Problem 32 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{0}^{2} \frac{d x}{\sqrt{|x-1|}} $$

Step-by-Step Solution

Answer

The evaluated integral is 4.

Step 1: Analyze the Integral

The given integral is \( \int_{0}^{2} \frac{dx}{\sqrt{|x-1|}} \). The absolute value function \(|x - 1|\) is important to consider, as it creates different cases depending on whether \(x < 1\) or \(x \geq 1\).

Step 2: Split the Integral

Since \(|x - 1|\) changes behavior at \(x = 1\), split the integral into two parts: \( \int_{0}^{1} \frac{dx}{\sqrt{1-x}} \) for \(x < 1\) and \( \int_{1}^{2} \frac{dx}{\sqrt{x-1}} \) for \(x \geq 1\).

Step 3: Evaluate \( \int_{0}^{1} \frac{dx}{\sqrt{1-x}} \)

Perform a substitution to evaluate this integral: let \( u = 1-x \), then \( du = -dx \). The limits change from \( x = 0 \) to \( u = 1 \) and \( x = 1 \) to \( u = 0 \). Thus, the integral becomes \( -\int_{1}^{0} \frac{du}{\sqrt{u}} = \int_{0}^{1} u^{-1/2} du \). This evaluates to \( [2u^{1/2}]_{0}^{1} = 2(1 - 0) = 2 \).

Step 4: Evaluate \( \int_{1}^{2} \frac{dx}{\sqrt{x-1}} \)

Use a substitution similar to Step 3: let \( v = x-1 \), then \( dv = dx \). The limits change from \( x = 1 \) to \( v = 0 \) and \( x = 2 \) to \( v = 1 \). The integral thus becomes \( \int_{0}^{1} v^{-1/2} dv \), which evaluates to \( [2v^{1/2}]_{0}^{1} = 2(1 - 0) = 2 \).

Step 5: Combine Results

Add the evaluated results from steps 3 and 4: \(2 + 2 = 4\). Thus, the value of the original integral is \(4\).

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique used to simplify complex integrals by transforming variables. This method is often called the "reverse chain rule." When we perform substitution, we replace a variable with a new one to make the integral easier to solve.

In our exercise, the given integral is split into different intervals owing to the absolute value function present in the integrand.

• For the interval from 0 to 1, we use the substitution \(u = 1-x\). This choice of substitution reverses the limits of integration as the variable changes from \(x\) to \(u\). With this transformation, the integrand reduces to \(u^{-1/2}\).
• Similarly, for the interval from 1 to 2, we use \(v = x-1\), which transforms the integrand into \(v^{-1/2}\). The choice of substitution simplifies the integral for direct evaluation.

By substituting appropriately and changing the limits of integration, the integral becomes much more direct to solve. This technique underscores the importance of selecting strategic substitutions, ensuring the process is streamlined, and effectively executing integration without excessive complication.

Absolute Value Integrals

Integrals involving absolute values often require special attention. The absolute value function modifies the behavior of the integrand based on the sign of its argument.

In our case,

• The function \(|x - 1|\) splits the integration process because \(x - 1\) transitions from negative to positive around \(x = 1\). This division creates two distinct integrals, each easier to handle.
• For \(x < 1\), the expression \(|x-1| = 1-x\) holds true, while for \(x \geq 1\), \(|x-1| = x-1\). Consequently, the original integral is solved over two intervals and summed together afterward.

Handling absolute value functions in integrals necessitates an understanding of their piecewise nature; always consider the points where the absolute value expression changes sign. This understanding helps decompose the integral into manageable parts and ultimately obtain the correct solution.

Mathematical Analysis

Mathematical analysis provides tools and methods to rigorously explore the behavior of functions and integrals. In this context, understanding function behavior helps in better problem-solving approaches.

• The integral \( \int \frac{dx}{\sqrt{|x-1|}} \) showcases how splitting integrals based on function behavior enhances comprehension and accuracy in results.
• Through analytical techniques like substitution, it’s easier to evaluate seemingly complex integrals by breaking them into simpler components.

Mathematical analysis assists in discerning patterns and properties in functions, which can be particularly useful in identifying and resolving absolute value integrals. Learning through this analytical lens aids in tackling a broad range of problems, strengthening one's foundation in calculus and integral calculus.