Problem 50 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int \frac{x}{(2 x-1)^{2 / 3}} d x$$

Step-by-Step Solution

Answer

The integral is \( \frac{3}{4} (2x-1)^{1/3} + \frac{3}{16} (2x-1)^{4/3} + C \).

Step 1: Choose a Suitable Substitution

To solve the integral \( \int \frac{x}{(2x-1)^{2/3}} \, dx \), we use substitution. Let's set \( u = 2x - 1 \). Hence, \( du = 2 \, dx \) or \( dx = \frac{du}{2} \). Also, rewrite \( x \) in terms of \( u \): \( x = \frac{u+1}{2} \) because \( 2x = u + 1 \). This makes substitution easier for the integral.

Step 2: Substitute and Simplify the Integral

Substitute \( x \) and \( dx \) in terms of \( u \) into the integral. The integral becomes\[ \int \frac{\frac{u+1}{2}}{u^{2/3}} \cdot \frac{1}{2} \, du = \frac{1}{4} \int \frac{u+1}{u^{2/3}} \, du. \]Split the integral into two separate integrals:\[ \frac{1}{4} \left( \int u^{-2/3} \, du + \int u^{1 - 2/3} \, du \right). \]

Step 3: Evaluate Each Integral Separately

Evaluate \( \int u^{-2/3} \, du \):The result is \( \frac{u^{1/3}}{1/3} = 3u^{1/3} \).Next, evaluate \( \int u^{1/3} \, du \):The result is \( \frac{u^{4/3}}{4/3} = \frac{3}{4}u^{4/3} \).

Step 4: Combine and Simplify the Results

Combine the two integrals from Step 3:\[ \frac{1}{4} \left( 3u^{1/3} + \frac{3}{4}u^{4/3} \right). \]Simplify further:\[ \frac{3}{4} u^{1/3} + \frac{3}{16} u^{4/3}. \]

Step 5: Substitute Back to Original Variable

Replace \( u \) with \( 2x - 1 \) to get the final solution:\[ \frac{3}{4} (2x-1)^{1/3} + \frac{3}{16} (2x-1)^{4/3} + C, \]where \( C \) is the constant of integration.

Key Concepts

Integration by Substitution

Integration by substitution is a technique used to simplify integrals by making a substitution that reduces the function to a more familiar form. This technique is similar to the chain rule used in differentiation but is applied in reverse. In essence, you are changing variables to make the integration process easier.

Here's how the process works:

• Select a substitution: Identify part of the integrand (the function within the integral) to replace with a new variable. In our problem, we chose to set \( u = 2x - 1 \).
• Differentiate the substitution: Compute \( du \) and express \( dx \) in terms of \( du \). For instance, \( du = 2 \, dx \), meaning \( dx = \frac{du}{2} \).
• Replace and simplify: Substitute all occurrences of the part selected and \( dx \) in the integral with \( u \) and \( du \), respectively. The transformation of the original integral simplifies the structure and makes integration easier.

This substitution method is especially useful for integrals involving complex algebraic expressions, making them manageable and solvable.

Definite and Indefinite Integrals

In calculus, integrals are categorized into two types: definite and indefinite integrals. Each type serves a particular purpose in mathematical analysis.

First, let's discuss the indefinite integrals. An indefinite integral represents the family of all antiderivatives of a function. It is expressed without upper or lower limits of integration, like \( \int f(x) \, dx \). The result includes a constant \( C \), representing an arbitrary constant value, since differentiation of a constant is zero. In our exercise, the integration process yielded \( \frac{3}{4} (2x-1)^{1/3} + \frac{3}{16} (2x-1)^{4/3} + C \).

On the other hand, definite integrals calculate the actual area under the curve of a function, between specific bounds \( a \) and \( b \). The result is a numerical value, not a function. However, our problem only involves an indefinite integral.

• Indefinite: General antiderivative form, including \( C \).
• Definite: Involves evaluation between bounds \( a \) and \( b \).

While indefinite integrals provide a general solution, definite integrals offer precise values, often used in practical applications.

Calculus Techniques

Calculus offers a variety of techniques to solve complex mathematical problems, especially in finding derivatives and integrals.

For the integration process, several techniques exist, such as:

• Integration by Substitution: Substitute a part of the integral to simplify the expression, as discussed earlier.
• Integration by Parts: Useful when integrating products of functions, applying the formula \( \int u \, dv = uv - \int v \, du \).
• Partial Fraction Decomposition: Breaks down rational functions into simpler fractions that are easier to integrate.
• Trigonometric Substitution: Substitutes trigonometric expressions for variable terms in integrals involving square roots.

Specifically, in the given problem, the substitution technique was pivotal. It transformed the integral involving a radical expression into a simpler polynomial form that could be approached with standard power rule integration. Mastery of these techniques is essential for solving diverse and challenging problems in integral calculus.