Problem 50 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \frac{\ln v d v}{v}=\) (A) \(\ln |\ln v|+C\) (B) \(\ln \frac{v^{2}}{2}+C\) (C) \(\frac{1}{2}(\ln v)^{2}+C\) (D) \(\frac{1}{2} \ln v^{2}+C\)

Step-by-Step Solution

Answer

The correct answer is (C) \(\frac{1}{2}(\ln v)^{2}+C\).

Step 1: Identify the Form of the Integral

The integral we have to solve is \(\int \frac{\ln v \, dv}{v}\). This integral resembles the form \(\int u \, du\) if we perform a substitution where the inside function \(u\) is \(\ln v\).

Step 2: Perform the Substitution

Let \(u = \ln v\). Then take the derivative of \(u\) with respect to \(v\), which gives \(\frac{du}{dv} = \frac{1}{v}\). This implies \(du = \frac{1}{v}\, dv\). Thus, the integral becomes \(\int u \, du\).

Step 3: Integrate \(\int u \, du\)

The antiderivative of \(u\) with respect to \(u\) is \(\frac{1}{2}u^2 + C\). Therefore, \(\int u \, du = \frac{1}{2}u^2 + C\).

Step 4: Substitute Back \(u = \ln v\)

Replace \(u\) with \(\ln v\) in the integrated function. So, \(\frac{1}{2}u^2 + C\) becomes \(\frac{1}{2}(\ln v)^2 + C\).

Step 5: Match with Given Options

Comparing the result \(\frac{1}{2}(\ln v)^2 + C\) with the options provided: (A) \(\ln |\ln v|+C\), (B) \(\ln \frac{v^{2}}{2}+C\), (C) \(\frac{1}{2}(\ln v)^{2}+C\), and (D) \(\frac{1}{2} \ln v^{2}+C\), we see that it matches option (C).

Key Concepts

Logarithmic Integration

Logarithmic integration is a technique used in integral calculus where you encounter a term involving a logarithmic function, such as \( \ln v \). This method often requires substitution to simplify the integral into a more manageable form. It's relevant in problems where the function inside the integral (often the integrand) can be expressed in terms of a logarithm, making it easier to integrate.

In our original exercise, the integral \( \int \frac{\ln v \, dv}{v} \) is solved using substitution by recognizing the logarithmic component. The substitution \( u = \ln v \) allows us to transform the integral into a simpler form.

• First, identify the logarithmic function within the integral.
• Use the substitution \( u = \ln v \) to simplify the integrand.
• Perform the integration on the transformed expression.

Understanding logarithmic integration is crucial because it frequently appears in calculus problems involving exponential and logarithmic functions. This method reduces complex integrals into basic polynomials, which are easier to work with.

Calculus Step by Step

The process of solving calculus problems step by step ensures clarity and breaks down complex tasks into simpler actions. This method is especially useful for beginners who are looking to understand the underlying principles of calculus.

To tackle the integral \( \int \frac{\ln v \, dv}{v} \), we employed a structured approach:

• Identify the form of the integral: Recognize that it suits a typical substitution scenario resembling \( \int u \, du \).
• Perform substitution: Substitute \( u = \ln v \), which simplifies the differential \( dv \) into \( du \).
• Integrate: After substitution, integrate the simplified form \( \int u \, du \). The antiderivative is \( \frac{1}{2}u^2 + C \).
• Substitute back: Replace \( u \) with \( \ln v \) to return to the original variable.
• Match results: Compare your solution with given options to confirm accuracy.

Using a step-by-step method allows students to verify their own work at each phase, identify mistakes easily, and achieve a deeper understanding of the integration process.

Integral Calculus Problems

Integral calculus problems encompass a variety of scenarios where finding the antiderivative or the area under curves is required. This type of calculus is pivotal for solving real-world problems such as calculating the net change, area, volume, and other quantities that accumulate over intervals.

In integral calculus, it's common to solve problems like the given exercise where you deal with integrals involving functions such as \( \ln v \). These problems illustrate the significance of substitution methods to simplify complex integrands.

• Identify parts of the integrand suitable for substitution.
• Transform the integral into a simpler form you can integrate directly.
• Ensure you perform limits or substitutions accurately to return to the original problem context.

The diversity of integral calculus problems requires skills like recognizing suitable methods for integration, handling logarithmic functions, applying creative substitutions, and understanding the implications of your calculations. Mastering these problems equips students with the capability to tackle extensive questions in mathematics and applied sciences.