Problem 41 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(29-50.\) $$\int \frac{e^{1 / x}}{x^{2}} d x$$

Step-by-Step Solution

Answer

The integral is \(-e^{1/x} + C\).

Step 1: Identify the substitution

First, notice that inside the integral, there is a composite function, \(e^{1/x}\). A suitable substitution is needed to simplify the integral. Let \( u = \frac{1}{x} \). This is a standard technique for simplifying integrals involving composite functions.

Step 2: Differentiate the substitution

Differentiate the substitution equation to find \(du\). The derivative of \(u = \frac{1}{x}\) with respect to \(x\) is \(\frac{du}{dx} = -\frac{1}{x^2}\). Thus, \(dx = -x^2 \, du\).

Step 3: Substitute into the integral

Substitute the expressions for \(u\) and \(dx\) back into the integral \(\int \frac{e^{1/x}}{x^2} \, dx\). Substitute \(e^{1/x} = e^u\), \(\frac{1}{x^2}\) remains as \(\frac{1}{x^2}\), and \(dx = -x^2 \, du\). The integral becomes \(-\int e^u \, du\).

Step 4: Integrate with respect to \(u\)

The integral \(-\int e^u \, du\) is straightforward. The integral of \(e^u\) is itself, so \(-\int e^u \, du = -e^u + C\), where \(C\) is the constant of integration.

Step 5: Substitute back to original variable

We need to express the final answer in terms of the original variable \(x\). Since \(u = \frac{1}{x}\), replace \(u\) with \(\frac{1}{x}\) in the integrated result: \(-e^u + C = -e^{1/x} + C\).

Key Concepts

Definite and Indefinite Integrals

Understanding integrals is key to solving many problems in calculus. Integrals can be either definite or indefinite.

• An indefinite integral is expressed without limits and represents a family of functions. It includes a constant of integration, usually denoted by \( C \), to account for all possible vertical shifts of the antiderivative.
• Conversely, a definite integral is calculated over an interval \([a, b]\) and will give a fixed number that represents the area under a curve within those bounds.

In our problem, we found the indefinite integral of \( \int \frac{e^{1/x}}{x^2} \, dx \). By using substitution, we evaluated the expression, ending with an antiderivative that includes the constant \( C \). This integral tells us about the comprehensive set of functions that, when differentiated, result in the original integrand.

Exponential Functions

Exponential functions are crucial in mathematics, especially within integrals and derivatives. They have the form \( e^x \), where \( e \) is approximately equal to 2.718.

• One feature of exponential functions is that they grow rapidly. The derivative of \( e^x \) is itself, \( e^x \), which is unique to the exponential function.
• When dealing with exponential functions in integrals, particularly those involving a transformation like \( e^{1/x} \), it requires special techniques such as substitution to simplify the expression.

In our example, \( e^{1/x} \) presents a nested structure that complicates direct integration, necessitating the use of substitution to manage the complexity. This substitution simplifies integration into a more straightforward form, thereby allowing us to solve it easily.

Substitution Method

The substitution method is a powerful tool in calculus to solve integrals of complex or composite functions. The idea is simple: to make integration easier, we transform the integral into a more manageable form.

• The first step involves identifying an inner function within the integral that can be substituted. In our problem, we selected \( u = \frac{1}{x} \).
• Next, differentiate \( u \) with respect to \( x \) to find \( du \). Then, re-write \( dx \) in terms of \( du \). This is crucial for performing the substitution.
• Transform the entire integral in terms of \( u \) and integrate. Often, this reduces the integral to a familiar form.
• Finally, replace \( u \) back with the original expression in terms of \( x \).

For our problem, once \( u \) was redefined, the integral simplified to \(-\int e^u \, du\), which eased the calculation significantly. This method is particularly valuable for integrating functions that cannot be handled by standard techniques.