Problem 16
Question
use the method of substitution to find each of the following indefinite integrals. $$ \int \sqrt[3]{2 x-4} d x $$
Step-by-Step Solution
Verified Answer
\( \frac{3(2x - 4)^{4/3}}{8} + C \)
1Step 1: Choose a substitution
Identify a substitution that simplifies the integrand. Let \( u = 2x - 4 \). This substitution means \( du = 2 \, dx \), or rearranging, \( dx = \frac{du}{2} \).
2Step 2: Substitute in the integral
Substitute \( u = 2x - 4 \) and \( dx = \frac{du}{2} \) into the integral. The integral becomes: \( \int \sqrt[3]{u} \cdot \frac{du}{2} \).
3Step 3: Simplify the integral
Factor out the constant \( \frac{1}{2} \). This simplifies the integral to \( \frac{1}{2} \int u^{1/3} \, du \).
4Step 4: Integrate using power rule
Use the power rule for integrating \( u^{n} \): \( \int u^{n} \, du = \frac{u^{n+1}}{n+1} + C \), where \( n = \frac{1}{3} \). The integral becomes \( \frac{1}{2} \cdot \left( \frac{u^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} \right) + C \).
5Step 5: Simplify the integrated result
Evaluate the expression: \( \frac{1}{2} \cdot \frac{3u^{4/3}}{4} + C = \frac{3u^{4/3}}{8} + C \).
6Step 6: Back-substitute for original variable
Replace \( u \) with \( 2x - 4 \) to express the solution in terms of \( x \). The solution is \( \frac{3(2x - 4)^{4/3}}{8} + C \).
7Step 7: Conclusion
The indefinite integral \( \int \sqrt[3]{2x-4} \, dx \) evaluated by substitution is \( \frac{3(2x - 4)^{4/3}}{8} + C \).
Key Concepts
Substitution MethodIntegrand SimplificationPower Rule Integration
Substitution Method
The **substitution method** is a vital technique in calculus that helps simplify integrals. This method involves changing variables to turn a complex integral into a simpler one that is easier to solve. In our problem, we have the integral \( \int \sqrt[3]{2x-4} \, dx \).
To use substitution, we first choose a new variable to replace a part of the integrand. Here, let \( u = 2x - 4 \). This substitution captures the expression inside the cube root. Next, we differentiate \( u \) to find \( dx \). The differentiation gives \( du = 2 \, dx \), leading to \( dx = \frac{du}{2} \). This relationship allows us to replace \( dx \) in the integral.
With \( u \) and \( dx \) expressed in terms of \( du \), the original integral simplifies significantly to \( \int \sqrt[3]{u} \cdot \frac{du}{2} \). Substitution transforms the problem, turning the integral into a form that is more manageable for further calculations. This process is often the first step in solving complex integrals effectively.
To use substitution, we first choose a new variable to replace a part of the integrand. Here, let \( u = 2x - 4 \). This substitution captures the expression inside the cube root. Next, we differentiate \( u \) to find \( dx \). The differentiation gives \( du = 2 \, dx \), leading to \( dx = \frac{du}{2} \). This relationship allows us to replace \( dx \) in the integral.
With \( u \) and \( dx \) expressed in terms of \( du \), the original integral simplifies significantly to \( \int \sqrt[3]{u} \cdot \frac{du}{2} \). Substitution transforms the problem, turning the integral into a form that is more manageable for further calculations. This process is often the first step in solving complex integrals effectively.
Integrand Simplification
**Integrand simplification** is a crucial step when solving integrals, especially after using substitution. Once we substitute the variables, the goal is to simplify the new integral. The complexity of the original integrand, \( \sqrt[3]{2x-4} \), is reduced after substitution to \( \sqrt[3]{u} \). During this process, constants outside the integral can be factored out to make further integration steps easier.
In our example, the integral \( \int \sqrt[3]{u} \cdot \frac{du}{2} \) can be rewritten as \( \frac{1}{2} \int u^{1/3} \, du \). Factoring out \( \frac{1}{2} \) is important because it simplifies the expression, allowing us to focus solely on integrating \( u^{1/3} \). By successfully simplifying the integrand, we make subsequent integration steps more straightforward and less prone to errors.
In our example, the integral \( \int \sqrt[3]{u} \cdot \frac{du}{2} \) can be rewritten as \( \frac{1}{2} \int u^{1/3} \, du \). Factoring out \( \frac{1}{2} \) is important because it simplifies the expression, allowing us to focus solely on integrating \( u^{1/3} \). By successfully simplifying the integrand, we make subsequent integration steps more straightforward and less prone to errors.
Power Rule Integration
**Power rule integration** is one of the fundamental techniques for solving integrals, especially after simplification. It applies to functions of the form \( u^n \), making it highly useful in our example.
Using the power rule, \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), we can integrate \( u^{1/3} \). In this scenario, \( n = \frac{1}{3} \). Applying the power rule results in \( \frac{u^{1/3+1}}{1/3+1} + C \). This simplifies to \( \frac{u^{4/3}}{4/3} \) after combining like terms.
Finally, we multiply by the factor \( \frac{1}{2} \) that we factored out previously, yielding \( \frac{1}{2} \cdot \frac{3u^{4/3}}{4} \), which reduces to \( \frac{3u^{4/3}}{8} + C \). This process highlights how applying the power rule maintains the methodical approach needed to solve integrals effectively. Once completed, the last step is to back-substitute the original expression for \( u \), resulting in the final solution expressed in terms of \( x \).
Using the power rule, \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), we can integrate \( u^{1/3} \). In this scenario, \( n = \frac{1}{3} \). Applying the power rule results in \( \frac{u^{1/3+1}}{1/3+1} + C \). This simplifies to \( \frac{u^{4/3}}{4/3} \) after combining like terms.
Finally, we multiply by the factor \( \frac{1}{2} \) that we factored out previously, yielding \( \frac{1}{2} \cdot \frac{3u^{4/3}}{4} \), which reduces to \( \frac{3u^{4/3}}{8} + C \). This process highlights how applying the power rule maintains the methodical approach needed to solve integrals effectively. Once completed, the last step is to back-substitute the original expression for \( u \), resulting in the final solution expressed in terms of \( x \).
Other exercises in this chapter
Problem 16
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