Problem 83 - Calculus and its applications

Question

Solve each integral. Each can be found using rules developed in this section, but some algebra may be required. $$ \int(3 x-5)(2 x+1)^{2} d x $$

Step-by-Step Solution

Answer

The integral of \\( (3x-5)(2x+1)^2 dx \\\) is \\( 3x^4 - \frac{8}{3}x^3 - \frac{17}{2}x^2 - 5x + C \\\).

Step 1: Expand the Expression

To begin solving the integral, first expand the expression \( (3x-5)(2x+1)^2 \). Expand \( (2x+1)^2 \) to get \( 4x^2 + 4x + 1 \). Next, distribute \( (3x-5) \) into \( 4x^2 + 4x + 1 \). You will get: \( 3x(4x^2 + 4x + 1) - 5(4x^2 + 4x + 1) \). This simplifies to \: - \( 12x^3 + 12x^2 + 3x - 20x^2 - 20x - 5 \). - Combine like terms to get \( 12x^3 - 8x^2 - 17x - 5 \).

Step 2: Set Up the Integral

Rewrite the integral with the expanded polynomial: \( \int (12x^3 - 8x^2 - 17x - 5) \, dx \). Split this into separate integrals: \( \int 12x^3 \, dx - \int 8x^2 \, dx - \int 17x \, dx - \int 5 \, dx \).

Step 3: Integrate Each Term

Now, integrate each term using power rule for integration \(( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C )\):- \( \int 12x^3 \, dx = 12 \cdot \frac{x^4}{4} = 3x^4 \)- \( \int 8x^2 \, dx = 8 \cdot \frac{x^3}{3} = \frac{8}{3}x^3 \)- \( \int 17x \, dx = 17 \cdot \frac{x^2}{2} = \frac{17}{2}x^2 \)- \( \int 5 \, dx = 5x \).

Step 4: Combine and Simplify the Result

Combine the results from each integral to form the complete antiderivative: \( 3x^4 - \frac{8}{3}x^3 - \frac{17}{2}x^2 - 5x + C \). Remember the constant of integration \( C \) is included here.

Key Concepts

Polynomial Expansion

Polynomial expansion is an essential initial step when facing the integration of a product of polynomials. In the given exercise, we're tasked with expanding \((3x-5)(2x+1)^2\).
To clarify this process, think of it as unpacking layers of simpler terms:

• First, expand \((2x+1)^2\) to obtain \(4x^2 + 4x + 1\).
• Next, distribute each term of \(3x-5\) across the expanded polynomial, applying the distributive property generously.
• Multiply \(3x\) by each term inside the parentheses, resulting in three terms: \(12x^3 + 12x^2 + 3x\).
• Similarly, distribute \(-5\) through the terms, yielding \(-20x^2 - 20x - 5\).
• Combine all these terms to simplify the expression to \(12x^3 - 8x^2 - 17x - 5\).

By carrying out these steps, we simplify the original polynomial into a manageable form to integrate.

Power Rule

The power rule is a fundamental tool for integrating polynomial terms. It helps simplify the process by applying a straightforward formula: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
This rule fits seamlessly into our exercise, taking each individual term from the polynomial \(12x^3 - 8x^2 - 17x - 5\):

• For \(12x^3\), the integration results in \(3x^4\).
• For \(-8x^2\), after integrating, you get \(-\frac{8}{3}x^3\).
• For \(-17x\), this becomes \(-\frac{17}{2}x^2\).
• Finally, the constant term \(-5\) transforms into \(-5x\).

The systematic use of the power rule makes integrating polynomials swift and efficient, paving the way to find the antiderivative.

Antiderivative

An antiderivative is the reverse process of differentiation, essentially retrieving an original function from its derivative. In the context of integral calculus, this means finding a function whose derivative is the integrand.
After expanding and applying the power rule, we combine terms: \(3x^4 - \frac{8}{3}x^3 - \frac{17}{2}x^2 - 5x\).

• These terms collectively form what is called the antiderivative of the original expression.
• It's crucial to remember that when dealing with indefinite integrals, without specific limits of integration, the result is a class of functions.
• This class is denoted by adding \(+ C\), where \(C\) represents an arbitrary constant.

The construction of the antiderivative enables us to generalize the solution over a spectrum of equivalent functions.

Constant of Integration

The constant of integration, denoted as \(C\), plays a key role in expressing the full scope of solutions provided by indefinite integrals.
In infinite mathematics, an indefinite integral does not just offer one function but rather a family of functions:

• This is because differentiation eliminates constants present in a function, making solutions to indefinite integrals not unique.
• \(C\) reflects the potential for various vertical shifts along the function's graph, representing distinct yet valid solutions.
• In practical terms, any specific scenario or boundary condition can determine the exact value of \(C\).

By incorporating \(C\), we account for every possible version of the antiderivative, ensuring comprehensive understanding of integrational outcomes.