Problem 372 - Calculus Volume 1

Question

In the following exercises, \(f(x) \geq 0\) for \(a \leq x \leq b\) . Find the area under the graph of \(f(x)\) between the given values \(a\) and \(b\) by integrating. $$f(x)=\frac{\log _{10}(x)}{x} ; a=10, b=100$$

Step-by-Step Solution

Answer

The area under the curve is approximately 3.453 square units.

Step 1: Understand the Problem

We are asked to find the area under the curve of the function \( f(x) = \frac{\log_{10}(x)}{x} \) between \( x = 10 \) and \( x = 100 \). This involves calculating a definite integral from \( a \) to \( b \).

Step 2: Set Up the Integral

The area under the curve from \( a \) to \( b \) is given by the definite integral \( \int_{a}^{b} f(x) \, dx \). In this case, we need to evaluate the integral \( \int_{10}^{100} \frac{\log_{10}(x)}{x} \, dx \).

Step 3: Simplify the Integrand

Convert \( \log_{10}(x) \) to the natural logarithm using the formula \( \log_{10}(x) = \frac{\ln(x)}{\ln(10)} \). Thus, the integral becomes \( \int_{10}^{100} \frac{\ln(x)}{x \cdot \ln(10)} \, dx \).

Step 4: Integrate the Simplified Expression

Recognize that \( \frac{\ln(x)}{x} \) is the derivative of \( \frac{1}{2}²(x) \). Therefore, the integral is \( \frac{1}{\ln(10)} \int \frac{\ln(x)}{x} \, dx = \frac{1}{\ln(10)} \cdot \frac{1}{2} \ln^2(x) + C \).

Step 5: Evaluate the Definite Integral

Evaluate the indefinite integral from Step 4 at \( a = 10 \) and \( b = 100 \) to find: \[ \frac{1}{2\ln(10)} [\ln^2(100) - \ln^2(10)] \] Calculate \( \ln(100) = 2\ln(10) \) and \( \ln(10) = \ln(10) \) to get: \[ = \frac{1}{2\ln(10)} [(2\ln(10))^2 - (\ln(10))^2] = \frac{1}{2\ln(10)} [4\ln^2(10) - \ln^2(10)] \] \[ = \frac{3\ln^2(10)}{2\ln(10)} \] Simplify to get \( \frac{3\ln(10)}{2} \approx 3(2.302)/2 = 3.453 \). Thus, the numerical value is approximately 3.453.

Key Concepts

Area Under a Curve

Finding the area under a curve is a fundamental concept in calculus. It involves integrating a function over a specific interval. In this exercise, we are looking at the function \( f(x) = \frac{\log_{10}(x)}{x} \) and its area from \( x = 10 \) to \( x = 100 \). Integration between these limits gives us the area under the curve, which can be interpreted as the total accumulation of the function's values within this range.

• The function \( f(x) \geq 0 \) ensures that the area calculated is non-negative, a basic requirement to represent geometric area.
• To calculate the area, we set up the definite integral \( \int_{10}^{100} \frac{\log_{10}(x)}{x} \, dx \).

Understanding the area under a curve helps in various fields such as physics for calculating work done, in probability for total probability under a curve, and more. Calculating these definite integrals allows for practical applications and deeper understanding.

Natural Logarithm Integration

Integrating functions that involve logarithms can sometimes be tricky, especially when dealing with logarithms with bases other than \( e \). For our function \( f(x) = \frac{\log_{10}(x)}{x} \), it becomes easier once we convert the logarithm base to the natural logarithm, commonly denoted as \( \ln \).

• The conversion from base 10 to base \( e \) is done using \( \log_{10}(x) = \frac{\ln(x)}{\ln(10)} \).
• This transforms our integral into a more manageable form: \( \int_{10}^{100} \frac{\ln(x)}{x \cdot \ln(10)} \, dx \).

Recognizing \( \frac{\ln(x)}{x} \) as a part of the derivative \( \ln^2(x)/2 \) allows us to integrate more easily. Using such transformations and recognitions is crucial in simplifying and solving complex integrals. This understanding paves the way for handling more advanced problems in calculus and analysis.

Simplifying Integrands

Simplification of integrands can significantly ease the process of integration, turning complex functions into standard integrate-friendly forms. In this exercise, we began with \( f(x) = \frac{\log_{10}(x)}{x} \), which might seem daunting at first.

• Through simplification, using the relationship between different logarithmic bases, we arrived at an integrand \( \frac{\ln(x)}{x \cdot \ln(10)} \).
• This step paved the way for recognizing it as a derivative form: \( \frac{1}{2\ln(10)} \ln^2(x) \).

Simplifying integrands often involves looking for patterns or parts that resemble common derivatives or integrals. This step is a vital part of calculus, making problems more approachable and often less time-consuming to solve. It's a technique that helps ensure efficiency and accuracy in problem-solving, allowing students to handle a wider range of integrals effectively.