Problem 43 - College Algebra and Calculus: An Applied Approach

Question

(a) perform the integration in two ways: once using the Simple Power Rule and once using the General Power Rule. (b) Explain the difference in the results. (c) Which method do you prefer? Explain your reasoning. $$ \int(x-1)^{2} d x $$

Step-by-Step Solution

Answer

The simple and general power rule give the same result for the integral of \((x-1)^{2}\), albeit in different formats. The Simple Power Rule provides result \(\frac{1}{3}x^3 - x^2 + x + C\) while the General Power Rule provides shorter, more compact result \(\frac{1}{3}(x - 1)^3 + C\). Although typical preference may be towards Simple Power Rule for its simplicity and broad usability, in specific situations the General Power Rule may be more suitable and quicker to apply.

Step 1: Apply the Simple Power Rule

The Simple Power Rule states that the integral of a function \(x^n\) is \(\frac{1}{n+1}x^{n+1}\). First you must expand \((x-1)^{2}\) to \(x^2 - 2x + 1\). Then apply the Simple Power Rule to each term separately. The integral \(\int x^2 dx\) becomes \(\frac{1}{3}x^3\), the integral \(\int 2x dx\) becomes \(x^2\) and finally, \(\int 1 dx\) becomes \(x\). Thus, the integral of \((x-1)^2\) is \(\frac{1}{3}x^3 - x^2 + x + C\), where \(C\) is the constant of integration.

Step 2: Apply the General Power Rule

The General Power Rule states that the integral of a function \((u(x))^n u'(x)\) is \(\frac{1}{n+1}(u(x))^{n+1}\). So identify \(u(x)\) as \(x-1\) and look for the derivative of \(u(x)\) in the integral. Here, no extra \(u'(x)\) is visible. Therefore, write the integral as \(\int (u(x))^2 dx\) which is equal to \(\frac{1}{3}(u(x))^3\). Substituting \(u(x) = x - 1\) back into the equation gives the integral \(\frac{1}{3}(x - 1)^3 + C\).

Step 3: Compare the results

The different methods of integration have delivered different results. The Simple Power Rule gave \(\frac{1}{3}x^3 - x^2 + x + C\), while the General Power Rule gave \(\frac{1}{3}(x - 1)^3 + C\). The results may look different, but in fact they are equivalent. The equation of the General Power Rule is surely more compact, but the expanded formula from the Simple Power Rule reveals the parabolic nature of the integral more clearly.

Step 4: Personal preference

Preparing to state a preferred method. There's no definitive correct choice. It could be said that the simplicity and versatility of the Simple Power Rule may make it a generally preferred option. However, in situations where the function to be integrated is already in a 'ready-made' format for the General Power Rule, it might be quicker and easier to use that rule instead.

Key Concepts

Simple Power Rule

The Simple Power Rule is a basic yet powerful technique in calculus for integrating polynomial functions. When you have a term like \(x^n\), where \(n\) is any real number except \(-1\), you can apply the rule to find the integral in a straightforward manner. The rule states that the integral of \(x^n\) is \(\frac{1}{n+1}x^{n+1} + C\). The value \(n + 1\) becomes the new exponent, and you divide by this number to scale the term correctly. It's crucial to remember to add \(C\), the constant of integration, since integration is an inverse operation of differentiation and we must account for any constant that might have been lost.

Let's break it down step by step:

• Expand the function if needed: For example, \((x-1)^2\) becomes \(x^2 - 2x + 1\).
• Apply the Simple Power Rule to each term: Integrate \(x^2\) to get \(\frac{1}{3}x^3\), \(2x\) to get \(x^2\), and \(1\) to get \(x\).
• Add the terms together: The final result is \(\frac{1}{3}x^3 - x^2 + x + C\).

This method provides a clear view of each component's contribution to the integral, making it excellent for detailed work.

General Power Rule

The General Power Rule is particularly useful when you encounter a composite function of the form \(((u(x)))^n\), where \(u(x)\) is a differentiable function and does not involve division by zero. This rule helps integrate complex expressions without needing to expand them, making calculations more compact and elegant. The principle behind it is to identify an inner function \(u(x)\), calculate its derivative \(u'(x)\), and then apply the rule.

• Identify \(u(x)\) in the expression: In \((x-1)^2\), \(u(x) = x-1\).
• Consider that \(u'(x)\) is needed: Here, \(u'(x) = 1\), which doesn't change the integral.
• Perform the integration: The integral of \((u(x))^n\) is \(\frac{1}{n+1}(u(x))^{n+1} + C\).

You substitute back \(u(x) = x-1\), resulting in \(\frac{1}{3}(x-1)^3 + C\).

This method keeps the expression compact, especially when dealing with higher powers or more intricate functions.

Constant of Integration

The Constant of Integration, typically denoted by \(C\), is an essential component of indefinite integrals. When you integrate, you are essentially reversing differentiation, and since a derivative of a constant is zero, any constant added while differentiating a function can be lost. To account for this, when integrating, we include \(C\) to represent this 'missing' constant.

The presence of \(C\) has several important implications:

• Reflects all possible antiderivatives: Without \(C\), you only represent one specific antiderivative.
• Essential for solving differential equations: Provides the general solution that represents a family of functions, not just a single one.
• Aligns solutions with initial conditions: In physics and engineering, \(C\) helps tailor solutions to specific scenarios.

Ultimately, including \(C\) reminds us that the process of integration yields a set of functions, not just one fixed answer. This understanding is crucial when working in fields that rely on precise initial conditions or boundary values.