Problem 15 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int_{1}^{2} \frac{\ln x}{x^{2}} d x\)

Step-by-Step Solution

Answer

\( \frac{1}{2} - \frac{\ln 2}{2} \)

Step 1: Identify Integration Method

The integral \( \int \frac{\ln x}{x^{2}} \, dx \) suggests the use of integration by parts, as it contains a product of two functions: \( \ln x \) and \( \frac{1}{x^2} \). Integration by parts formula is \( \int u \, dv = uv - \int v \, du \).

Step 2: Choose u and dv

Choose \( u = \ln x \) and \( dv = \frac{1}{x^2} \, dx \). This allows us to differentiate \( \ln x \) easily and integrate \( \frac{1}{x^2} \).

Step 3: Compute du and v

Differentiating \( u = \ln x \) gives \( du = \frac{1}{x} \, dx \). Integrating \( dv = \frac{1}{x^2} \, dx \) gives \( v = -\frac{1}{x} \).

Step 4: Apply Integration by Parts

Apply the integration by parts formula \( \int u \, dv = uv - \int v \, du \). Substitute the values to get \( \int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} - \int -\frac{1}{x} \cdot \frac{1}{x} \, dx \).

Step 5: Simplify and Solve Last Integral

Simplify the expression to \( -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx \). Integrate \( \frac{1}{x^2} \) to get \( -\frac{1}{x} \). Therefore, the complete integral is \( -\frac{\ln x}{x} - \frac{1}{x} + C \).

Step 6: Evaluate the Definite Integral

Evaluate from 1 to 2: \( \left[-\frac{\ln x}{x} - \frac{1}{x}\right]_{1}^{2} = \left(-\frac{\ln 2}{2} - \frac{1}{2}\right) - \left(-\ln 1 - 1\right) = -\frac{\ln 2}{2} - \frac{1}{2} + 1 = 1 - \frac{\ln 2 + 1}{2} \).

Step 7: Final Calculation

Simplify the expression: \( 1 - \frac{1 + \ln 2}{2} = 1 - \frac{1}{2} - \frac{\ln 2}{2} = \frac{1}{2} - \frac{\ln 2}{2} \).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that help us find the accumulated value of a function over a specific interval. The notation \( \int_{a}^{b} f(x) \, dx \) denotes a definite integral of the function \( f(x) \) from \( x = a \) to \( x = b \). This results in a single numerical value that represents the area under the curve of the function \( f(x) \) between those two points.
When evaluating definite integrals, we often deal with real-life applications such as calculating areas, volumes, and other physical quantities. In the context of the exercise, we evaluated the integral of \( \frac{\ln x}{x^2} \) from 1 to 2. This involved finding the indefinite integral first and then applying specific limits.
**Steps to Evaluate a Definite Integral:**- Find the indefinite integral of the function.- Evaluate the resulting expression at the upper and lower bounds.- Subtract the lower bound evaluation from the upper bound evaluation.
This process provides an exact measure of the desired quantity, as demonstrated in our solution step regarding the definite integral from 1 to 2.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is a logarithm to the base \( e \), where \( e \approx 2.718 \). It is a crucial function in calculus, often appearing in integration and differentiation problems.
The derivative of \( \ln x \) is straightforward: \( \frac{d}{dx} [\ln x] = \frac{1}{x} \). This property makes it easier to handle in calculus problems, especially when using techniques like integration by parts.
**Key Properties of Natural Logarithms:** - \( \ln 1 = 0 \) because \( e^0 = 1 \).- \( \ln(e^x) = x \) because of the inverse relationship with exponentials.In the given problem, \( \ln x \) was selected as part of the integration by parts process. Its derivative is simple, allowing us to solve the integral systematically.

Integration Techniques

Integration techniques are methods used to calculate integrals that do not have straightforward antiderivatives. One such technique is integration by parts, which is particularly useful for integrals involving products of functions. This is expressed by the formula:\[ \int u \, dv = uv - \int v \, du \]In our problem, we tackled the integral \( \int \frac{\ln x}{x^2} \, dx \) using integration by parts. We selected \( u = \ln x \) and \( dv = \frac{1}{x^2} \, dx \) to simplify the integration.
**Steps in Integration by Parts:**- Choose \( u \) and \( dv \) wisely, such that the resulting \( du \) and \( v \) are manageable.- Differentiate \( u \) to get \( du \), and integrate \( dv \) to find \( v \).- Apply the integration by parts formula.- Simplify and solve the resulting integrals if necessary.
This technique showcases the power of breaking down complex expressions into simpler parts to find a solution, and is an essential tool in a student's integration skills arsenal.