Problem 38 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int \sqrt{\frac{x-1}{x^{5}}} d x$$

Step-by-Step Solution

Answer

\(\int \sqrt{\frac{x-1}{x^5}} \, dx = \frac{2}{3}(x-1)^{3/2} - (x-1)^{5/2} + C\) + higher order terms from the series.

Step 1: Simplify the Integrand

Let's start by rewriting the integrand in terms of powers of x. We have \(\sqrt{\frac{x-1}{x^5}} = \frac{\sqrt{x-1}}{x^{5/2}}\). This will make it easier for us to work with.

Step 2: Use Substitution Method

To evaluate the integral \(\int \frac{\sqrt{x-1}}{x^{5/2}} \, dx\), we'll use substitution. Set \(u = x - 1\), then \(du = dx\) and \(x = u + 1\). This changes our integral to \(\int \frac{\sqrt{u}}{(u+1)^{5/2}} \, du\).

Step 3: Integrate Using a Power Series Expansion

Since direct integration is complex, expand \((u + 1)^{-5/2}\) as a binomial series: \[(1 + u)^{-5/2} = 1 - \frac{5}{2}u + \frac{35}{8}u^2 - \frac{315}{48}u^3 + \cdots\]Multiplying with \(\sqrt{u}\), integrate term by term: \(\int u^{1/2}(1 - \frac{5}{2}u + \frac{35}{8}u^2 - \cdots) \, du\).

Step 4: Integrate Term by Term

Now integrate each term separately:\(\int u^{1/2} \, du = \frac{2}{3}u^{3/2},\)\(\int u^{3/2}(rac{-5}{2}) \, du = -\frac{5}{2} \times \frac{2}{5}u^{5/2} = -u^{5/2},\)Continue this process for the series expansion as needed.

Step 5: Combine the Integrated Terms

The integral becomes a sum of terms:\(\int u^{1/2} - u^{3/2} + \cdots \, du = \frac{2}{3}u^{3/2} - u^{5/2} + \cdots + C\), where \(C\) is the constant of integration.

Step 6: Substitute Back to Original Variable

Replace \(u\) with \(x - 1\) to return to the variable \(x\):\[\frac{2}{3}(x-1)^{3/2} - (x-1)^{5/2} + \cdots + C\]. This is the indefinite integral of the given function.

Key Concepts

Substitution Method

The Substitution Method is a fundamental technique in Integral Calculus, particularly useful for making integration simpler when dealing with complex functions. By introducing a new variable, the substitution transforms the integral into a more manageable form.

In the exercise you provided, we use the substitution method to evaluate the integral \( \int \sqrt{\frac{x-1}{x^5}} \, dx \).

• We start with substituting \( u = x - 1 \), simplifying our variable of integration. This implies that \( du = dx \), and we can express \( x \) as \( x = u + 1 \).
• This substitution transforms the original integral into the form \( \int \frac{\sqrt{u}}{(u+1)^{5/2}} \, du \).

Through this process, we transform a challenging integration problem into a potentially simpler one, allowing us to tackle the problem by focusing on the newly defined variable \( u \). This method is particularly advantageous when the transformed integral is in a standard form that is easier to solve.

Binomial Series Expansion

The Binomial Series Expansion allows us to expand expressions of the form \((1 + u)^n\) into an infinite series. It is an essential tool when functions in integrals are not easily integrable by standard methods.

In our exercise, once we have applied the substitution method, we encountered the function \((u + 1)^{-5/2}\).

• This function is expanded using the binomial series as:\[(1 + u)^{-5/2} = 1 - \frac{5}{2}u + \frac{35}{8}u^2 - \frac{315}{48}u^3 + \cdots\]
• The series expansion helps us express \( (u + 1)^{-5/2} \) as an infinite series, making it possible to integrate term by term.

Applying the Binomial Series Expansion is particularly useful in integration as it allows us to transform complex expressions into a series of simpler terms, each of which can be integrated separately.

Definite and Indefinite Integrals

Integrals come in two primary forms: definite and indefinite. Understanding both is crucial in solving calculus problems. Here’s a closer look into these concepts based on the problem you provided.

• An **indefinite integral** represents a family of functions and is expressed with a constant of integration \( C \) (e.g., \( \int f(x) \, dx = F(x) + C \)). In the given exercise, solving \( \int \sqrt{\frac{x-1}{x^5}} \, dx \) yields an expression without fixed bounds, so it results in an indefinite integral. This means the solution represents a general form of antiderivative for any constant value.
• An **definite integral** differs in that it calculates the net area between the function and the x-axis over a specified interval. While this was not part of the provided exercise, it's vital to understand that a definite integral would involve evaluation of the antiderivative at the bounds to provide a numeric answer.

In solving indefinite integrals, as demonstrated, the aim is often to simplify the integrand using techniques like substitution, and then integrate term by term using expansions like the binomial series. This process provides the groundwork for tackling more complex integration scenarios, whether dealing with definite or indefinite integrals.