Chapter 5: Problem 15
Evaluate. $$ \int \frac{8 x-5}{\sqrt[3]{x}} d x $$
Short Answer
Expert verified
\( \int \frac{8x-5}{\sqrt[3]{x}} \, dx = \frac{24}{5}x^{5/3} - \frac{15}{2}x^{2/3} + C \).
Step by step solution
01
Rewrite the Integral in a Simplified Form
Begin by rewriting the integral to make it easier to evaluate. We know that \( \sqrt[3]{x} = x^{1/3} \). So, the integral \( \int \frac{8x - 5}{\sqrt[3]{x}} \, dx \) can be rewritten as \( \int (8x^{1-1/3} - 5x^{-1/3}) \, dx \). Simplifying gives us \( \int (8x^{2/3} - 5x^{-1/3}) \, dx \).
02
Integrate Each Term Separately
Integrate the terms separately. For \( \int 8x^{2/3} \, dx \), increase the power by 1 to get \( x^{5/3} \) and multiply by the reciprocal to get \( \frac{8}{5/3}x^{5/3} = \frac{24}{5}x^{5/3} \). For \( \int -5x^{-1/3} \, dx \), increase the power by 1 to get \( x^{2/3} \) and multiply by the reciprocal to get \( \frac{-5}{2/3}x^{2/3} = \frac{-15}{2}x^{2/3} \).
03
Combine the Integrated Terms
Combine the results from step 2. The antiderivative will be \( \frac{24}{5}x^{5/3} - \frac{15}{2}x^{2/3} + C \), where \( C \) 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.
definite and indefinite integrals
Integrals are a fundamental concept in calculus, often referred to as the area under a curve. There are two main types of integrals:
- Indefinite integrals: Represent the antiderivative of a function and do not have specific limits of integration. Their solution includes an arbitrary constant, typically noted as \( C \). This constant represents the family of functions that differ by only a vertical shift.
- Definite integrals: Calculate the total area under a curve within a specified interval \([a, b]\). The result of a definite integral is a number and it does not include the constant \( C \).
integration by substitution
Integration by substitution is a powerful technique used to simplify integrals, especially when the integral involves complex expressions. The method relies on simplifying the integral by substituting part of the equation with a single variable \( u \), essentially transforming the integral into an easier form to evaluate.
This technique is similar to the chain rule used in differentiation. Here’s a simple guide:
This technique is similar to the chain rule used in differentiation. Here’s a simple guide:
- Select a part of the integral: Choose a substitution \( u = g(x) \) that simplifies the integral.
- Differentiate to find \( du \): Calculate \( du/dx \) to express \( du \) in terms of \( dx \).
- Rewrite the integral: Substitute \( u \) and \( du \) back into the integral, which should now be in a simpler form.
- Integrate with respect to \( u \): Solve the simplified integral in terms of \( u \).
- Back-substitute: Replace \( u \) with the original expression in terms of \( x \).
algebraic manipulation in calculus
Algebraic manipulation plays a crucial role in both simplifying expressions for integration and solving integrals effectively.
In the given exercise, we started by rewriting the integrand \( \frac{8x-5}{\sqrt[3]{x}} \) as \( 8x^{2/3} - 5x^{-1/3} \). This step involved breaking down complex expressions into simpler parts by using algebraic identities and properties of exponents.Key steps in algebraic manipulation:
In the given exercise, we started by rewriting the integrand \( \frac{8x-5}{\sqrt[3]{x}} \) as \( 8x^{2/3} - 5x^{-1/3} \). This step involved breaking down complex expressions into simpler parts by using algebraic identities and properties of exponents.Key steps in algebraic manipulation:
- Substitute equivalent expressions: Replace terms with their algebraic equivalents to simplify.
- Apply exponent rules: Use laws of indices to rewrite terms (e.g., from roots to fractional exponents).
- Simplify each term: Break down the expression so that each term can be integrated independently.