Problem 34 - Applied Calculus

Question

Use integration by parts to find each integral. \(\int \ln x^{2} d x\)

Step-by-Step Solution

Answer

The integral is \( x \ln x^2 - 2x + C \).

Step 1: Identify Parts for Integration by Parts

In integration by parts, we need to identify parts for the formulas, \( u \) and \( dv \). Here, let \( u = \ln x^2 \), so that \( du \) becomes the derivative of \( \ln x^2 \), and \( dv = dx \), making \( v = x \).

Step 2: Compute Derivative and Integral

The derivative of \( u \), which is \( \ln x^2 \), is \( du = \frac{2}{x} dx \). The integral of \( dv \), where \( dv = dx \), is \( v = x \).

Step 3: Apply Integration by Parts Formula

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

Step 4: Simplify the Integral

Simplify the integral \( \int x \cdot \frac{2}{x} dx \) to \( \int 2 \, dx \). The integral of \( 2 \) with respect to \( x \) is \( 2x \).

Step 5: Combine and Simplify

Now substitute back and simplify the expression from step 3: \( x \ln x^2 - 2x + C \), where \( C \) is the constant of integration.

Key Concepts

Definite Integral

The concept of a definite integral can be quite fascinating. It essentially represents the accumulation of quantities and gives the net area under a curve. When we talk about a definite integral of a function, say over the interval \([a, b]\), it provides a numerical value. This value corresponds to the area between the graph of the function and the x-axis.The definite integral \( \int_a^b f(x) \, dx \) is denoted using limits of integration, \(a\) and \(b\). These limits define where the integration starts and stops.Understanding definite integrals helps with:

• Calculating areas under curves.
• Solving real-world problems involving accumulation, like distance traveled over time.
• Providing solutions to differential equations.

Unlike indefinite integrals, the definite integral of a continuous function provides a specific value rather than a family of functions with an added constant.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a specific logarithm with several distinct features. It uses the mathematical constant \(e\) as its base. This constant, known as Euler's number, is approximately equal to 2.71828.The natural logarithm \(\ln(x)\) is pivotal in mathematics because of its simplicity when it comes to differentiation and integration. For instance:

• The derivative of \(\ln(x)\) is \(1/x\), which is straightforward to handle in calculus.
• It usually appears in problems involving growth processes or decay processes because of its unique properties.

When you encounter functions like \(\ln(x^2)\), remember that properties of logarithms allow us to simplify calculations. Here, \(\ln(x^2) = 2 \ln(x)\), making some calculus problems more manageable. Mastering the natural logarithm can bolster your problem-solving ability in calculus and other mathematics applications.

Integration Techniques

Integration is the reverse process of differentiation, but it can sometimes be challenging. Luckily, there exist various integration techniques to help find an integral, especially when the direct approach isn't feasible.One powerful technique is "Integration by Parts," useful when dealing with the product of functions. Derived from the product rule of differentiation, it's expressed as:\[\int u \, dv = uv - \int v \, du\]Here’s a quick guide on applying integration by parts:

• Identify parts of the function that can be split into \(u\) and \(\frac{dv}{dx}\).
• Differentiate \(u\) to get \(du\) and integrate \(dv\) to find \(v\).
• Substitute into the integration by parts formula.

Other techniques include substitution, partial fraction decomposition, and trigonometric identities. Learning these techniques helps tackle integrals that seem complex at first glance.By systematically applying the appropriate technique, integrating complex functions becomes manageable and often rewarding.