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In Exercises 1 through 38 , find the antiderivatives. $$ \int \frac{1}{2 y^{5}} d y $$

Short Answer

Expert verified
The antiderivative is \(-\frac{1}{8} y^{-4} + C\).

Step by step solution

01

Identify the integral form

The given exercise asks us to find the antiderivative of \( \int \frac{1}{2 y^{5}} \ dy \). This can also be rewritten in a more workable form as \( \int \frac{1}{2} y^{-5} \ dy \).
02

Use the power rule for integration

The power rule for integrating functions says that \( \int y^n \ dy = \frac{y^{n+1}}{n+1} + C \), where \( n eq -1 \). Here, \( n = -5 \).
03

Apply the power rule

Using the power rule, integrate the function. We have \( n = -5 \), so adding 1 gives us \( n + 1 = -4 \). So, the integration of \( y^{-5} \) is \( \frac{y^{-4}}{-4} \), which simplifies to \( -\frac{1}{4} y^{-4} \).
04

Include the constant coefficient

We originally factored out a \( \frac{1}{2} \) from the integral. Therefore, multiply the result by \( \frac{1}{2} \), resulting in \( -\frac{1}{8} y^{-4} \).
05

Add the constant of integration

Finally, add the constant of integration \( C \) to the result, giving the final answer \( -\frac{1}{8} y^{-4} + C \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule for Integration
The power rule for integration is a fundamental concept when finding antiderivatives or indefinite integrals. It simplifies the process of integrating power functions. The basic form of the rule is: \[ \int y^n \ dy = \frac{y^{n+1}}{n+1} + C, \quad \text{where } n eq -1. \]To apply this rule, follow these steps:
  • Identify the exponent of the variable in the integrand (the function you’re integrating).
  • Add one to the exponent. This gives us the new exponent for the variable.
  • Divide the term by this new exponent, effectively multiplying by the reciprocal.
  • Don’t forget to add the constant of integration, denoted as \( C \).
Using the power rule, complex functions with variables raised to a power become manageable, transforming them into simpler expressions that include the constant of integration.
Constant of Integration
When finding the antiderivative of a function, it's important to include the constant of integration. This constant, typically denoted by \( C \), accounts for all the possible constants that could be added to a function without altering its derivative.Why is the constant of integration necessary? Consider that the derivative of a constant is zero, so when we find an antiderivative, we must reflect any constant that might have been part of the original function. That's why you’ll see this \( C \) at the end of indefinite integrals.For example, if you integrate \( f(y) = \frac{1}{2} y^{-5} \), you'll add \( C \) to indicate the most general form of the antiderivative, which could be \(-\frac{1}{8} y^{-4} + C\). This signifies an infinite number of functions all differing by a constant term.
Mathematical Integration Process
The mathematical integration process is the technique of finding antiderivatives. It is a systematic method of determining the original function that a given derivative would come from. In simple terms, integration is the reverse of differentiation.Here’s a step-by-step breakdown of the process:
  • **Rewriting the Function:** Adjust complex functions into easier-to-integrate forms. For the expression \( \int \frac{1}{2 y^{5}} \ dy \), rewrite it as \( \int \frac{1}{2} y^{-5} \ dy \) to simplify.
  • **Apply the Power Rule:** Use the rule to transform the expression, as in the exercise where \( y^{-5} \) was changed into \( \frac{y^{-4}}{-4} \).
  • **Include Constants:** Factor out constants, such as \( \frac{1}{2} \), and multiply them throughout the integration process. In our case, this involved multiplying the result by \( \frac{1}{2} \), leading to \( -\frac{1}{8} y^{-4} \).
  • **Add the Constant of Integration:** The final essential step is adding \( C \), making the final result \( -\frac{1}{8} y^{-4} + C \).
By following this process, you can reliably find antiderivatives for a wide range of functions, simplifying complex mathematical problems.

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