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Chapter 5: Problem 24

Find each indefinite integral. $$ \int\left(\frac{1}{z^{2}}+\frac{1}{\sqrt[3]{z}}\right) d z $$

Short Answer

Expert verified
The indefinite integral is \(-z^{-1} + \frac{3}{2}z^{2/3} + C\).

Step by step solution

01

Separate the Integral

The given integral is \( \int \left( \frac{1}{z^{2}} + \frac{1}{\sqrt[3]{z}} \right) \, dz \). We can separate this into two separate integrals: \( \int \frac{1}{z^{2}} \, dz \) and \( \int \frac{1}{\sqrt[3]{z}} \, dz \).
02

Rewrite the Terms in Power Form

Rewriting each term in a more usable form, we have \( \int z^{-2} \, dz \) for the first term and \( \int z^{-1/3} \, dz \) for the second term.
03

Integrate the First Term

To integrate \( \int z^{-2} \, dz \), we use the power rule of integration: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1\). Applying this, we get: \( \frac{z^{-1}}{-1} + C_1 = -z^{-1} + C_1 \).
04

Integrate the Second Term

For \( \int z^{-1/3} \, dz \), we also use the power rule. This gives us: \( \frac{z^{(2/3)}}{\frac{2}{3}} + C_2 = \frac{3}{2}z^{2/3} + C_2 \).
05

Combine the Results and Add the Constant of Integration

Combining the two results from Steps 3 and 4, we have the indefinite integral: \( -z^{-1} + \frac{3}{2}z^{2/3} + C \), where \( C = C_1 + C_2 \) is the constant of integration.

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Key Concepts

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

Power Rule in Integration
The Power Rule is a fundamental concept when solving indefinite integrals in calculus. It provides an easy way to find the antiderivative of functions of the form \(x^n\). The Power Rule states:
  • If \(n eq -1\), then the integral \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.
This rule is particularly useful in calculus as it transforms complex expressions into simpler ones. In the exercise, terms like \(\frac{1}{z^2}\) and \(\frac{1}{\sqrt[3]{z}}\) are rewritten as \(z^{-2}\) and \(z^{-1/3}\) respectively to apply the Power Rule easily.
By converting these expressions into a power form, it becomes straightforward to integrate each term individually, making it a powerful tool in calculus integration problems.
Common Integration Techniques
Integration techniques are methods applied to solve integrals that don鈥檛 have a straightforward antiderivative. In this exercise, a couple of techniques were used:
  • Rewriting Terms: The initial step was to rewrite each term into a more recognizable form for integration, such as changing radicals or fractions to powers of variables (power notation).
  • Separation of Integrals: By breaking down the integral of a sum into separate parts, it allows each part to be tackled individually, as seen with the functions \(\int \frac{1}{z^2} \, dz\) and \(\int \frac{1}{\sqrt[3]{z}} \, dz\).
These techniques simplify the process and ensure easy application of the Power Rule and other integration methods. Mastery of these techniques enables students to solve more complex calculus problems efficiently.
Calculus Problem Solving
Solving calculus problems, such as finding indefinite integrals, requires a methodical approach. The given problem involves finding the antiderivative of a composite function:
  • The first step is often through simplifying the integrand to a form where known rules like the Power Rule can be directly applied.
  • Careful separation of integrals ensures that each component can be managed without interference, leading to accurate results.
  • Finally, the combination of results emphasizes the importance of consistency and verification in the final solution, ensuring that any constants of integration are accounted for.
Effective problem solving in calculus hinges on recognizing applicable rules and techniques, breaking down problems into manageable steps, and maintaining accuracy throughout the process. This enhances understanding and builds confidence in handling a variety of calculus challenges.

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