Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{1}{2 y} d y$$

Short answer

Question: Find the indefinite integral of the function \(f(y) = \frac{1}{2y}\) and verify the result by differentiation. Answer: The indefinite integral of the given function is \(\int \frac{1}{2y} dy = \frac{1}{2} \ln |y| + C\).

Step-by-step solution

Identify the function to integrate

Here, the given function is \(f(y) = \frac{1}{2y}\). Our goal is to compute the indefinite integral of this function, which is represented as: $$\int f(y) dy$$

Integrate the function

To find the indefinite integral of the function, we will apply the power rule for integration. The power rule states that for any constant \(n\), and a function \(y^n\), the integral is: $$\int y^n dy = \frac{1}{n+1} y^{n+1} + C$$ In our case, we need to rewrite the function as a power of \(y\) to apply the power rule: $$f(y) = \frac{1}{2y} = \frac{1}{2} y^{-1}$$ Now, we apply the power rule: $$\begin{aligned} \int f(y) dy &= \int \frac{1}{2} y^{-1} dy \\ &= \frac{1}{2} \int y^{-1} dy \\ &= \frac{1}{2} \cdot \frac{1}{-1 + 1} y^{-1 + 1} + C \\ \end{aligned}$$

Simplify the result

Now, we simplify the resulting expression: $$\begin{aligned} \int f(y) dy &= \frac{1}{2} \cdot \frac{1}{0} y^0 + C \\ &= \frac{1}{2} \cdot 1 + C \\ &= \frac{1}{2} + C \\ \end{aligned}$$ So, the indefinite integral of the given function is: $$\int \frac{1}{2y} dy = \frac{1}{2} + C$$

Check the result by differentiation

Now, we will differentiate the result to verify our answer. Specifically, we want to compute the derivative of our indefinite integral with respect to \(y\): $$\frac{d}{dy} \left( \frac{1}{2} + C \right)$$ Since both \(\frac{1}{2}\) and \(C\) are constants, their derivatives with respect to \(y\) are zero: $$\frac{d}{dy} \left( \frac{1}{2} + C \right) = 0 + 0 = 0$$ However, this result is incorrect, as the derivative should be equal to the original function \(\frac{1}{2y}\). This implies that there was an error in our integration.

Correct the integration error

We realize that when applying the power rule in step 2, we didn't actually have a valid power (the rule is invalid for \(n=-1\)). To correctly integrate this function, we should recognize that the integral of \(\frac{1}{y}\) is a natural logarithm: $$\int y^{-1} dy = \ln |y| + C$$ Then, taking the constant \(\frac{1}{2}\) into account, the correct indefinite integral is: $$\int \frac{1}{2} y^{-1} dy = \frac{1}{2} \ln |y| + C$$

Check the corrected result by differentiation

Now, we will differentiate the corrected result to verify our answer. Specifically, we want to compute the derivative of our indefinite integral with respect to \(y\): $$\frac{d}{dy} \left( \frac{1}{2} \ln |y| + C \right)$$ Using the chain rule and noticing that the derivative of \(C\) is zero as before, we get: $$\frac{d}{dy} \left( \frac{1}{2} \ln |y| + C \right) = \frac{1}{2} \cdot \frac{1}{y} = \frac{1}{2y}$$ This confirms that our corrected indefinite integral is correct. Hence, the correct answer is: $$\int \frac{1}{2y} dy = \frac{1}{2} \ln |y| + C$$

Key concepts

Integration Techniques

Integration techniques refer to various methods used to find integrals, both definite and indefinite. When faced with an integral, choosing the right technique can simplify the process and provide accurate results.

Here are some common integration techniques:

• Power Rule: Used for integrating functions of the form \( y^n \), where the integral is \( \int y^n \, dy = \frac{1}{n+1} y^{n+1} + C \) for \( n eq -1 \).
• U-Substitution: Useful when dealing with composite functions, allowing simplification by substituting a part of the function.
• By Parts: Applied to integrals of products of functions and relies on differentiation along with integration.

Understanding these techniques helps tackle various integrals by simplifying complex expressions or recognizing familiar forms. For example, in our exercise, the recognition of a logarithmic function is crucial for correctly solving the integral \( \int \frac{1}{2y} \ dy \).

Logarithmic Integration

Logarithmic integration occurs when integrating functions that involve the reciprocal of a variable, such as \( \frac{1}{y} \). Unlike other powers, the power rule does not apply to \( y^{-1} \). Instead, the integral is directly related to the natural logarithm.

Understanding Natural Logarithm in Integration

When you integrate \( \int \frac{1}{y} \, dy \), the result is \( \ln |y| + C \). This stems from the definition of the natural log function and its differentiation, where \( \frac{d}{dy}(\ln |y|) = \frac{1}{y} \). Such recognition was key in correcting the error in our exercise.

The presence of a constant coefficient, such as \( \frac{1}{2} \) in the exercise, means that it should be factored out and multiplied by the logarithmic result. Thus, \( \int \frac{1}{2y} \, dy \) results in \( \frac{1}{2} \ln |y| + C \). Recognizing this pattern is vital in integrating rational functions correctly.

Differentiation

Differentiation is the process of finding the derivative of a function, which calculates the rate of change or the slope at any point on the function's graph. In the context of integration, differentiation serves as a powerful tool to verify the results of an indefinite integral.

The Role of Differentiation in Verifying Integrals

After integrating a function, like in our exercise, you can differentiate the result to ensure its correctness. The derivative of an indefinite integral should repurchase the original integrand. For instance, differentiating \( \frac{1}{2} \ln |y| + C \) should yield \( \frac{1}{2y} \), confirming the integral is correct. This process reinforces the understanding that differentiation and integration are inverse operations.

Using differentiation as a check helps catch errors, as seen when the exercise mistakenly applied an incorrect rule initially. It prompts reconsideration of the problem, aiding in mastering both differentiation and integration techniques.