Problem 27
Question
In Exercises \(17-54\) , find the most general antiderivative or indefinite integral. Check your answers by differentiation. $$ \int(\sqrt{x}+\sqrt[3]{x}) d x $$
Step-by-Step Solution
Verified Answer
The most general antiderivative is \( \frac{2}{3}x^{3/2} + \frac{3}{4}x^{4/3} + C \).
1Step 1: Identify the Integral Components
We are given the integral \( \int (\sqrt{x} + \sqrt[3]{x}) \, dx \). This consists of two terms: \( \sqrt{x} \) and \( \sqrt[3]{x} \). We can write these in power form as \( x^{1/2} \) and \( x^{1/3} \), respectively.
2Step 2: Integrate Each Term Separately
Using the power rule for integration, which is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), we integrate each term separately:- For \( \int x^{1/2} \, dx \), the result is \( \frac{x^{1/2 + 1}}{1/2 + 1} = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2} \).- For \( \int x^{1/3} \, dx \), the result is \( \frac{x^{1/3 + 1}}{1/3 + 1} = \frac{x^{4/3}}{4/3} = \frac{3}{4}x^{4/3} \).
3Step 3: Combine the Results and Add Constant of Integration
Combine the integrated results of both terms:\[ \frac{2}{3}x^{3/2} + \frac{3}{4}x^{4/3} + C \]Here, \( C \) is the constant of integration, representing the family of antiderivatives.
4Step 4: Verify the Solution by Differentiation
To verify, differentiate \( \frac{2}{3}x^{3/2} + \frac{3}{4}x^{4/3} + C \):- The derivative of \( \frac{2}{3}x^{3/2} \) is \( \frac{2}{3} \cdot \frac{3}{2}x^{1/2} = x^{1/2} \).- The derivative of \( \frac{3}{4}x^{4/3} \) is \( \frac{3}{4} \cdot \frac{4}{3}x^{1/3} = x^{1/3} \).Adding them gives \( x^{1/2} + x^{1/3} \), which matches the integrand, confirming the solution is correct.
Key Concepts
Indefinite IntegralPower Rule for IntegrationIntegration VerificationConstant of Integration
Indefinite Integral
An indefinite integral, also known as an antiderivative, is a fundamental concept in calculus. It represents the reverse process of differentiation and is used to find a function whose derivative is the given function. When we integrate, we are essentially finding a function that describes the area under the curve of the given function. This process does not provide a specific solution but rather a general form that includes a constant of integration.
In mathematical notation, the indefinite integral of a function \( f(x) \) is written as \( \int f(x) \, dx \). This indicates that we are integrating the function \( f(x) \) with respect to \( x \). In our exercise, the integral \( \int (\sqrt{x} + \sqrt[3]{x}) \, dx \) is an example where we seek the antiderivative of the given function. This operation helps us understand how the derivative of a function works in reverse, allowing flexibility in manipulating equations.
In mathematical notation, the indefinite integral of a function \( f(x) \) is written as \( \int f(x) \, dx \). This indicates that we are integrating the function \( f(x) \) with respect to \( x \). In our exercise, the integral \( \int (\sqrt{x} + \sqrt[3]{x}) \, dx \) is an example where we seek the antiderivative of the given function. This operation helps us understand how the derivative of a function works in reverse, allowing flexibility in manipulating equations.
Power Rule for Integration
The power rule for integration is a handy tool when dealing with polynomial expressions or terms raised to a power. It is expressed as \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \). This formula tells us how to handle integration when the function is a simple power of \( x \).
In the given problem, we are integrating \( \sqrt{x} \) and \( \sqrt[3]{x} \), which translate to \( x^{1/2} \) and \( x^{1/3} \) in power form. Applying the power rule separately:
In the given problem, we are integrating \( \sqrt{x} \) and \( \sqrt[3]{x} \), which translate to \( x^{1/2} \) and \( x^{1/3} \) in power form. Applying the power rule separately:
- For \( x^{1/2} \), the integral becomes \( \frac{x^{1/2 + 1}}{1/2 + 1} = \frac{2}{3}x^{3/2} \).
- For \( x^{1/3} \), the integral becomes \( \frac{x^{1/3 + 1}}{1/3 + 1} = \frac{3}{4}x^{4/3} \).
Integration Verification
After finding the indefinite integral, it is crucial to verify the solution by differentiation. This step ensures that the integration process has been executed correctly. If we correctly integrated the function, differentiating it should return us to the original function.
For our example, we found the antiderivative \( \frac{2}{3}x^{3/2} + \frac{3}{4}x^{4/3} + C \). To verify:
For our example, we found the antiderivative \( \frac{2}{3}x^{3/2} + \frac{3}{4}x^{4/3} + C \). To verify:
- Differentiate \( \frac{2}{3}x^{3/2} \). The result is \( \frac{2}{3} \times \frac{3}{2}x^{1/2} = x^{1/2} \).
- Differentiate \( \frac{3}{4}x^{4/3} \). This results in \( \frac{3}{4} \times \frac{4}{3}x^{1/3} = x^{1/3} \).
Constant of Integration
When dealing with indefinite integrals, the constant of integration \( C \) plays a pivotal role. This constant arises because the process of differentiation does not give us the information about any constants that were present in the original function.
Without \( C \), we would only have one particular solution from the infinite set of possible solutions. For example, if \( F(x) \) is any antiderivative of \( f(x) \), then \( F(x) + C \) is also an antiderivative for any constant value of \( C \). Each value of \( C \) gives a different function, illustrating the infinite nature of solutions to an indefinite integral.
Without \( C \), we would only have one particular solution from the infinite set of possible solutions. For example, if \( F(x) \) is any antiderivative of \( f(x) \), then \( F(x) + C \) is also an antiderivative for any constant value of \( C \). Each value of \( C \) gives a different function, illustrating the infinite nature of solutions to an indefinite integral.
- The presence of \( C \) reflects all possible vertical shifts of the antiderivative.
- When differentiating, the constant disappears, which is why starting integration from a known value or initial condition can determine \( C \).
Other exercises in this chapter
Problem 26
In Exercises \(15-30\) , find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on th
View solution Problem 27
Curves that are nearly flat at the root Some curves are so flat that, in practice, Newton's method stops too from the root to give a useful estimate. Try Newton
View solution Problem 27
L'Hopital's Rule does not help with the limits in Exercises \(27-30 .\) Try it; you just keep on cycling. Find the limits some other way. $$ \lim _{x \rightarro
View solution Problem 27
Constructing cones A right triangle whose hypotenuse is \(\sqrt{3} \mathrm{m}\) long is revolved about one of its legs to generate a right circular cone. Find t
View solution