Problem 45 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(f^{\prime \prime}(x)\) 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. $$ f^{\prime \prime}(x)=\frac{x^{4}+1}{x^{3}} $$

Step-by-Step Solution

Answer

\(f(x) = \frac{x^3}{6} + \frac{1}{2x} + C_1 x + C_2\) with constants \(C_1\) and \(C_2\).

Step 1: Simplify the Second Derivative

First, let's simplify the expression given for the second derivative. We have\[ f''(x) = \frac{x^4 + 1}{x^3} \]This can be separated into two terms:\[ f''(x) = \frac{x^4}{x^3} + \frac{1}{x^3} = x + x^{-3} \]

Step 2: Find the First Derivative

To find the first derivative \(f'(x)\), we need to integrate \(f''(x)\). We integrate each term separately:\[ \int (x + x^{-3}) \, dx = \int x \, dx + \int x^{-3} \, dx \]The integrals are:- \(\int x \, dx = \frac{x^2}{2} + C_1\)- \(\int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2} + C_2\)Combining these results, we obtain:\[ f'(x) = \frac{x^2}{2} - \frac{1}{2x^2} + C_1 \]

Step 3: Find the Original Function

Next, we integrate \(f'(x)\) to find \(f(x)\). Integrate each term in \(f'(x)\):\[ \int \left(\frac{x^2}{2} - \frac{1}{2x^2} + C_1\right) \, dx \]The integrals are:- \(\int \frac{x^2}{2} \, dx = \frac{x^3}{6}\)- \(\int -\frac{1}{2x^2} \, dx = \frac{1}{2x}\)- \(\int C_1 \, dx = C_1 x\)Combining these results and adding a second constant of integration \(C_2\), we get:\[ f(x) = \frac{x^3}{6} + \frac{1}{2x} + C_1 x + C_2 \]

Step 4: Final Solution

The original function \(f(x)\) is:\[ f(x) = \frac{x^3}{6} + \frac{1}{2x} + C_1 x + C_2 \]where \(C_1\) and \(C_2\) are arbitrary constants.

Key Concepts

Integration

Integration is an essential process in calculus that involves finding a function whose derivative matches a given function. In this particular exercise, we are given a second derivative, denoted as \(f''(x)\), and our task is to find the original function \(f(x)\), which requires integrating twice.
This process is also known as antidifferentiation or indefinite integration.
by integrating twice, we work backwards through the derivative chain to reconstruct the original function.The importance of integration in calculus cannot be overstated. It is used to determine the area under a curve, solve differential equations, and in various physical applications such as finding velocity from acceleration.
When integrating, we must pay attention to the power rule, which states how to integrate functions of the form \(x^n\):

• When \(n eq -1\), the integral is \(\frac{x^{n+1}}{n+1}\)
• For \(n = -1\), the integral is the natural logarithm, \(\ln|x|\)

In the context of our exercise, integrating \(f''(x) = x + x^{-3}\) allows us to determine \(f'(x)\) and subsequently \(f(x)\). By properly applying the integration rules and recognizing patterns, we gain a deeper understanding of the relationships between functions and their derivatives.

Arbitrary Constants

During the process of antidifferentiation, we introduce arbitrary constants, often denoted as \(C_1\), \(C_2\), and so forth.
These constants arise because the derivative of a constant is zero, making it indistinguishable from the rest of the integrated function unless initial conditions or additional information are provided to evaluate them.
Including these constants ensures the solution is as general as possible.In our problem, when integrating the second derivative to find the first derivative \(f'(x)\), we add an arbitrary constant \(C_1\).
Repeating the integration process to find \(f(x)\), we introduce another arbitrary constant \(C_2\).
This second constant makes the final expression for \(f(x)\) more versatile by encompassing all functions that have the same derivative \(f''(x)\).Arbitrary constants are crucial in integration because:

• They represent a family of solutions, allowing for all possible shifts or vertical translations of a function.
• They indicate that multiple functions can share the same derivative, distinguished only by these constants.

Understanding this concept is vital for both pure mathematical applications and real-world problems, as it helps us grasp the nature of indefinite integration.

Calculus Problem Solving

Solving calculus problems often involves understanding the relationships between derivatives, integrals, and constants.
In this exercise, comprehensive problem-solving skills are necessary to transition from the second derivative function, \(f''(x)\), to the original function, \(f(x)\), through a series of well-calculated steps.One key aspect is simplifying complex expressions. Before integrating, it's often useful to simplify the derivative, as seen in separating \(f''(x) = \frac{x^4 + 1}{x^3}\) into simpler terms like \(x + x^{-3}\).
This type of simplification can make the integration process more straightforward and reduce the complexity of the calculations.Steps to effectively solve calculus problems:

• Break down complex expressions into manageable parts.
• Apply appropriate theorems and rules, such as the power rule for integration.
• Introduce arbitrary constants and understand their significance.
• Thoroughly check your work and verify each step.

By following these practices, you not only solve specific problems but also enhance your overall calculus problem-solving aptitude. This can be particularly helpful in dealing with real-world problems, where calculus is frequently used to model phenomena and solve equations.