Chapter 4: Problem 12
Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(\frac{1}{3 x}\) b. \(\frac{2}{5 x}\) c. \(1+\frac{4}{3 x}-\frac{1}{x^{2}}\)
Short Answer
Expert verified
a. \( \frac{1}{3} \ln |x| + C \); b. \( \frac{2}{5} \ln |x| + C \); c. \( x + \frac{4}{3} \ln |x| + \frac{1}{x} + C \).
Step by step solution
01
Identify the Antiderivative for (a)
To find the antiderivative for the function \( \frac{1}{3x} \), we first rewrite it as \( \frac{1}{3} \cdot \frac{1}{x} \). A known antiderivative for \( \frac{1}{x} \) is \( \ln |x| \), so by the constant multiple rule, the antiderivative for \( \frac{1}{3} \cdot \frac{1}{x} \) is \( \frac{1}{3} \ln |x| + C \), where \( C \) is the constant of integration.
02
Differentiate to Check (a)
Differentiate \( \frac{1}{3} \ln |x| + C \) with respect to \( x \) to verify. The derivative of \( \ln |x| \) is \( \frac{1}{x} \), so the derivative is \( \frac{1}{3} \cdot \frac{1}{x} = \frac{1}{3x} \), which confirms the antiderivative is correct.
03
Identify the Antiderivative for (b)
For the function \( \frac{2}{5x} \), rewrite it as \( \frac{2}{5} \cdot \frac{1}{x} \). By the same rule as before, the antiderivative is \( \frac{2}{5} \ln |x| + C \).
04
Differentiate to Check (b)
Differentiating \( \frac{2}{5} \ln |x| + C \) gives \( \frac{2}{5} \cdot \frac{1}{x} = \frac{2}{5x} \), confirming it is correct.
05
Identify the Antiderivative for (c)
The function is \( 1 + \frac{4}{3x} - \frac{1}{x^{2}} \). Separate into terms: the antiderivative of \( 1 \) is \( x \), of \( \frac{4}{3x} \) is \( \frac{4}{3} \ln |x| \), and of \( -\frac{1}{x^2} \), rewrite as \(-x^{-2}\), which has an antiderivative of \( x^{-1} \) or \(-\frac{1}{x} \). So, the entire antiderivative is \( x + \frac{4}{3} \ln |x| + \frac{1}{x} + C \).
06
Differentiate to Check (c)
Differentiating \( x + \frac{4}{3} \ln |x| + \frac{1}{x} + C \) results in \( 1 + \frac{4}{3x} - \frac{1}{x^2} \), which matches the original function. This confirms the solution is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration Techniques
Integration is a fundamental process in calculus, often used to find an antiderivative of a function. Essentially, you're "reversing" differentiation. When aimed at various functions, different integration techniques can simplify the work.
- Power Rule: For a function of the form \(x^{n}\), the antiderivative is \(\frac{x^{n+1}}{n+1}\), \(n eq -1\).
- Logarithmic Rule: The function \(\frac{1}{x}\) has a special antiderivative, which is \(\ln |x|\). This is often encountered in integration problems.
- Constant Multiple Rule: When you have a constant multiple of a function, \(k \cdot f(x)\), the antiderivative is \(k \cdot F(x) + C\), where \(F(x)\) is the antiderivative of \(f(x)\).
Logarithmic Functions
Logarithmic functions play a crucial role in integration, especially because of their relationship with exponential functions. The integral of \(\frac{1}{x}\) gives the natural logarithm: \(\int \frac{1}{x} \, dx = \ln |x| + C\). This transformation is a result of differentiation where \(\frac{d}{dx}\ln|x| = \frac{1}{x}\).When dealing with expressions involving \(\frac{1}{x}\), it's beneficial to rewrite them in a form that makes the antiderivative obvious, as seen in the original problem:
- The function \(\frac{1}{3x}\) is rewritten as \(\frac{1}{3}\)\( \frac{1}{x}\) to use the Logarithmic Rule directly.
- Similarly, \(\frac{2}{5x}\) becomes \(\frac{2}{5}\)\(\frac{1}{x}\), making it straightforward to see the antiderivative is \(\frac{2}{5}\ln|x| + C\).
Constant of Integration
When finding antiderivatives, it's important to include a constant of integration, \(C\). This constant accounts for all possible vertical shifts of a function’s graph, since differentiation "loses" constants.In practical terms, here’s why the constant of integration matters:
- Without \(C\), you'd only find one of the many possible antiderivatives for a given function. Each \(C\) represents a parallel function, infinitely many functions share the same derivative.
- In problems where initial conditions are known, this constant can be determined, providing a unique solution. For example, if you know a specific point on the original function, you can solve for \(C\) to ensure the antiderivative fits perfectly.