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Chapter 4: Problem 44

\(f^{\prime \prime}(x)\) is given. Find \(f(x)\) by antidifferentiating twice. Note that in this case your answer should involve two arbitrary constants, one from each antidifferentiation. For example, if \(f^{\prime \prime}(x)=x\), then \(f^{\prime}(x)=x^{2} / 2+C_{1}\) and \(f(x)=\) \(x^{3} / 6+C_{1} x+C_{2} .\) The constants \(C_{1}\) and \(C_{2}\) cannot be combined because \(C_{1} x\) is not a constant. $$ f^{\prime \prime}(x)=x^{4 / 3} $$

Short Answer

Expert verified
\(f(x) = \frac{9}{70}x^{10/3} + C_1 x + C_2\)

Step by step solution

01

Find the first derivative

Given that the second derivative of the function is \(f''(x) = x^{4/3}\), we start by finding the first derivative \(f'(x)\) by integrating \(f''(x)\). The integral of \(x^{4/3}\) is obtained using the power rule for integration: \[ \int x^{4/3} \,dx = \frac{x^{4/3+1}}{4/3+1} + C_1 = \frac{x^{7/3}}{7/3} + C_1 = \frac{3}{7}x^{7/3} + C_1 \]Thus, \(f'(x) = \frac{3}{7}x^{7/3} + C_1\), where \(C_1\) is an arbitrary constant of integration.
02

Find the original function

Next, find the original function \(f(x)\) by integrating \(f'(x)\) which is \(\frac{3}{7}x^{7/3} + C_1\).\[ f(x) = \int \left(\frac{3}{7}x^{7/3} + C_1\right) \,dx = \frac{3}{7} \int x^{7/3} \,dx + C_1 \int \,dx \] - Integrating \(x^{7/3}\) yields: \[ \frac{3}{7} \int x^{7/3} \,dx = \frac{3}{7} \left( \frac{x^{10/3}}{10/3} \right) = \frac{9}{70}x^{10/3} \] - Integrating \(C_1\) yields: \[ \int C_1 \,dx = C_1 x \]Combining these integrals, we have\[ f(x) = \frac{9}{70}x^{10/3} + C_1 x + C_2 \]where \(C_2\) is another arbitrary constant of integration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Introduction to Calculus
Calculus is a branch of mathematics that deals with continuous change. It encompasses various concepts, including differentiation and integration. These concepts help us understand how functions behave, allowing us to model real-world problems.

In the context of the given exercise, calculus involves working with derivatives and integrals to find a function from its second derivative.
  • This process, known as antidifferentiation, means taking the integral of the derivative multiple times to return to the original function.
  • Understanding differentiation and integration is essential for solving such problems in calculus, helping us move between functions and their rates of change.
Understanding the Power Rule in Calculus
The power rule is a vital tool in calculus, primarily used for differentiating and integrating functions that are power functions of the form \( x^n \). When applying the power rule for integration, the rule simplistically states:

\[ \int x^n \,dx = \frac{x^{n+1}}{n+1} + C \]
Where \( C \) is an arbitrary constant. This rule helps us find the antiderivative of power functions easily.

In the exercise, we used this rule to integrate \( x^{4/3} \) by increasing the power by one and dividing by the new power:
  • Integrated form becomes \( \frac{x^{7/3}}{7/3} \)
  • Adjusted to \( \frac{3}{7}x^{7/3} \) by simplifying
Understanding and applying the power rule correctly is crucial to performing antidifferentiation effectively.
Integration in Calculus
Integration is a fundamental concept in calculus that involves finding a function given its derivative. This reverse process of differentiation is often referred to as finding the antiderivative.
  • Integration can give us the original function if we know a higher-order derivative, such as the second derivative.
  • In our exercise, we integrated the second derivative twice to retrieve the original function.
By integrating step-by-step:
  • First, we find the first derivative \( f'(x) \) by integrating \( f''(x) \).
  • Next, we integrate \( f'(x) \) to find the original function \( f(x) \).
This stepwise approach is necessary in such problems, as it ensures that we calculate each part accurately, including the integration constants.
The Role of Arbitrary Constants
When performing integration, arbitrary constants play an essential role. Each time we integrate a function, we introduce a constant of integration because antidifferentiation is an indefinite process.

These constants represent any constant value that, when differentiated, would disappear. They are crucial because:
  • They account for the "initial conditions" or unknowns when solving differential equations.
  • The first constant \( C_1 \) appears after the first integration, while \( C_2 \) appears after the second integration in our problem.
  • Different constants correspond to different possible solutions or functions that fit the original derivative.
Carefully keeping track of these constants is important, as they ensure the completeness and accuracy of the solution.

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Most popular questions from this chapter

Brass is produced in long rolls of a thin sheet. To monitor the quality, inspectors select at random a piece of the sheet, measure its area, and count the number of surface imperfections on that piece. The area varies from piece to piece. The following table gives data on the area (in square feet) of the selected piece and the number of surface imperfections found on that piece. $$ \begin{array}{ccc} \hline \text { Piece } & \begin{array}{c} \text { Area in } \\ \text { Square Feet } \end{array} & \begin{array}{c} \text { Number of } \\ \text { Surface Imperfections } \end{array} \\ \hline 1 & 1.0 & 3 \\ 2 & 4.0 & 12 \\ 3 & 3.6 & 9 \\ 4 & 1.5 & 5 \\ 5 & 3.0 & 8 \\ \hline \end{array} $$ (a) Make a scatter plot with area on the horizontal axis and number of surface imperfections on the vertical axis. (b) Does it look like a line through the origin would be a good model for these data? Explain. (c) Find the equation of the least-squares line through the origin. (d) Use the result of part (c) to predict how many surface imperfections there would be on a sheet with area \(2.0\) square feet

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First find the general solution (involving a constant \(\mathrm{C}\) ) for the given differential equation. Then find the particular solution that satisfies the indicated condition. $$ \frac{d z}{d t}=t^{2} z^{2} ; z=1 / 3 \text { at } t=1 $$

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