Chapter 4: Problem 83
Solve each integral. Each can be found using rules developed in this section, but some algebra may be required. $$ \int(3 x-5)(2 x+1)^{2} d x $$
Short Answer
Expert verified
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 by step solution
01
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 \( 12x^3 - 8x^2 - 17x - 5 \).
02
Set Up the Integral
Rewrite the integral with the expanded polynomial: \( \int (12x^3 - 8x^2 - 17x - 5) \, dx \). Split this into separate integrals: \( \int 12x^3 \, dx - \int 8x^2 \, dx - \int 17x \, dx - \int 5 \, dx \).
03
Integrate Each Term
Now, integrate each term using power rule for integration \(( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C )\):- \( \int 12x^3 \, dx = 12 \cdot \frac{x^4}{4} = 3x^4 \)- \( \int 8x^2 \, dx = 8 \cdot \frac{x^3}{3} = \frac{8}{3}x^3 \)- \( \int 17x \, dx = 17 \cdot \frac{x^2}{2} = \frac{17}{2}x^2 \)- \( \int 5 \, dx = 5x \).
04
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.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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 \(12x^3 - 8x^2 - 17x - 5\).
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 \(12x^3 - 8x^2 - 17x - 5\):
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 \(-5x\).
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: \(3x^4 - \frac{8}{3}x^3 - \frac{17}{2}x^2 - 5x\).
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.
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:
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 \(C\).