Problem 33 - Calculus and its applications

Question

Evaluate using integration by parts. $$ \int_{2}^{0} \ln (x+8) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(-2 \ln(10) - 2 + 8 \ln\left( \frac{5}{4} \right) \).\)

Step 1: Integration by Parts Formula

We begin by recalling the integration by parts formula for definite integrals:\[ \int_{a}^{b} u \, dv = \left[ u \cdot v \right]_{a}^{b} - \int_{a}^{b} v \, du \]In this problem, we need to choose expressions for \( u \) and \( dv \).

Step 2: Choose u and dv

For the integral \( \int_{2}^{0} \ln(x+8) \, dx \), a good choice is \( u = \ln(x+8) \) since its derivative simplifies nicely, and \( dv = dx \). We then have:\[ u = \ln(x+8) \quad \text{and} \quad dv = dx \]

Step 3: Differentiate u and Integrate dv

Differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \):\[ du = \frac{1}{x+8} \, dx \]\[ v = x \]

Step 4: Apply Integration by Parts Formula

Substitute the expressions for \( u \), \( du \), and \( v \) into the integration by parts formula:\[ \int_{2}^{0} \ln(x+8) \cdot dx = \left[ \ln(x+8) \cdot x \right]_{2}^{0} - \int_{2}^{0} x \cdot \frac{1}{x+8} \, dx \]

Step 5: Evaluate the Boundary Term

Calculate the boundary term \( \left[ \ln(x+8) \cdot x \right]_{2}^{0} \):\[ \left[ \ln(x+8) \cdot x \right]_{2}^{0} = \left[ 0 \cdot \ln(0+8) \right] - \left[ 2 \cdot \ln(2+8) \right] = 0 - 2 \ln(10) = -2 \ln(10) \]

Step 6: Simplify the Remaining Integral

Focus on simplifying the remaining integral: \( \int_{2}^{0} \frac{x}{x+8} \, dx \). Rewrite it as:\[ \int_{2}^{0} \left( 1 - \frac{8}{x+8} \right) \, dx \]Now split the integral into two simpler integrals:\[ \int_{2}^{0} dx - 8 \int_{2}^{0} \frac{1}{x+8} \, dx \]

Step 7: Evaluate Each Integral

Compute each integral separately:1. The first integral: \( \int_{2}^{0} dx = [x]_{2}^{0} = 0 - 2 = -2 \).2. The second integral: \(-8 \int_{2}^{0} \frac{1}{x+8} \, dx = -8 [\ln|x+8|]_{2}^{0} = -8 \cdot ( \ln(8) - \ln(10)) \).Simplify using logarithm properties:\(-8 (\ln(8) - \ln(10)) = 8 \ln\left( \frac{10}{8} \right) = 8 \ln\left( \frac{5}{4} \right) \).

Step 8: Combine Results

Combine all the results to find the final answer:\[ \int_{2}^{0} \ln(x+8) \, dx = -2 \ln(10) - 2 + 8 \ln\left( \frac{5}{4} \right) \].

Key Concepts

Definite Integrals

Definite integrals are a central concept in calculus with essential applications in various fields such as physics, engineering, and economics. They are used to calculate the net area under a curve over a specific interval.
\[\int_{a}^{b} f(x) \, dx\]
In the provided exercise, we aim to find the integral of the natural logarithm function \( \ln(x+8) \) from 2 to 0. One special thing about definite integrals is that they have limits of integration, also known as boundaries, which are applied to provide a precise value, as opposed to an indefinite integral that represents a family of functions.

• Upper and Lower Limits: A definite integral evaluates the area between two points on the x-axis, in this case, between \(x = 2\) and \(x = 0\).
• Evaluation: After integrating, the result is evaluated at these limits to provide a definite number. In our example, the expression \(\left[ \ln(x+8) \cdot x \right]_{2}^{0}\) shows how to apply these boundaries.

Understanding definite integrals requires appreciating that it measures the total accumulation, like total distance traveled or total area under a curve, providing invaluable insights across mathematical and physical applications.

Integration Techniques

When faced with complex functions to integrate, like \( \ln(x+8) \), simple integration rules often don't suffice. That's where integration techniques like integration by parts come in handy. Integration by parts is derived from the product rule for differentiation and allows us to transform difficult integrals into simpler ones.

• Integration by Parts Formula: \[ \int u \, dv = uv - \int v \, du \]
• Choosing \( u \) and \( dv \): It's crucial to select \( u \) such that its derivative \( du \) simplifies the problem. Here, we picked \( u = \ln(x+8) \) because its derivative \( du = \frac{1}{x+8} \, dx \) simplifies nicely, and \( dv = dx \).
• Transforming the Problem: This step turns \( \int_{2}^{0} \ln(x+8) \, dx \) into simpler components, leading to \( \left[ \ln(x+8) \cdot x \right]_{2}^{0} - \int_{2}^{0} x \cdot \frac{1}{x+8} \, dx \).

Mastering different integration techniques such as integration by parts enables solving a wider range of problems, making this skill essential for any calculus student.

Natural Logarithm Integral

The integration of natural logarithm functions can initially seem intimidating, but with clever use of techniques like integration by parts, they become manageable. In the example \( \int_{2}^{0} \ln(x+8) \, dx \), the expression \( \ln(x+8) \) needs special handling.

• Reason for \( u = \ln(x+8) \): Natural logarithms often form part of integrals because their derivatives are straightforward. With \( du = \frac{1}{x+8} \, dx \), the integral becomes easier to handle.
• Simplifying Integrals: After applying integration by parts, simplifying \( \int_{2}^{0} x \cdot \frac{1}{x+8} \, dx \) further to \( \int_{2}^{0} 1 - \frac{8}{x+8} \, dx \) allows us to work with simpler and more straightforward integrals.

Understanding the properties and derivatives of logarithmic functions is essential since it often appears in calculus problems across different contexts. With practice, the integration of natural logarithm expressions becomes less daunting and more systematic.