Problem 327 - Calculus Volume 1

Question

In the following exercises, compute each indefinite integral. $$\int \frac{1}{\sqrt{x}} d x$$

Step-by-Step Solution

Answer

The indefinite integral is \( 2\sqrt{x} + C \).

Step 1: Identify the Integral

The given integral is \( \int \frac{1}{\sqrt{x}} \, dx \). Here, the integrand is \( \frac{1}{\sqrt{x}} \), which can be rewritten as \( x^{-1/2} \).

Step 2: Determine the Antiderivative

We use the power rule for integration which states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \). In our case, \( n = -1/2 \).

Step 3: Apply the Power Rule

Using the power rule, the integral becomes \( \int x^{-1/2} \, dx = \frac{x^{-1/2+1}}{-1/2+1} + C = \frac{x^{1/2}}{1/2} + C \).

Step 4: Simplify the Expression

The expression \( \frac{x^{1/2}}{1/2} \) can be simplified by multiplying by the reciprocal: \( 2x^{1/2} \). Thus, the integral becomes \( 2x^{1/2} + C \).

Step 5: Rewrite in Terms of Square Roots

Finally, notice that \( x^{1/2} \) is equivalent to \( \sqrt{x} \). Therefore, our solution is \( 2\sqrt{x} + C \).

Key Concepts

Power Rule for Integration

The power rule for integration is a fundamental tool in calculus and helps us find the antiderivatives of polynomial expressions. If you have an integrand in the form of \( x^n \), you can apply the power rule, which states:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

Here, \( n \) represents any real number except \( -1 \).
This rule simplifies the process by adding one to the exponent and dividing by the new exponent. The constant \( C \) represents the constant of integration, which is crucial because it accounts for any constant term that might have been in the original function before differentiation.
In our specific problem, the integrand \( \frac{1}{\sqrt{x}} \) can be rewritten as \( x^{-1/2} \), allowing us to directly utilize the power rule for an easy solution.

Antiderivative

An antiderivative is essentially the reverse process of differentiation.

• While differentiation provides the rate of change of a function, the antiderivative retraces the steps back to the original function.

In other words, finding an antiderivative restores a function from its derivative form.
For instance, in our exercise with the integral \( \int \frac{1}{\sqrt{x}} \, dx \), identifying \( \frac{1}{\sqrt{x}} \) as \( x^{-1/2} \) helps us find the antiderivative using integration rules.
Our resulting antiderivative \( 2x^{1/2} + C \) (or \( 2\sqrt{x} + C \)) represents the family of original functions from which \( \frac{1}{\sqrt{x}} \) could have been derived.

Simplifying Expressions

Simplifying expressions is an important step in both calculus and algebra, allowing us to work with more manageable forms of complex equations. Once we find the antiderivative using the power rule, we often need to simplify the expression further for clarity or application.
In the given exercise, we simplified \( \frac{x^{1/2}}{1/2} \) by recognizing that dividing by \( 1/2 \) is the same as multiplying by 2:

• \( \frac{x^{1/2}}{1/2} = 2x^{1/2} \)

This simplification makes it easier to interpret the expression, especially when applying it to real-world scenarios or gaining insights from graphical representations. Always look for opportunities to simplify your expressions to make further calculations less prone to errors.

Calculus Step-by-Step Solution

Solving calculus problems step by step is key to understanding not just the answer, but the reasoning behind it.
Let's review our exercise from start to finish:

• First, recognize the integrand \( \frac{1}{\sqrt{x}} \) and rewrite it as \( x^{-1/2} \).
• Next, apply the power rule to find the antiderivative. This gives us \( \int x^{-1/2} \, dx = \frac{x^{1/2}}{1/2} + C \).
• Then, simplify \( \frac{x^{1/2}}{1/2} \) to \( 2x^{1/2} \).
• Finally, rewrite \( x^{1/2} \) as \( \sqrt{x} \), resulting in \( 2\sqrt{x} + C \).

This methodical approach ensures we understand each transformation and adjustment in the steps and how they contribute to the final solution. Practicing such step-by-step solutions helps build confidence and accuracy in solving more complex calculus problems.