Problem 43 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int x(x-1)^{10} d x$$

Step-by-Step Solution

Answer

\(\frac{(x-1)^{12}}{12} + \frac{(x-1)^{11}}{11} + C\).

Step 1: Explore Substitution Possibility

We can simplify the integration by realizing that the function inside the integral can benefit from substitution. Let us set \( u = x - 1 \), which implies \( du = dx \) and \( x = u + 1 \). Thus, the integral becomes \( \int (u+1)u^{10} du \).

Step 2: Expand the Integrand

Using the distributive property, expand \((u+1)u^{10}\) to get \(u^{11} + u^{10}\). Now the integral becomes \( \int u^{11} + u^{10} \, du \).

Step 3: Integrate Term by Term

Integrate each term separately. The integral of \(u^{11}\) is \( \frac{u^{12}}{12} \) and the integral of \(u^{10}\) is \( \frac{u^{11}}{11} \). Thus, the integral becomes \( \frac{u^{12}}{12} + \frac{u^{11}}{11} + C \).

Step 4: Substitute Back in Terms of x

Replace \(u\) back with \(x-1\) to return to the original variable. Thus, the integral expression is \( \frac{(x-1)^{12}}{12} + \frac{(x-1)^{11}}{11} + C \).

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus to simplify the process of finding integrals. It involves substituting a part of the integral with a new variable, which can often make the integral easier to solve. In the given exercise, the expression

• \((x - 1)^{10}\)

is identified as a candidate for substitution. By setting:

• \(u = x - 1\)

we change the problem into terms of \(u\), which allows for a simpler function to integrate. Alongside this substitution, we must also change \(dx\) to \(du\), which in this case remains the same since \(du = dx\). This step transforms the expression and grants us a path to proceed with solving the integral in a more straightforward manner. The substitution method is akin to changing perspectives, simplifying our view of the problem so we can tackle it more efficiently.

Integration Techniques

Understanding different integration techniques is crucial for mastering calculus. Once the substitution is applied in the exercise, we are left with the integral:

• \(\int (u+1)u^{10} \, du\)

At this point, the expression becomes easier to handle using standard integration techniques. One common technique is to handle each term of a polynomial separately after expanding it. Integration of simple polynomials is straightforward because each term can be integrated using the basic power rule. This rule states that:

• \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\)

This technique allows us to address complicated expressions term by term, ensuring even complex polynomial expressions are managed in a clear, structured way.

Algebraic Expansion

Algebraic expansion is a basic yet powerful technique that helps in rearranging expressions into forms that are more manageable for integration. After substitution, the expression

• \((u+1)u^{10}\)

needs to be expanded. Apply the distributive property to get:

• \(u^{11} + u^{10}\)

This step is crucial as it unpacks a complex expression into simpler parts. Each resulting term can then be integrated using simple rules of integration. Here, algebraic expansion plays the role of preparing the groundwork, making integration tasks less cumbersome by reducing complexity before actually integrating the terms. Through careful expansion, we ensure that each part of the expression is accessible and ready for straightforward integration.