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Chapter 4: Problem 20

Finding antiderivatives Find all the antiderivatives of the following functions. Check your work by taking derivatives. $$h(y)=y^{-1}$$

Short Answer

Expert verified
Question: Find the antiderivative of the function h(y) = y^(-1). Solution: The antiderivative of the function h(y) = y^(-1) is H(y) = ln|y| + C.

Step by step solution

01

Integrating h(y) to find the antiderivative

To find the antiderivative of h(y), we need to integrate the function with respect to y: $$\int h(y) dy$$ We know that: $$h(y) = y^{-1}$$ So, substitute it into the integral: $$\int y^{-1} dy$$
02

Calculate the integral

Now, we will calculate the integral of the given function: $$\int y^{-1} dy$$ Recall the integral formula for power functions: $$\int y^n dy = \frac{y^{n+1}}{n+1} + C$$ But for n= -1, we use the following formula: $$\int y^{-1} dy = \ln{|y|} + C$$ Now, we apply this formula: $$\int y^{-1} dy = \ln{|y|} + C$$
03

Taking the derivative to check the result

Next, we will check if this antiderivative is correct or not: $$h'(y) = \frac{d(\ln{|y|} + C)}{dy},$$ where h'(y) is the derivative of the antiderivative of h(y). Since the derivative of a constant term C is 0, we have: $$h'(y) = \frac{d(\ln{|y|}))}{dy} = \frac{1}{y}$$ As we can see, h'(y) = h(y) = y^(-1), which means that our solution is correct. So, the antiderivative of h(y) is: $$H(y) = \ln{|y|} + C$$

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Key Concepts

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

Integration
Integration is the process of finding the antiderivative or integral of a function. It's essentially the reverse operation of differentiation. When we integrate a function, we're looking for another function whose derivative is the given function. This new function is known as the antiderivative. In the exercise, we started by identifying the function \( h(y) = y^{-1} \). We then set up the integral as \( \int h(y) \, dy = \int y^{-1} \, dy \). This integral represents the process of determining how much the function accumulates as \( y \) changes. Each function has infinitely many antiderivatives, differing by a constant \( C \) because the derivative of a constant is zero.
Derivatives
Derivatives measure how a function changes as its input changes. It's a fundamental concept in calculus, capturing the rate of change of quantities. In our exercise, the derivative was used to verify the antiderivative we found. We took the derivative of \( \ln{|y|} + C \) and checked if it returned the original function \( y^{-1} \). Using derivative rules, we computed \( \frac{d}{dy}(\ln{|y|}) = \frac{1}{y} \), confirming that our antiderivative was correct. Therefore, derivatives not only tell us how functions behave but also validate our integration results.
Natural Logarithm
The natural logarithm, represented as \( \ln \), is the inverse operation of the exponential function with base \( e \). It plays a vital role in calculus, especially when integrating functions like \( y^{-1} \). In the solution, recognizing that \( \int y^{-1} \, dy = \ln{|y|} + C \) is essential because it's a special case. The formula used here doesn't fit the typical power rule, showcasing the unique nature of logarithmic functions in calculus. Understanding logarithms helps unravel complex mathematical relationships and solve problems involving exponential growth and decay.
Integral Formulas
Integral formulas are essential tools in calculus that provide solutions to integrals we often encounter. They simplify the process of finding antiderivatives. In this exercise, we used a specific integral formula: \( \int y^{-1} \, dy = \ln{|y|} + C \). This is a pivotal example because it doesn't follow the usual power rule \( \int y^n \, dy = \frac{y^{n+1}}{n+1} + C \) when \( n = -1 \). Knowing when to apply this special formula allows us to compute integrals effectively. Integral formulas are like shortcuts, enabling us to solve integration problems swiftly and correctly.

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Most popular questions from this chapter

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