integrate due to its resemblance to the derivative of a natural logarithm. Simplifying integrands in this manner paves the way for smoother integral calculus, easing the process of solving the problem.

Antiderivative

The antiderivative is the reverse process of differentiation, uncovering a function whose derivative is the integrand. Once we've simplified the integrand to \(-\left(\frac{1}{u-1}\right)\) in the previous section, we can focus on finding its antiderivative.
Recognizing patterns in calculus is crucial, and here, the expression \( \frac{1}{u-1} \) resembles the derivative of a natural logarithm function. Specifically, the derivative of \( \ln|u-1| \) is \( \frac{1}{u-1} \), guiding us to conclude that:

• The antiderivative of \( \frac{1}{u-1} \) is \( \ln|u-1| \).
• Given our integrand is \(-\left(\frac{1}{u-1}\right)\), its antiderivative is \(-\ln|u-1| \).

Finding antiderivatives involves recognizing these familiar derivatives and reversing them correctly, ensuring that you have a function whose derivative yields the original integrand.

Fundamental Theorem of Calculus

One of the cornerstone principles in calculus is the Fundamental Theorem of Calculus. This theorem bridges the processes of differentiation and integration, providing a methodical way of evaluating definite integrals. Specifically, it tells us that:

• The definite integral of a function over an interval is the difference between the values of an antiderivative at the endpoints of the interval.
• Mathematically, for function \( f(x) \) and its antiderivative \( F(x) \), \( \int_a^b f(x)\, dx = F(b) - F(a) \).

In our problem, we've identified that the antiderivative of \(-\left(\frac{1}{u-1}\right)\) is \(-\ln|u-1| \).
Applying the Fundamental Theorem, we substitute the bounds into the antiderivative:

• At \( -1 \), the antiderivative \(-\ln|-1-1| = -\ln| -2 | \).
• At \( -2 \), the antiderivative \(-\ln|-2-1| = -\ln| -3 | \).
• The definite integral thus becomes \(-\ln(2) + \ln(3) = \ln\left(\frac{3}{2}\right)\).

This section of the theorem simplifies the evaluation of the integral, highlighting the beauty and utility of calculus in connecting seemingly complex integrals to simple arithmetic calculations at their boundaries.