Problem 58 - Thomas Calculus

Question

Evaluate the integrals. \(\int x^{\sqrt{2}-1} d x\)

Step-by-Step Solution

Answer

\( \frac{x^{\sqrt{2}}}{\sqrt{2}} + C \)

Step 1: Identify the Integral Form

The integral given is \( \int x^{\sqrt{2}-1} \, dx \). This is a simple power rule integration where the integrand is in the form \( x^n \).

Step 2: Apply the Power Rule for Integration

For the integral \( \int x^n \, dx \), the antiderivative is \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. Here, \( n = \sqrt{2} - 1 \).

Step 3: Calculate \( n+1 \)

To use the power rule, calculate \( n+1 = (\sqrt{2} - 1) + 1 = \sqrt{2} \).

Step 4: Integrate the Function

Apply the power rule to find the integral: \( \int x^{\sqrt{2}-1} \, dx = \frac{x^{\sqrt{2}}}{\sqrt{2}} + C \).

Step 5: Write the Final Solution

The evaluated integral is \( \frac{x^{\sqrt{2}}}{\sqrt{2}} + C \).

Key Concepts

Understanding Antiderivative

An antiderivative, also known as an integral, is a function that reverses differentiation. In simpler terms, finding an antiderivative of a function is equivalent to finding what the original function was before it was differentiated. In calculus, when you take the derivative of a function and then find an antiderivative, you essentially retrieve a family of functions that originally could have given you that derivative. This process is crucial for solving various types of problems in calculus, particularly when working with integrals. Here is an easy example:

• If you know that the derivative of the function \( x^2 \) is \( 2x \), then the antiderivative of \( 2x \) is \( x^2 \), with an added constant of integration.

This constant is important because differentiation removes it, hence it needs inclusion to cover all potential original forms.

The Role of Constant of Integration

The concept of the constant of integration, often represented by \( C \), is crucial in the process of finding an antiderivative. When integrating a function, the result includes an arbitrary constant because there are infinitely many functions whose derivatives are the same. Consider this essential aspect:

• Whenever you find an indefinite integral, the set of all possible antiderivatives is signified by adding \( C \). This is because differentiation does not retain constant values, thus to describe the general form, we include \( C \).

In our example, after applying the power rule, we found the result to be \( \frac{x^{\sqrt{2}}}{\sqrt{2}} + C \). The \( C \) signifies that there could be an infinite set of solutions shifted vertically on the graph.

Integrand Basics

The term "integrand" refers to the function being integrated in an integral. In our problem, the integrand is \( x^{\sqrt{2}-1} \). Understanding the form of the integrand is key to choosing the correct integration technique. When integrating with the power rule, the integrand typically takes the form \( x^n \). The given problem's integrand fits perfectly into this pattern, making it straightforward to apply the power rule. To integrate effectively:

• Identify \( n \) in your integrand \( x^n \). Here, \( n = \sqrt{2} - 1 \).

With this information, you can accurately apply the technique to solve for the antiderivative.

Mastering Calculus Problem Solving

Solving calculus problems requires a solid grasp of various rules and concepts, such as the power rule. Here's how the power rule helped us solve this particular problem:

• The power rule for integration indicates that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
• In applying this, we calculated \( n+1 \) for our specific function, resulting in \( n+1 = \sqrt{2} \).

By establishing a step-by-step approach, starting from identifying the type of integral to applying the rule consistently, anyone can tackle similar problems.Keep these problem-solving tips in mind:

• Always simplify the integrand if possible before integrating.
• Check your results by differentiating your result to see if it matches the original integrand.

With practice, integrating functions becomes a straightforward exercise in applying these learned techniques.