Problem 22 - Applied Calculus

Question

Find each indefinite integral. \(\int\left(e^{2 x}-\frac{2}{x}\right) d x\)

Step-by-Step Solution

Answer

\( \frac{1}{2}e^{2x} - 2\ln|x| + C \)

Step 1: Break Down the Integral

Identify the structure of the given integral: \( \int\left(e^{2 x}-\frac{2}{x}\right) d x \). Notice that this is a sum of two separate functions: \( e^{2x} \) and \( -\frac{2}{x} \). We will integrate these functions separately.

Step 2: Integrate the First Term

Consider the first term \( e^{2x} \). Recall that the integral of \( e^{ax} \) with respect to \( x \) is \( \frac{1}{a}e^{ax} + C \). For \( e^{2x} \), let \( a = 2 \), so the integral is \( \frac{1}{2}e^{2x} \).

Step 3: Integrate the Second Term

Consider the second term \( -\frac{2}{x} \). Recall that the integral of \( \frac{1}{x} \) with respect to \( x \) is \( \ln|x| + C \). Therefore, the integral of \( -\frac{2}{x} \) is \( -2\ln|x| \).

Step 4: Combine the Integrals

Combine the results from the integration of both terms to form the indefinite integral: \( \int\left(e^{2 x}-\frac{2}{x}\right) d x = \frac{1}{2}e^{2x} - 2\ln|x| + C \).

Step 5: Write Final Solution

The final indefinite integral is \( \int\left(e^{2 x}-\frac{2}{x}\right) d x = \frac{1}{2}e^{2x} - 2\ln|x| + C \), where \( C \) is the constant of integration.

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used to solve integrals where the standard rules for integration do not directly apply. The method is derived from the product rule of differentiation; \[ \int u \, dv = uv - \int v \, du \]where \( u \) and \( dv \) are chosen parts of the original integral.When dealing with complex integrals, such as the sum found in our original problem, breaking down the integral into a form of \( u \, dv \) can simplify the solution. Let's consider when you may need to apply integration by parts:

• When dealing with products of algebraic and transcendental functions, like \( x e^x \).
• When facing integrals involving logarithmic or inverse trigonometric functions.
• If a straightforward application of standard integration rules is complex or impossible.

By carefully selecting \( u \) and \( dv \), the goal is to create a simpler integral \( \int v \, du \), which is easier to solve. In many cases, choosing \( u \) as the function that simplifies upon differentiation, and \( dv \) as the remainder, helps reduce the problem efficiently.

Exponential Functions

Exponential functions are a core component of calculus, often appearing in problems involving growth and decay, as well as in engineering and the sciences.An exponential function is of the form \( f(x) = e^{ax} \), where \( e \) is Euler's number, approximately 2.718. We encounter these functions frequently because of their distinct properties: their derivatives (and integrals) are proportional to the original function.Integration of an exponential function is relatively straightforward if you remember the essential rule:

• The indefinite integral of \( e^{ax} \) with respect to \( x \) is \( \frac{1}{a}e^{ax} + C \), where \( C \) is the constant of integration.

In the original problem, the integration of \( e^{2x} \) required identifying \( a = 2 \) and applying the rule to find \( \frac{1}{2}e^{2x} \). This ease with which we solve an integral is one of the reasons exponential functions are so prevalent.

Natural Logarithm

Natural logarithms, denoted as \( \ln(x) \), are logarithms with the base \( e \). They are essential in the realm of calculus due to their connection with growth processes and inverse calculations.One particularly notable property of natural logarithms is how they behave under integration:

• The indefinite integral of \( \frac{1}{x} \) with respect to \( x \) is \( \ln|x| + C \).

This relationship is crucial when handling integrals of rational functions, as seen in the original exercise where the term \( -\frac{2}{x} \) translates to an integral involving the natural logarithm. Specifically, the solution reveals that integrating \( -\frac{2}{x} \) results in \( -2\ln|x| \).Understanding the natural logarithm's integral helps in various applications, from calculating time in continuous growth scenarios to solving equations involving complex logarithmic functions. It's a robust tool that simplifies the integration process in situations involving division and fractions.