Problem 64 - 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, the results differ by a constant, which is normal for indefinite integrals.

Step 1: Expand the Expression

First, let's multiply out the expression \((x+5)^2\). Use the expansion formula \((a+b)^2 = a^2 + 2ab + b^2\) to expand: \[(x+5)^2 = x^2 + 10x + 25\]

Step 2: Integrate the Expanded Expression

Now, integrate each term of the expanded expression separately with respect to \(x\):\[\int (x^2 + 10x + 25) \, dx = \int x^2 \, dx + \int 10x \, dx + \int 25 \, dx\]Calculating each integral:\[\int x^2 \, dx = \frac{x^3}{3}\]\[\int 10x \, dx = 10 \cdot \frac{x^2}{2} = 5x^2\]\[\int 25 \, dx = 25x\]Combining these results, the integral is:\[\frac{x^3}{3} + 5x^2 + 25x + C\]

Step 3: Substitution Method Setup

To integrate using substitution, let \(w=x+5\), then \(dw=dx\). Thus, the original integral becomes:\[\int (w)^2 \, dw\]

Step 4: Integrate Using Substitution

Integrate \(w^2\) with respect to \(w\):\[\int w^2 \, dw = \frac{w^3}{3} + C\]Substitute back \(w = x+5\):\[\frac{(x+5)^3}{3} + C\]

Step 5: Simplify and Compare Results

Simplify the result of the substitution method by expanding \((x+5)^3\) and compare it to the result from Step 2:\[(x+5)^3 = x^3 + 15x^2 + 75x + 125\]The simplified integral result (dividing each term by 3):\[\frac{x^3}{3} + 5x^2 + 25x + \frac{125}{3} + C\]Both methods yield results that differ by a constant (in this case, \(\frac{125}{3} - 125\)). However, both are valid solutions for indefinite integrals as they differ only by a constant term, which does not affect the derivative.

Key Concepts

Substitution Method

The substitution method is an essential technique for solving integrals. It involves changing the variable of integration to simplify the integral. This method is especially useful when the integral takes a form where one part is the derivative of another.Imagine you have an integral with a complicated term. By setting a new variable equal to this term, you can transform the integral into a simpler form. In our example, we took the integral \( \int (x+5)^2 \, dx \). We chose \( w = x+5 \) to simplify it. When we do this, our integral turns into \( \int w^2 \, dw \). This substitution makes the process straightforward.Here are some critical steps to remember:

• Choose a substitution: Set \( w \) such that it makes the integral easier.
• Express \( dw \) in terms of \( dx \) (or vice versa).
• Rewrite the integral in the new variable.
• Integrate with respect to the new variable.
• Substitute back to the original variable.

This method not only simplifies the integral but also cleverly uses variable transformation to link algebra with calculus. By practicing various problems, students can better identify when substitution is the ideal choice.

Polynomial Expansion

Polynomial expansion is a valuable algebraic technique. It allows transformation of expressions like powers of binomials into a sum of simpler terms. This technique is handy when dealing with integrals because it transforms a complex integral into a set of easier integrals.In our case, with the expression \((x+5)^2\), we used the formula \((a+b)^2 = a^2 + 2ab + b^2\). This expanded our expression to \(x^2 + 10x + 25\). Once in this expanded form, each part can be integrated separately.Why is polynomial expansion helpful?

• It breaks down complex expressions into simpler parts.
• Makes integration straightforward as you handle one polynomial term at a time.

With practice, polynomial expansion becomes an intuitive process, enabling you to handle more challenging integrals with ease. It's like untangling a complicated knot by taking it apart one strand at a time.

Indefinite Integrals

Indefinite integrals represent the family of functions that, when differentiated, give the original function. They are essential in calculus as they offer insights into the accumulation of quantities.When calculating an indefinite integral, you generally end with a function \( F(x) \) plus a constant \( C \). This constant is crucial because indefinite integrals don't specify limits; thus, different functions can vary by a constant value.In our example, by integrating both the expanded form and using substitution, we arrive at slightly different expressions due to a constant difference. Despite this, both approaches are valid since they only differ by a constant.Key things to note:

• Indefinite integrals don't have specific upper and lower limits.
• They include an arbitrary constant \( C \).
• Multiple forms of the integral can exist, differing only by a constant.

Understanding indefinite integrals requires recognizing that they encapsulate all possible antiderivatives of a function. As you become comfortable with them, they unlock new insights into the behavior and properties of functions in a broader sense.