Problem 23 - Calculus: Early Transcendental Functions

Question

Evaluate the integral. $$\int \frac{e^{2 / x}}{x^{2}} d x$$

Step-by-Step Solution

Answer

The integral of the function \( \frac{e^{2 / x}}{x^{2}} \) is \( -\frac{e^{2 / x}}{2} + C \).

Step 1: Identify Function for Substitution

Here, the function \(g(x) = \frac{2}{x}\) is suitable for substitution. Its derivative is \(-\frac{2}{x^2}\), which is present in the integrand.

Step 2: Apply Substitution

Let \( u = g(x) = \frac{2}{x}\). Therefore, \( du = -\frac{2}{x^2} dx \) and \( -\frac{1}{2} du = \frac{1}{x^2} dx \). Replacing these in the integral, we get: \[ \int -\frac{e^u}{2} du \].

Step 3: Solve the New Integral

This new integral is fairly straightforward to solve. It is in the standard form, \[ \int e^u du = e^u + C \]. Therefore, \[ \int -\frac{e^u}{2} du = -\frac{e^u}{2} + C \].

Step 4: Replace the Substituted Variable

Substituting back \( u = \frac{2}{x} \) into the above result, the final answer will be: \[ -\frac{e^{2 / x}}{2} + C \].

Key Concepts

Integration Techniques

Integration can often feel like solving a puzzle, with each piece requiring different techniques to fit into the overall picture. Mastering a variety of integration techniques is crucial for anyone tackling calculus problems. Among the most common methods are u-substitution, integration by parts, partial fractions, and dealing with trigonometric integrals. The key to deciphering which technique to use lies in recognizing specific patterns and forms within the integrand. For instance, if you have a product of functions or a function whose derivative also appears in the integrand, you might consider u-substitution or integration by parts, respectively.

Exponential Functions

Exponential functions, such as e^x, are remarkable in calculus due to their self-derivative property; the derivative of e^x is itself. This characteristic makes them naturally appear in many scientific fields, from radioactive decay to compound interest calculations. When integrating exponential functions, the process usually simplifies because differentiation and integration of e^x are straightforward, leading to a direct antiderivative. However, when the exponential function is composed with another function, as in e^{g(x)}, you may need a technique like u-substitution to find the integral.

Definite Integrals

Definite integrals quantify the area under a curve between two points along the x-axis. They are not merely pure mathematical constructs; they relate directly to physical concepts like distance, area, and probability. Calculating a definite integral involves finding the antiderivative of the function, then evaluating it at the upper and lower bounds of integration. The process is epitomized by the Fundamental Theorem of Calculus, which bridges the concepts of differentiation and integration, thus providing a method to compute areas under curves and related quantities rigorously.

Indefinite Integrals

In contrast to definite integrals, indefinite integrals represent families of functions with an added constant of integration, often denoted as C. These integrals give the antiderivative of a function without upper and lower bounds, implying a general formula for all possible anti-derivatives. When we encounter an integrand, we seek a function whose derivative is the integrand. The constant of integration C represents the unknown offset that could be present in the original function before differentiation.

U-Substitution

The u-substitution technique is a powerful tool to simplify complex integrands by transforming them into more familiar forms. The idea is akin to an algebraic 'change of variables' to make the integral more manageable. The method involves selecting a substitution that makes the integrand a function of u instead of x, and then replacing dx with du in the integral. This technique is particularly effective when part of the integrand's derivative is also present within the integrand. Care must be taken to completely convert all x terms into u terms, and finally, after integrating with respect to u, substituting back to return to the original variable x. In practice, finding the right substitution can transform an intimidating integral into a straightforward problem, as illustrated by the solution to the given exercise.