Understanding the Cubic Polynomial

Oct 13, 2025 by recoveryshake.com 31 views

Introduction

In this article, we will explore the concept of integration and its application to a cubic polynomial. The cubic polynomial $f(x)$ is defined by $f(x)=x^3-19x+30$ . Our goal is to find the exact value of $\int_{-5}^3 f(x) \, dx$ using integration.

Understanding the Cubic Polynomial

A cubic polynomial is a polynomial of degree three, which means it has the general form $ax^3+bx^2+cx+d$ , where $a$ , $b$ , $c$ , and $d$ are constants. In this case, the cubic polynomial $f(x)$ is defined by $f(x)=x^3-19x+30$ . This polynomial has a degree of three, which means it will have three roots.

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) states that differentiation and integration are inverse processes. In other words, if we have a function $f(x)$ and we integrate it from $a$ to $b$ , we get the definite integral $\int_{a}^{b} f(x) \, dx$ . The FTC also states that if we have a function $f(x)$ and we differentiate it, we get the derivative $f'(x)$ .

Integration of the Cubic Polynomial

To find the exact value of $\int_{-5}^3 f(x) \, dx$ , we need to integrate the cubic polynomial $f(x)=x^3-19x+30$ . We can do this by using the power rule of integration, which states that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ .

Step 1: Integrate the Cubic Term

The first term in the cubic polynomial is $x^3$ . We can integrate this term using the power rule of integration.

\int x^3 \, dx = \frac{x^4}{4} + C

Step 2: Integrate the Linear Term

The second term in the cubic polynomial is $-19x$ . We can integrate this term using the power rule of integration.

\int -19x \, dx = -\frac{19x^2}{2} + C

Step 3: Integrate the Constant Term

The third term in the cubic polynomial is $30$ . We can integrate this term by treating it as a constant.

\int 30 \, dx = 30x + C

Combining the Terms

Now that we have integrated each term, we can combine them to get the final result.

\int f(x) \, dx = \frac{x^4}{4} - \frac{19x^2}{2} + 30x + C

Evaluating the Definite Integral

To find the exact value of $\int_{-5}^3 f(x) \, dx$ , we need to evaluate the definite integral. We can do this by plugging in the limits of integration and simplifying.

\int_{-5}^3 f(x) \, dx = \left[ \frac{x^4}{4} - \frac{19x^2}{2} + 30x \right]_{-5}^3

Simplifying the Expression

Now that we have evaluated the definite integral, we can simplify the expression.

\int_{-5}^3 f(x) \, dx = \left( \frac{3^4}{4} - \frac{19(3)^2}{2} + 30(3) \right) - \left( \frac{(-5)^4}{4} - \frac{19(-5)^2}{2} + 30(-5) \right)

Simplifying Further

Now that we have simplified the expression, we can simplify further.

\int_{-5}^3 f(x) \, dx = \left( \frac{81}{4} - \frac{171}{2} + 90 \right) - \left( \frac{625}{4} - \frac{475}{2} - 150 \right)

Final Answer

Now that we have simplified the expression, we can find the final answer.

\int_{-5}^3 f(x) \, dx = \left( \frac{81}{4} - \frac{171}{2} + 90 \right) - \left( \frac{625}{4} - \frac{475}{2} - 150 \right)

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150

\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac<br/> **Q&A: The Cubic Polynomial and Integration** ============================================= **Q: What is a cubic polynomial?** ------------------------------ A: A cubic polynomial is a polynomial of degree three, which means it has the general form $ax^3+bx^2+cx+d$, where $a$, $b$, $c$, and $d$ are constants. **Q: How do you integrate a cubic polynomial?** ---------------------------------------------- A: To integrate a cubic polynomial, you can use the power rule of integration, which states that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$. You can also use substitution or integration by parts to integrate the polynomial. **Q: What is the Fundamental Theorem of Calculus?** ---------------------------------------------- A: The Fundamental Theorem of Calculus (FTC) states that differentiation and integration are inverse processes. In other words, if we have a function $f(x)$ and we integrate it from $a$ to $b$, we get the definite integral $\int_{a}^{b} f(x) \, dx$. The FTC also states that if we have a function $f(x)$ and we differentiate it, we get the derivative $f'(x)$. **Q: How do you evaluate a definite integral?** ---------------------------------------------- A: To evaluate a definite integral, you need to plug in the limits of integration and simplify the expression. You can use the power rule of integration, substitution, or integration by parts to simplify the expression. **Q: What is the final answer to the problem?** ---------------------------------------------- A: The final answer to the problem is $\int_{-5}^3 f(x) \, dx = \frac{81}{4} - \frac{171}{2} + 90 - \frac{625}{4} + \frac{475}{2} + 150$. **Q: Can you explain the steps to solve the problem?** ------------------------------------------------ A: Yes, here are the steps to solve the problem: 1. Integrate the cubic polynomial $f(x)=x^3-19x+30$ using the power rule of integration. 2. Evaluate the definite integral $\int_{-5}^3 f(x) \, dx$ by plugging in the limits of integration and simplifying the expression. 3. Simplify the expression to get the final answer. **Q: What are some common mistakes to avoid when integrating a cubic polynomial?** -------------------------------------------------------------------------------- A: Some common mistakes to avoid when integrating a cubic polynomial include: * Forgetting to use the power rule of integration * Not plugging in the limits of integration when evaluating the definite integral * Not simplifying the expression to get the final answer * Using the wrong formula or method to integrate the polynomial **Q: Can you provide more examples of integrating cubic polynomials?** ---------------------------------------------------------------- A: Yes, here are some more examples of integrating cubic polynomials: * $\int (x^3+2x^2-3x+1) \, dx = \frac{x^4}{4} + \frac{2x^3}{3} - \frac{3x^2}{2} + x + C$ * $\int (x^3-4x^2+3x-1) \, dx = \frac{x^4}{4} - \frac{4x^3}{3} + \frac{3x^2}{2} - x + C$ * $\int (x^3+5x^2-2x-1) \, dx = \frac{x^4}{4} + \frac{5x^3}{3} - \frac{2x^2}{2} - x + C$ **Q: Can you explain the concept of integration by parts?** --------------------------------------------------- A: Yes, integration by parts is a method of integration that involves differentiating one function and integrating the other function. The formula for integration by parts is $\int u \, dv = uv - \int v \, du$. You can use integration by parts to integrate polynomials, trigonometric functions, and other types of functions. **Q: Can you provide more information on the Fundamental Theorem of Calculus?** ------------------------------------------------------------------------- A: Yes, the Fundamental Theorem of Calculus (FTC) states that differentiation and integration are inverse processes. In other words, if we have a function $f(x)$ and we integrate it from $a$ to $b$, we get the definite integral $\int_{a}^{b} f(x) \, dx$. The FTC also states that if we have a function $f(x)$ and we differentiate it, we get the derivative $f'(x)$. The FTC is a fundamental concept in calculus and is used to solve a wide range of problems in mathematics and science.