Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-36, use integration by parts to evaluate each integral. $$ \int \ln \left(7 x^{5}\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( x \ln(7x^5) - 5x + C \).

Step 1: Identify the parts for integration by parts

For the integration by parts method, we choose parts to represent the formula \( \int u \, dv = uv - \int v \, du \). Here, let \( u = \ln(7x^5) \) and \( dv = dx \). This choice leads to differentiating the logarithm function.

Step 2: Differentiate and integrate components

Now compute \( du \) and \( v \):1. Differentiate \( u \): \( du = \frac{d}{dx}(\ln(7x^5))dx = \frac{1}{7x^5} \cdot 35x^4 \, dx = \frac{5}{x}dx \).2. Integrate \( dv \): \( v = \int dx = x \).

Step 3: Apply the integration by parts formula

Substitute \( u \), \( v \), \( du \), and the newly found \( v = x \) into the integration by parts formula:\[ \int \ln(7x^5) \, dx = x \ln(7x^5) - \int x \cdot \frac{5}{x} \, dx \].Simplify the integral: \( \int x \cdot \frac{5}{x} \, dx = \int 5 \, dx \).

Step 4: Evaluate the remaining simple integral

Perform the integral \( \int 5 \, dx \): this results in\[ 5x + C \], where \( C \) is the constant of integration.

Step 5: Write the final antiderivative

Substitute back into the integration by parts result to get the final antiderivative:\[ x \ln(7x^5) - 5x + C \].This expression represents the integral of \( \ln(7x^5) \).

Key Concepts

Logarithmic Functions

Logarithmic functions are a key concept in calculus, particularly when dealing with integration challenges. The logarithm function, such as the natural logarithm \( \ln(x) \), is the inverse of the exponential function. This means that if you know \( a = \ln(b) \), then \( b = e^a \). Logarithmic functions have distinct properties that can simplify complex expressions and integrals.

• One key property of logarithms is that they transform multiplication into addition: \( \ln(ab) = \ln(a) + \ln(b) \).
• Likewise, division can be simplified as well: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).
• Finally, powers of a number can be rewritten: \( \ln(a^b) = b \ln(a) \).

Understanding these properties helps tremendously when applying calculus techniques such as integration by parts, as they allow us to manipulate the expression into a more manageable form. In the original exercise, recognizing how to differentiate \( \ln(7x^5) \) was crucial, as it set up the correct expression for \( du \).

Definite Integrals

Definite integrals are used to compute the total area under a curve within specified limits. It turns the indefinite integral into a precise numerical value, typically representing the accumulated quantity of some continuous process over a specific interval.
In calculus, the definite integral is given by\[\int_a^b f(x) \, dx = F(b) - F(a),\]where \( F \) is the antiderivative or the indefinite integral of \( f(x) \).

• The bounds \( a \) and \( b \) indicate the interval over which the integration is performed.
• This measure allows the computation of quantities such as total distance, total area, or total mass within those bounds.

In our exercise, even though only an indefinite integral is presented, understanding this concept is essential for grasping the full range of possible applications of integrals.

Antiderivatives

Antiderivatives play a central role in calculus. Essentially, an antiderivative is a function \( F(x) \) whose derivative is the original function \( f(x) \), expressed as\[F'(x) = f(x).\]Calculating the antiderivative involves integrating the function, which is the reverse process of differentiation.
For example, if given \( f(x) = 5 \), its antiderivative would be \( F(x) = 5x + C \), where \( C \) represents the constant of integration.

• In the exercise, the integration by parts method yielded the antiderivative \( x \ln(7x^5) - 5x + C \).
• This solution showcases the process of assembling these integral parts to reach the final antiderivative expression.

Thus, finding the antiderivative is a critical step in solving integral problems and understanding the behavior of functions over continuous intervals.

Calculus Techniques

Calculus techniques for integration are essential tools in solving mathematical problems. Among these techniques, integration by parts is particularly useful for integrating products of functions.
Based on the formula\[\int u \, dv = uv - \int v \, du,\]this method helps break down the integral into simpler parts that are easier to manage.
The integration by parts process involves selecting \( u \) and \( dv \) appropriately so that \( du \) and \( v \) can be easily calculated.

• Selecting \( u = \ln(7x^5) \) allowed us to differentiate it straightforwardly to find \( du \).
• The \( dv = dx \) component was integrated to find \( v = x \).
• These parts were then combined to simplify the integral of \( \ln(7x^5) \).

Understanding and practicing such calculus techniques not only enhances problem-solving skills but also reinforces the foundational principles of calculus.