Question

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

Short answer

The area under the curve is approximately 3.453 square units.

Step-by-step solution

Identify the Integral

To find the area under the curve of a function over a specific interval, we need to calculate the definite integral of the function. The integral we need to solve is:\[\int_{10}^{100} \frac{\log_{10}(x)}{x} \, dx\]

Simplify the Function

Before integration, it's often helpful to simplify the integrand. We can use the change of base formula, which states: \(\log_{10}(x) = \frac{\ln(x)}{\ln(10)}\). Therefore, \[\frac{\log_{10}(x)}{x} = \frac{1}{\ln(10)} \cdot \frac{\ln(x)}{x}\]making the integral: \[\frac{1}{\ln(10)} \int_{10}^{100} \frac{\ln(x)}{x} \, dx\]

Determine the Integration Method

The integral \(\int \frac{\ln(x)}{x} \, dx\) can be solved using integration by parts. Let \(u = \ln(x)\) and \(dv = \frac{1}{x} \, dx\). Then, \(du = \frac{1}{x} \, dx\) and \(v = \ln(x)\).

Apply Integration by Parts Formula

Using the integration by parts formula \(\int u \, dv = u \, v - \int v \, du\), we have:\[\int \frac{\ln(x)}{x} \, dx = \ln(x) \cdot \ln(x) - \int \ln(x) \cdot \frac{1}{x} \, dx\]However, observe that integrating \(\frac{d}{dx}\left(\frac{1}{2}(\ln(x))^2\right) = \frac{\ln(x)}{x}\) gives us the solution, so:\[ \int \frac{\ln(x)}{x} \, dx = \frac{1}{2}(\ln(x))^2 + C\].

Evaluate the Definite Integral

Now let's evaluate \(\frac{1}{\ln(10)}\left[\frac{1}{2}(\ln(x))^2\right]\) from 10 to 100:\[\frac{1}{\ln(10)} \left[ \frac{1}{2}(\ln(100))^2 - \frac{1}{2}(\ln(10))^2 \right]\]Compute this as:\[\ln(100) = 2\ln(10) \quad \text{and} \quad \ln(10) = \ln(10)\]So we have:\[\frac{1}{\ln(10)} \left[ \frac{1}{2}(2\ln(10))^2 - \frac{1}{2}(\ln(10))^2 \right] = \frac{1}{\ln(10)}\left[ 2(\ln(10))^2 - \frac{1}{2}(\ln(10))^2 \right]\]\[= \frac{1}{\ln(10)} \left[ \frac{3}{2}(\ln(10))^2 \right]\]\[= \frac{3}{2} \ln(10)\]

Calculate and Interpret the Result

Substitute \(\ln(10)\) to get the numerical value and interpret the final result. \(\ln(10) \approx 2.302\), so:\[\frac{3}{2} \times 2.302 \approx 3.453\]Thus, the area under the curve of \(f(x)\) from \(x = 10\) to \(x = 100\) is approximately 3.453 square units.

Key concepts

Integration by Parts

Integration by parts is a powerful technique used to solve integrals, particularly those involving products of functions. It is derived from the product rule for differentiation and is expressed as:\[ \int u \, dv = uv - \int v \, du \]This method allows us to break down complex integrals into simpler parts by choosing parts of the integrand to set as \( u \) and \( dv \).

In the exercise above, the integrand \( \frac{\ln(x)}{x} \) suggests integration by parts because it is a product of the natural logarithm function and \( \frac{1}{x} \). Here, we set \( u = \ln(x) \) (since the derivative \( du/dx = \frac{1}{x} \) is straightforward) and \( dv = \frac{1}{x}dx \), resulting in \( v = \ln(x) \). Applying integration by parts simplifies our calculations and helps in finding the area under the curve, as it efficiently handles the complexity of logarithmic functions.

Logarithmic Integration

Logarithmic integration involves integrals with logarithmic functions. Functions that include logarithms can be challenging to integrate directly, but they often become manageable with the right techniques. For example, in our exercise, \( \frac{\ln(x)}{x} \) was transformed into a simpler form using the change of base property of logarithms:
\[ \log_{10}(x) = \frac{\ln(x)}{\ln(10)} \]This transformation simplifies the integrand to
\[ \frac{1}{\ln(10)} \cdot \frac{\ln(x)}{x} \]
enabling us to apply integration by parts.

It’s crucial to understand how logarithmic identities help simplify expressions and make integration tractable. Recognizing when to use logarithmic integration is an important skill, particularly in solving definite integrals where bounds are defined.

Area Under the Curve

Calculating the area under the curve is a common application of definite integrals in mathematics. In our exercise, the goal was to find the area under the curve of \( f(x) = \frac{\log_{10}(x)}{x} \) on the interval from \( x = 10 \) to \( x = 100 \). This task involves integrating the function over a specified range, capturing the total accumulation of the area. The definite integral set up for this task was: \[ \int_{10}^{100} \frac{\log_{10}(x)}{x} \, dx \] Evaluating the definite integral facilitates measuring the area, providing insights into the total value represented by the function over the interval. As seen, the methods of integration by parts and logarithmic simplification played a critical role in solving this problem.

Understanding the relationship between integrals and the area under the curve is vital in various domains, including physics and economics, where such calculations model real-world phenomena.