Problem 10 - Thomas Calculus

Question

In Exercises \(1-16,\) find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(\frac{1}{2} x^{-1 / 2} \quad\) b. \(-\frac{1}{2} x^{-3 / 2} \quad\) c. \(-\frac{3}{2} x^{-5 / 2}\)

Step-by-Step Solution

Answer

The antiderivatives are: a) \(x^{1/2} + C\); b) \(x^{-1/2} + C\); c) \(x^{-3/2} + C\).

Step 1: Understanding the Power Rule for Antiderivatives

The power rule for antiderivatives states that for any function of the form \( ax^n \), the antiderivative is \( \frac{a}{n+1}x^{n+1} + C \) where \( C \) is an arbitrary constant. This applies for any \( n eq -1 \). We will apply this rule to each part of the exercise.

Step 2: Calculate the Antiderivative for \( \frac{1}{2} x^{-1/2} \)

Using the power rule, we add 1 to the exponent \( -\frac{1}{2} + 1 = \frac{1}{2} \) and divide the coefficient by the new exponent: \( \frac{1/2}{1/2} x^{1/2} + C = x^{1/2} + C \).

Step 3: Calculate the Antiderivative for \( -\frac{1}{2} x^{-3/2} \)

We add 1 to the exponent \( -\frac{3}{2} + 1 = -\frac{1}{2} \) and divide the coefficient by the new exponent: \( \frac{-1/2}{-1/2} x^{-1/2} + C = x^{-1/2} + C \).

Step 4: Calculate the Antiderivative for \( -\frac{3}{2} x^{-5/2} \)

Add 1 to the exponent: \( -\frac{5}{2} + 1 = -\frac{3}{2} \), and divide the coefficient by the new exponent: \( \frac{-3/2}{-3/2} x^{-3/2} + C = x^{-3/2} + C \).

Step 5: Verify by Differentiation

Differentiate each calculated antiderivative to verify correctness.- For \( x^{1/2} + C \), differentiating yields \( \frac{1}{2} x^{-1/2} \), which matches the original function.- For \( x^{-1/2} + C \), differentiating yields \( -\frac{1}{2} x^{-3/2} \), which matches the original function.- For \( x^{-3/2} + C \), differentiating yields \( -\frac{3}{2} x^{-5/2} \), which matches the original function.

Key Concepts

Power Rule

The power rule is an essential technique for finding antiderivatives, especially when dealing with polynomial functions of the form \( ax^n \). It helps us determine the antiderivative by using a simple process:

• Increase the exponent \( n \) by 1, resulting in \( n+1 \).
• Divide the coefficient \( a \) by the new exponent \( n+1 \).

This gives us the antiderivative \( \frac{a}{n+1}x^{n+1} + C \), where \( C \) represents an arbitrary constant. Remember, this rule can't be used if the exponent \( n = -1 \), as it leads to a logarithm instead of a power form. This method is particularly useful because of its simplicity, allowing quick mental computation for common functions.

Differentiation

Differentiation is a process that finds the derivative, which is essentially the rate at which a function changes. It works as a dual to integration, or finding antiderivatives, as it can check our antiderivative work. If applied correctly, differentiating the antiderivative should yield back the original function:

• The rule used is, if \( F(x) \) is an antiderivative of \( f(x) \), then the derivative \( F'(x) = f(x) \).
• Checking via differentiation confirms whether the derived antiderivative is correct and aligns with the original function.

This rigorous method ensures that our antiderivative calculations are accurate. By verifying the provided solution, any discrepancy or mistake in integration can be identified and corrected.

Exponents

Exponents provide a way to express repeated multiplication of a base number. In antiderivatives, exponents are crucial as they guide how we apply the power rule. Some key points to remember include:

• When adding 1 to an exponent during integration, it's crucial to adjust both the exponent itself and the coefficient correctly.
• Negative exponents indicate reciprocals. For example, \( x^{-n} = \frac{1}{x^n} \).

Proper manipulation of exponents is central to applying antiderivative rules accurately, as they determine the shape and characteristics of the resulting function.

Arbitrary Constant

In integration, the arbitrary constant, often represented as \( C \), is a crucial component. It's important to include \( C \) because:

• The process of differentiation eliminates constants; hence adding \( C \) accounts for any constant value that existed before.
• For any given function, there are infinitely many antiderivatives, each differing by just a constant, due to \( C \).

By representing the family of all possible solutions with this constant, you ensure completeness in your antiderivative result. This principle reflects the general nature of integration, bridging all potential unique functions back to their derivatives seamlessly.