Problem 25
Question
In Exercises \(17-54\) , find the most general antiderivative or indefinite integral. Check your answers by differentiation. $$ \int x^{-1 / 3} d x $$
Step-by-Step Solution
Verified Answer
\( \frac{3}{2} x^{2/3} + C \) is the most general antiderivative of \( x^{-1/3} \).
1Step 1: Identify the Integral Expression
We are given the integral \( \int x^{-1/3} \, dx \). This expression represents an indefinite integral that needs to be solved to find the most general antiderivative.
2Step 2: Apply the Power Rule for Integration
The power rule for integration states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \). For our problem, \( n = -1/3 \). Applying the rule:\[ \int x^{-1/3} \, dx = \frac{x^{(-1/3)+1}}{(-1/3)+1} + C \]
3Step 3: Simplify the Expression
Simplify the exponent and the denominator obtained in the previous step:\(-1/3 + 1 = 2/3\), thus\[ \frac{x^{2/3}}{2/3} + C \]This simplifies further to \( \frac{3}{2} x^{2/3} + C \).
4Step 4: Verify the Solution by Differentiation
Differentiate the antiderivative \( \frac{3}{2} x^{2/3} + C \) to verify:\( \frac{d}{dx} [\frac{3}{2} x^{2/3} + C] = \frac{3}{2} \cdot \frac{2}{3} x^{-1/3} = x^{-1/3} \).Since the original function is retrieved, the solution is correct.
Key Concepts
AntiderivativePower Rule for IntegrationDifferentiationCalculus
Antiderivative
An antiderivative, often referred to as an indefinite integral, is a fundamental idea in calculus. When you find the antiderivative of a function, you are essentially determining a function whose derivative is the original function you began with. This operation seeks to reverse the process of differentiation.
This concept is crucial because:
For example, if you have a function representing speed, its antiderivative would give you the distance traveled over time. Antiderivatives are illustrated using the integral symbol \( \int \), signifying integration is being performed. Unlike definite integrals, which calculate an exact value, indefinite integrals yield a family of functions differing by a constant \( C \), as the process of differentiation loses information about constant terms.
This concept is crucial because:
- It allows us to determine the accumulation of quantities, such as area under curves.
- It plays a key role in solving differential equations.
- It helps in understanding the original function behind a rate of change.
For example, if you have a function representing speed, its antiderivative would give you the distance traveled over time. Antiderivatives are illustrated using the integral symbol \( \int \), signifying integration is being performed. Unlike definite integrals, which calculate an exact value, indefinite integrals yield a family of functions differing by a constant \( C \), as the process of differentiation loses information about constant terms.
Power Rule for Integration
The power rule for integration is a key tool when solving problems involving integrals, particularly those involving powers of \( x \). It's the most straightforward method for finding antiderivatives of polynomial functions.
The rule states that for a function \( x^n \), where \( n \) is any real number except \( -1 \), the antiderivative is given by:
The rule states that for a function \( x^n \), where \( n \) is any real number except \( -1 \), the antiderivative is given by:
- \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)
Differentiation
Differentiation is the counterpart to integration in calculus and deals with finding the rate at which one quantity changes relative to another. In simpler terms, it helps us find the slope of a function at any given point or the tangent's rate of change.
The primary goal of differentiation in the context of solving indefinite integrals is to verify the accuracy of a potential solution. By differentiating the resulting antiderivative, we can confirm that it matches the original function. This verification step is crucial:
In our exercise, after using the power rule, differentiating \( \frac{3}{2} x^{2/3} + C \) confirms that it is indeed the correct antiderivative, retrieving the original expression \( x^{-1/3} \).
The primary goal of differentiation in the context of solving indefinite integrals is to verify the accuracy of a potential solution. By differentiating the resulting antiderivative, we can confirm that it matches the original function. This verification step is crucial:
- If differentiating gives you back the original expression under the integral sign, your integration was performed correctly.
- It reinforces the relationship between integration and differentiation as inverse operations.
In our exercise, after using the power rule, differentiating \( \frac{3}{2} x^{2/3} + C \) confirms that it is indeed the correct antiderivative, retrieving the original expression \( x^{-1/3} \).
Calculus
Calculus, as a branch of mathematics, revolves around the concepts of limits, derivatives, integrals, and functions. It plays a pivotal role in understanding and modeling real-world phenomena involving change and motion.
This mathematical field is split into two main parts:
In the context of our task, calculus helps to establish the connection between antiderivatives and the geometric interpretation behind them. Understanding the process of integration via calculus allows us to discover functions related to growth, area, and more. Calculus provides the framework and tools needed to transition from observing a rate of change to understanding the broader picture of accumulation. This interplay between integration and differentiation reveals the dynamic nature of change in our world and aids us in solving complex problems across various disciplines.
This mathematical field is split into two main parts:
- Differential Calculus: Focused on finding instantaneous rates of change, like slopes and velocities.
- Integral Calculus: Concerned with accumulation, such as areas under curves and antiderivatives.
In the context of our task, calculus helps to establish the connection between antiderivatives and the geometric interpretation behind them. Understanding the process of integration via calculus allows us to discover functions related to growth, area, and more. Calculus provides the framework and tools needed to transition from observing a rate of change to understanding the broader picture of accumulation. This interplay between integration and differentiation reveals the dynamic nature of change in our world and aids us in solving complex problems across various disciplines.
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