Problem 43
Question
Find a polynomial whose derivative the given polynomial. \(x^{8}+6 x^{5}-3 x\)
Step-by-Step Solution
Verified Answer
\(\frac{1}{9}x^9 + x^6 - \frac{3}{2}x^2 + C\) is the required polynomial.
1Step 1: Understand the Problem
We are given the polynomial \(x^{8}+6x^{5}-3x\) and asked to find another polynomial whose derivative becomes the given polynomial. This means we need to find the original polynomial before differentiation took place.
2Step 2: Integrate the Polynomial
To find the original polynomial, we need to integrate the given polynomial. The integral of a polynomial \(ax^n\) is \(\frac{a}{n+1}x^{n+1}+C\), where \(C\) is the constant of integration.
3Step 3: Integrate Each Term Separately
Integrate each term of the polynomial \(x^8+6x^5-3x\) individually:- The integral of \(x^8\) is \(\frac{1}{9}x^9\).- The integral of \(6x^5\) is \(\frac{6}{6}x^6 = x^6\).- The integral of \(-3x\) is \(-\frac{3}{2}x^2\).
4Step 4: Write the Integrated Polynomial
Combine the integrated terms to write the final polynomial:\[\frac{1}{9}x^9 + x^6 - \frac{3}{2}x^2 + C\], where \(C\) is the constant of integration.
Key Concepts
Understanding DifferentiationIntegrating a PolynomialThe Importance of the Constant of Integration
Understanding Differentiation
Differentiation is a fundamental concept in calculus that focuses on finding the rate at which a function changes. Essentially, differentiation takes a function and produces another function, known as the derivative, which tells us the slope of the original function at any point. This process is particularly useful in various fields, such as physics, engineering, and economics.When we differentiate a polynomial, we apply the power rule, which states: for a term of the form \( ax^n \), the derivative is \( nax^{n-1} \). For instance, the derivative of \( x^8 \) is \( 8x^7 \). Similarly, for \( 6x^5 \), it would become \( 30x^4 \), and for \( -3x \), the derivative is simply \( -3 \). Each of these steps transforms the original polynomial into a new one that shows how each term is likely to change.Differentiation reverses the process of integration, which is why, by integrating the derivative polynomial, we can work backwards to find the original polynomial. Understanding differentiation as a process to find the slope helps in comprehending how integration serves to reconstruct the function from its slope.
Integrating a Polynomial
Integration is the reverse process of differentiation. When dealing with polynomials, integration allows us to retrieve the original function that gave rise to a given derivative. It is often described as finding the area under the curve for the function, but with polynomials, it identifies the original equation.For any term \( ax^n \) in a polynomial, the general formula for integration is \( \frac{a}{n+1}x^{n+1} + C \). This formula accounts for the constant of integration, which is especially crucial when performing indefinite integrals.Let’s consider the polynomial provided in the problem: \( x^8+6x^5-3x \). By integrating each term separately:
- For \( x^8 \), we get \( \frac{1}{9}x^9 \).
- For \( 6x^5 \), the result is \( \frac{6}{6}x^6 = x^6 \).
- For \( -3x \), we have \( -\frac{3}{2}x^2 \).
The Importance of the Constant of Integration
One often overlooked aspect of integration is the constant of integration, represented by \( C \). Whenever we integrate to find an indefinite integral, the result includes this constant. The reason for this is that when a function is differentiated, any constant term becomes zero. Hence, a specific derivative can correspond to many different original polynomials, each differing by a constant.In our example, when we integrate the polynomial \( x^8+6x^5-3x \), we arrive at the polynomial \[ \frac{1}{9}x^9 + x^6 - \frac{3}{2}x^2 + C \]. This \( C \) allows for all possible vertical shifts of the curve of this polynomial. It’s crucial for accurately describing families of functions that could have led to the same derivative.The constant of integration highlights the indefinite nature of integration, as it implies that multiple functions can share the same derivative. Understanding this concept is key to solving problems in calculus where initial conditions might later determine the exact value of \( C \). Including \( C \) ensures the solution reflects all possible original polynomials that could exist.
Other exercises in this chapter
Problem 43
A position \(p(t)\) is given. Calculate the acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ). \(p(t)=3 t^{2}+16 t \mathrm{~m}\)
View solution Problem 43
Use the specified value of \(c\) and the given information about \(f\) and \(g\) to compute \((g \circ f)^{\prime}(c)\). \(g(1 / 2)=6, g^{\prime}(1 / 2)=1 / 3,
View solution Problem 43
Calculate the derivative of each of the expressions in Exercises 39-44 by applying both the Product and Quotient Rules. $$ \sin ^{2}(x) \cdot \sec (x) $$
View solution Problem 43
If \(C(x)\) is the cost of producing \(x\) units of an item, then the marginal cost of the \(x^{\text {th }}\) item is defined to be \(C^{\prime}(x)\). Suppose
View solution