Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-16, perform the indicated integrations. \(\int x(1-x)^{2 / 3} d x\)

Step-by-Step Solution

Answer

The integral is \( \frac{3}{8} (1-x)^{8/3} - \frac{3}{5} (1-x)^{5/3} + C \).

Step 1: Select a suitable substitution

This integral can be solved by using substitution. Let's set \( u = 1 - x \). Thus, \( du = -dx \) and \( x = 1 - u \). This transforms the integral terms, simplifying the integration process.

Step 2: Transform the integral

After substitution, rewrite the original integral \( \int x (1-x)^{2/3} \, dx \) in terms of \( u \). Substitute \( x = 1 - u \) and \( dx = -du \), resulting in the new integral:\[\int (1-u) u^{2/3} (-1) \, du = \int (u - 1) u^{2/3} \, du\]

Step 3: Simplify and Expand the integrand

Expand the product inside the integral:\[\int (u u^{2/3} - u^{2/3}) \, du = \int (u^{5/3} - u^{2/3}) \, du\]

Step 4: Integrate term by term

Integrate each term separately:\[\int u^{5/3} \, du = \frac{u^{8/3}}{8/3} = \frac{3}{8} u^{8/3}\]\[\int u^{2/3} \, du = \frac{u^{5/3}}{5/3} = \frac{3}{5} u^{5/3}\]

Step 5: Combine results of integration

Combine the integration results into the original integral transformed in terms of \( u \):\[\frac{3}{8} u^{8/3} - \frac{3}{5} u^{5/3} + C\]

Step 6: Substitute back in terms of x

Replace \( u \) with \( 1-x \) to revert back to the variable \( x \):\[\frac{3}{8} (1-x)^{8/3} - \frac{3}{5} (1-x)^{5/3} + C\]

Step 7: Finalize and verify the solution

Ensure all transformations and integrations maintain the integrity of the equation. Verify if any errors in calculation are corrected and the final expression is consistent and correct.

Key Concepts

u-substitution

The u-substitution technique is a powerful method for solving integrals, where we aim to simplify the integrand by substituting part of the expression with a single variable, usually denoted as \( u \). In the problem, we chose \( u = 1 - x \). This choice simplifies the expression inside the integral, allowing more straightforward integration.

• First, identify a portion of the integrand that makes the integration complex or involves a composite function.
• Substitute \( u \) with this identified part and express \( du \), which is the derivative of \( u \) with respect to \( x \). Here, \( du = -dx \).
• Rewrite the entire integrand in terms of \( u \) and \( du \).

With this transformation, the integral is rephrased using simpler terms making it easier to solve. u-substitution helps in transforming difficult integrals into a form that matches basic integration rules.

definite integration

Definite integration involves evaluating the integral with upper and lower boundaries. This results in a numerical value representing the area under the curve for the given interval.

• It is performed in two primary steps: first, find the antiderivative using indefinite integration.
• Secondly, apply the Fundamental Theorem of Calculus by plugging in the boundary limits into the antiderivative function, and finding their difference.

Definite integrals are essential in calculating areas, volumes, and other physical quantities. Though the original problem does not specify bounds, the method outlined is crucial when such limits exist. Remember, with definite integration, constants of integration \( C \) are irrelevant as they cancel out when finding the difference between bounds.

indefinite integration

Indefinite integration, often referred to simply as "integration," does not involve limits of integration and results in the general antiderivative of a function plus an arbitrary constant \( C \). In this exercise, we found and combined the antiderivatives:

• The integral \( \int u^{5/3} \, du \) leading to \( \frac{3}{8} u^{8/3} \).
• The integral \( \int u^{2/3} \, du \) resulting in \( \frac{3}{5} u^{5/3} \).

Adding these, the result is expressed back in the original variable \( x \) after substituting \( u \) back to \( 1-x \). The constant \( C \) reflects that this is a family of solutions, depending on initial conditions. Indefinite integrals play a key role in understanding the relationship between functions and their rates of change.

integration by parts

Integration by parts is another technique used to integrate products of functions. While not used explicitly in this problem, it's valuable to understand its role. The idea is based on the formula:

• \( \int u \, dv = uv - \int v \, du \)
• This method decomposes the product into simpler parts, choosing \( u \) and \( dv \) such that the resulting \( du \) and \( v \) make the integral easier to compute.

To effectively apply, select \( u \) as a function that simplifies when derived, and \( dv \) as a function easy to integrate. Although integration by parts wasn't directly applied here, it’s a complementary technique to keep in mind when tackling more complex integrals, often used alongside u-substitution to simplify multi-layered integrals.