Problem 55 - AP Calculus with 8 Practice Tests 15th

Question

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

Step-by-Step Solution

Answer

The answer is (C): \(-\frac{1}{y}(\ln y+1)+C\).

Step 1: Substitution

Use a substitution to simplify the integral. Let \( u = \ln y \), so \( du = \frac{1}{y} dy \). The integral becomes \( \int \frac{u}{y^2} dy \). To express \( dy \) in terms of \( du \), rearrange to get \( dy = y du \). Since \( y = e^u \), now \( y = e^{\ln y} = y \) becomes \( y = e^u \). Substitute \( y = e^u \) and \( dy = e^u du \) into the integral to get \( \int \frac{u}{e^{2u}} e^u du \), simplifying to \( \int u e^{-u} du \).

Step 2: Integration by Parts

Use integration by parts, where \( \int u dv = uv - \int v du \). Let \( v = -e^{-u} \) and \( dv = e^{-u} du \), and \( du = du \), \( u = u \). Then \( uv = -ue^{-u} \), and \( \int v du = - \int e^{-u} du = \int e^{-u} du = e^{-u} \). Applying integration by parts: \( \int u e^{-u} du = - ue^{-u} + e^{-u} + C \).

Step 3: Back-Substitution

Recall that \( u = \ln y \). Substitute \( u \) back into the expression: \( - u e^{-u} + e^{-u} = - \ln y \cdot \frac{1}{y} + \frac{1}{y} \). Simplify this result: \(-\frac{\ln y}{y} + \frac{1}{y} + C = -\frac{\ln y + 1}{y} + C \).

Step 4: Compare with Options

Compare the final expression \( -\frac{\ln y + 1}{y} + C \) with the options given. It matches option (C): \(-\frac{1}{y}(\ln y+1)+C\).

Key Concepts

Substitution Method

The substitution method is a powerful technique in integration that simplifies the process by changing the variable of integration. Imagine you're trying to solve a complex integral, but it feels cumbersome. This is where substitution comes to the rescue!

• Identify a part of the integral that can be replaced with a single variable, often to simplify the integrand.
• In our example, we used the substitution \( u = \ln y \), which simplifies the integration process significantly.
• Substitution also involves changing the differential, in this case, \( du = \frac{1}{y} dy \), to relate the variables correctly.

By transforming the integral into a simpler form, it becomes easier to tackle using basic integration techniques or further advanced methods.

Integration by Parts

Integration by parts is another essential technique in calculus problem-solving, especially when dealing with products of functions. It is based on the product rule for differentiation and is expressed as:\[ \int u \, dv = uv - \int v \, du \]Here's how you use it:

• Choose \( u \) and \( dv \) wisely. In our problem, \( u = u \), and \( dv = e^{-u} du \).
• Differentiate to find \( du \) and integrate \( dv \) to find \( v \).
• Apply the formula, which combines differentiation and integration, to tackle the problem.

In our solution, this method transformed the problem into a manageable one, with results easily simplified thereafter. This demonstrates the strength of integration by parts.

Definite Integrals

While the example provided is an indefinite integral, understanding definite integrals is crucial. Definite integrals compute the area under the curve of the function over a specific interval. Here’s how it works:

• It involves calculating the difference in the antiderivative at the upper and lower bounds.
• The notation \( \int_{a}^{b} f(x) \, dx \) indicates that the area is being calculated from \( x = a \) to \( x = b \).
• This results in a numerical value representing the total area under the curve within these bounds.

Understanding how to evaluate definite integrals builds on the foundational skills of dealing with indefinite integrals and is invaluable for calculus problem-solving.

Calculus Problem Solving

Calculus offers a range of problem-solving techniques, and the key is knowing when to use each one. By comprehensively understanding methods like substitution and integration by parts, students can tackle more complex problems effectively.

• Identify the problem type: Recognition is crucial in determining which method to apply.
• Use appropriate techniques: Choose substitution for simplifying integrals or integration by parts for functions as products.
• Practice: Consistent practice sharpens these skills and aids in mastering the various techniques.

By building a strong foundation in these techniques, students enhance their problem-solving toolkit, allowing them to approach calculus problems with confidence and ease.