Question

Find the error Suppose you evaluate \(\int \frac{d x}{x}\) using integration by parts. With \(u=1 / x\) and \(d v=d x,\) you find that \(d u=-1 / x^{2} d x\) \(v=x,\) and $$\int \frac{d x}{x}=\left(\frac{1}{x}\right) x-\int x\left(-\frac{1}{x^{2}}\right) d x=1+\int \frac{d x}{x}$$ You conclude that \(0=1 .\) Explain the problem with the calculation.

Short answer

Question: Identify the mistake in the given integration by parts procedure for integrating \(1/x\) and explain why it led to the wrong conclusion. Answer: The mistake in the given procedure is the choice of \(u\) and \(dv\). Choosing \(u = 1/x\) and \(dv = dx\) did not simplify the integral after applying the integration by parts formula, as it left the same integral on the right-hand side. This faulty choice led to an incorrect conclusion, highlighting the importance of making wise choices for \(u\) and \(dv\) when using integration by parts.

Step-by-step solution

Recognize the Correct Formula for Integration by Parts

The correct formula for integration by parts is: $$\int u\, dv = u v - \int v\, du$$

Identify the Functions \(u\) and \(dv\) and Compute \(du\) and \(v\)

The given problem states that we should choose: $$ u = \frac{1}{x} \quad \text{and} \quad dv = dx $$ Differentiate \(u\) with respect to \(x\) to get \(du\): $$du = -\frac{1}{x^2} dx$$ Integrate \(dv\) with respect to \(x\) to get \(v\): $$v = x$$

Apply the Integration by Parts Formula

Plug the computed values of \(u, dv, du,\) and \(v\) into the integration by parts formula: $$\int \frac{1}{x} dx = \left( \frac{1}{x} \right) x - \int x \left(-\frac{1}{x^2} \right) dx$$

Simplify and Analyze the Resulting Integral

Simplify the expression: $$\int \frac{1}{x} dx = 1 + \int \frac{1}{x} dx$$ At this point, it seems that we've reached a fundamental error. The integral that we were trying to solve in the first place is still present on the right-hand side of the equation. This means that our method of integrating by parts did not lead to a simplification of the problem.

Identify the Error

The error in this calculation is not due to the integration by parts formula itself, but rather our choice of \(u\) and \(dv\). Here, we chose poorly because the resulting integral after applying integration by parts did not simplify the problem. Instead, it left the same integral on the right-hand side. In general, integration by parts can be an effective method for solving integrals, but it is essential to make wise choices for \(u\) and \(dv\). In this case, attempting to integrate \(\frac{1}{x}dx\) using integration by parts was not appropriate and led to an incorrect conclusion.

Key concepts

Indefinite Integral

An indefinite integral, also known as an antiderivative, is a function that reverses the process of differentiation. In other words, it finds a function whose derivative is the given function. For instance, if we want to compute the indefinite integral of \(\frac{1}{x}\), denoted as \(\int \frac{dx}{x}\), this is equivalent to finding a function \(F(x)\) such that \(F'(x) = \frac{1}{x}\).

In the case of \(\frac{1}{x}\), the indefinite integral is known as the natural logarithm function, specifically \(\ln|x|\) plus a constant of integration. It's critical to remember that the constant represents an infinite number of possible functions, all differing by a constant amount, as they all have the same derivative. When solving for indefinite integrals, especially in cases that seem straightforward, it's important to recall this fundamental concept, so as not to unnecessarily complicate the process by applying more advanced techniques like integration by parts without need.

u-Substitution

u-substitution is a technique used in integration that simplifies the integral by substituting part of the integral with a new variable, often denoted as \(u\). This method is analogous to the chain rule in differentiation and is highly effective when we have composite functions or when we aim to transform an integral to a more familiar form.

To conduct a u-substitution, one typically follows these steps:

• Identify a portion of the integral that you can substitute with a variable \(u\).
• Determine the differential \(du\) and substitute it along with \(u\) into the integral.
• Integrate with respect to \(u\).
• Substitute back in terms of the original variable \(x\) to get the final integral in terms of the original variable.

Applying u-substitution to \(\int \frac{dx}{x}\) might be redundant since the function is already in a basic form whose antiderivative is known.

Antiderivative

The antiderivative of a function is another function that, when differentiated, yields the original function. It is one of the key concepts in calculus, frequently referred to when solving integration problems. The antiderivative is integral (pun intended) when dealing with the problems involving indefinite integrals, such as finding the general solution of a differential equation or computing the area under a curve.

For example, the function \(F(x) = \ln|x| + C\), where \(C\) represents a constant, is the antiderivative of \(\frac{1}{x}\), because \(\frac{d}{dx}(\ln|x| + C) = \frac{1}{x}\). This basic relationship highlights the inverse nature of differentiation and integration. Understanding this concept is crucial, as misapplication of integration techniques like in the original exercise's solution can lead to an incorrect approach, where a simple antiderivative calculation would suffice.

Calculus Error Analysis

In calculus, error analysis is the process of reviewing solutions to problems to identify where mistakes may have occurred. It can involve verifying formulas, checking derivative and integral computations, or ensuring proper application of methods. In the context of the original exercise, the error occurred due to a misapplication of the integration by parts technique.

Integration by parts is a powerful tool, but it requires careful selection of \(u\) and \(dv\) to simplify the problem effectively. The erroneous conclusion that \(0 = 1\) is a classic sign that an error has been made. The function \(\frac{1}{x}\) has a straightforward antiderivative, hence the application of integration by parts added unnecessary complexity. Error analysis in this scenario reminds us to consider the most straightforward approach first and to validate that our chosen methods lead to progressive simplification of the integral at hand. Remembering this can save time and prevent the propagation of errors in more complex calculus problems.