Problem 21 - Applied Calculus

Question

Find each indefinite integral. \(\int\left(e^{3 x}-\frac{3}{x}\right) d x\)

Step-by-Step Solution

Answer

\( \frac{1}{3} e^{3x} - 3 \ln |x| + C \)

Step 1: Break Down the Integral

The integral to solve is \( \int \left( e^{3x} - \frac{3}{x} \right) \, dx \). We notice that this is the combination of two separate parts: \( \int e^{3x} \, dx \) and \( \int -\frac{3}{x} \, dx \). Therefore, we can solve each part individually.

Step 2: Solve \( \int e^{3x} \, dx \)

For the first integral \( \int e^{3x} \, dx \), we can use the substitution method. Let \( u = 3x \), which implies \( du = 3 \, dx \) or \( dx = \frac{du}{3} \). Therefore, the integral becomes \( \int e^u \cdot \frac{1}{3} \, du = \frac{1}{3} \int e^u \, du \). Integrating gives \( \frac{1}{3} e^u + C \), and substituting back for \( u \), we get \( \frac{1}{3} e^{3x} + C \).

Step 3: Solve \( \int -\frac{3}{x} \, dx \)

For the second integral \( \int -\frac{3}{x} \, dx \), notice that it is a standard form \( \int \frac{a}{x} \, dx = a \ln |x| + C \). So, integrating \( -\frac{3}{x} \), we get \( -3 \ln |x| + C \).

Step 4: Combine Results

Combine the results of the two integrals. The solution to the original integral is \( \frac{1}{3} e^{3x} - 3 \ln |x| + C \), where \( C \) is the constant of integration.

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus, especially useful for simplifying integrals by changing variables. In this exercise, we encountered an integral of the form \(\int e^{3x} \, dx\). Here, the substitution method helps us deal with the exponential component which might otherwise seem complicated.
To apply substitution:

• Select a substitution variable \(u\), such as \(u = 3x\), that simplifies the integral.
• Find \(du\) in terms of \(dx\), where \(du = 3 \, dx\), or equivalently, \(dx = \frac{du}{3}\).
• Replace \(dx\) and the function in terms of \(u\) to rewrite the integral, resulting in \(\int e^u \cdot \frac{1}{3} \, du\).

This simplification allows us to handle complex expressions by integrating with respect to \(u\) and then returning to the original variable \(x\) by substituting back. The process transforms an initially challenging integral into a more manageable form.

Integration Techniques

Understanding various integration techniques is essential in solving indefinite integrals effectively. One technique focuses on using basic integral forms and rules for integration. This involves recognizing patterns or standard forms within the integrand, allowing for straightforward integration.
In the original problem, besides using substitution for \( e^{3x} \), we also come across a standard integral, \(-\frac{3}{x}\), which can be directly integrated. This integral follows the formula \(\int \frac{a}{x} \, dx = a \ln |x| + C\), applied here as \(-3 \ln |x| + C\).
By breaking down the original problem, the integration techniques let us handle each part separately:

• Use substitution for complex expressions like \(e^{ax}\).
• Apply standard integration rules for simpler fractions like \(\frac{1}{x}\).

Integration techniques are strategies that reduce complex problems into known integrals, often existing in integral tables or learned through drill and practice.

Calculus

Calculus is the branch of mathematics that deals with rates of change and areas under curves. In this context, we're focused on integration, which is the reverse process of differentiation. This exercise required finding an indefinite integral, essentially determining a function whose derivative matches the given expression.
The overall solution reflects core calculus principles:

• Breaking down into manageable parts, solving them using known strategies.
• Applying mathematical rules of substitution and standard forms.
• Remembering that integration adds an arbitrary constant \(C\), as it represents a family of antiderivatives rather than a specific solution.

Calculus allows us to solve real-world problems involving areas, volumes, and functional transformations. Grasping these concepts deeply is crucial, as they form the groundwork for more advanced mathematical analysis, physics, and engineering challenges.