Problem 42 - Applied Calculus

Question

(a) Find \(\int(x+5)^{2} d x\) in two ways: (i) By multiplying out (ii) By substituting \(w=x+5\) (b) Are the results the same? Explain.

Step-by-Step Solution

Answer

Yes, both methods yield the same integral \(\frac{x^3}{3} + 5x^2 + 25x + C\).

Step 1: Expansion

First, expand the integrand \((x+5)^2\). By using the identity \((a+b)^2 = a^2 + 2ab + b^2\), we have \((x+5)^2 = x^2 + 2 \cdot 5 \cdot x + 5^2 = x^2 + 10x + 25\).

Step 2: Integration of Expanded Form

Now integrate each term separately. The integral becomes \(\int (x^2 + 10x + 25)\,dx\). This equals \(\frac{x^3}{3} + 5x^2 + 25x + C\) where \(C\) is the integration constant.

Step 3: Substitution

Now, use the substitution method. Let \(w = x + 5\), then \(dw = dx\). The integral \(\int (x+5)^2 dx\) becomes \(\int w^2 \, dw\).

Step 4: Integrate Using Substitution

Integrate \(w^2\) with respect to \(w\). We have \(\int w^2 \, dw = \frac{w^3}{3} + C\). Substitute back \(w = x + 5\) to get \(\frac{(x+5)^3}{3} + C\).

Step 5: Verify Equivalence of Results

Expand \(\frac{(x+5)^3}{3}\) to check if results are equivalent. Expanding gives \(\frac{x^3+15x^2+75x+125}{3}\), which matches \(\frac{x^3}{3} + 5x^2 + 25x + C\). The constants are already matched in the equations.

Key Concepts

Integration Techniques

When tackling integrals, two primary techniques often come into play: algebraic expansion and substitution. Each method has its strengths and is useful in different scenarios.

Algebraic expansion involves breaking down expressions into simpler terms, which allows for a straightforward integration of basic functions. This technique uses mathematical identities to simplify the process, making it particularly useful when dealing with binomial expressions.

The substitution method, on the other hand, simplifies an integral by transforming it into a more manageable form. This involves changing the variable of integration, which can often turn a complex integral into one that's much easier to solve. This technique is extremely useful when the integral contains compositions of functions, or when a substitution can reveal a hidden elementary form.

Each of these techniques can simplify the task of finding integrals, contributing to a deeper understanding of calculus through diverse problem-solving strategies.

Substitution Method

The substitution method is a powerful tool that transforms integrals into simpler forms by changing the variable of integration. It works much like changing the coordinates to simplify calculations.

• Start by identifying a suitable substitution, which often involves setting a part of the integrand to be equal to a new variable, such as setting \(w = x + 5\).
• This step requires finding the differential - meaning you calculate \(dw = dx\) when \(w = x + 5\).
• Replace the original variable and differential with the new ones in the integral, resulting in substituting \(\int (x+5)^2 dx\) with \(\int w^2 dw\).
• The transformed integral is usually more straightforward to solve. In this example, the calculation becomes integrating \(w^2\) to yield \(\frac{w^3}{3} + C\).
• Finally, replace the substituted variable back with the original terms, translating back \(w = x + 5\) to get the final integrated result.

By efficiently changing variables, substitution can convert intimidating integrals into familiar forms easier to evaluate.

Algebraic Expansion

Algebraic expansion involves using algebraic identities to break down complex expressions, making them easier to integrate. This method is particularly useful for polynomial expressions.

Consider the expression \((x+5)^2\). By using the identity \((a+b)^2 = a^2 + 2ab + b^2\), you simplify it to \((x^2 + 10x + 25)\).

• Expand the integrand by applying the algebraic identity to obtain a polynomial.
• Integrate each term separately within the polynomial. Here, integrating \(x^2\) results in \(\frac{x^3}{3}\), while \(10x\) becomes \(5x^2\), and the constant \(25\) becomes \(25x\).
• Combine all the integrated terms and add an integration constant, \(C\), to account for indefinite integration.

This approach uses basic operations to break down complicated expressions. It allows integration to become a matter of handling simpler, individual components.