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Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(\frac{1}{x}\) b. \(\frac{7}{x}\) c. \(1-\frac{5}{x}\)

Short Answer

Expert verified
a. \(\ln|x| + C\); b. \(7\ln|x| + C\); c. \(x - 5\ln|x| + C\).

Step by step solution

01

Understand Antiderivative Concept

An antiderivative of a function is a function whose derivative is the given function. For example, if the derivative of a function is \(f(x)\), any function whose derivative is \(f(x)\) is an antiderivative of \(f(x)\).
02

Antiderivative of \(\frac{1}{x}\)

The antiderivative of \(\frac{1}{x}\) is \(\ln|x| + C\), where \(C\) is the constant of integration. This is because the derivative of \(\ln|x|\) is \(\frac{1}{x}\).
03

Antiderivative of \(\frac{7}{x}\)

The antiderivative of \(\frac{7}{x}\) is \(7\ln|x| + C\). The constant multiple rule allows us to take the constant 7 outside the integral, and then we find that the antiderivative of \(\frac{1}{x}\) is used, scaling the result by 7.
04

Antiderivative of \(1 - \frac{5}{x}\)

To find the antiderivative, split the function into two parts and apply the rule separately: the antiderivative of 1 is \(x\) and the antiderivative of \(-\frac{5}{x}\) is \(-5\ln|x|\). Combining both results, the antiderivative of the function is \(x - 5\ln|x| + C\).
05

Verify by Differentiation

Differentiate each of your solutions to check if they are correct. For \(\ln|x| + C\), differentiate to get \(\frac{1}{x}\). For \(7\ln|x| + C\), differentiate and obtain \(\frac{7}{x}\). Lastly, the derivative of \(x - 5\ln|x| + C\) should result in \(1 - \frac{5}{x}\). If all match, your solutions are verified.

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Key Concepts

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

Understanding the Integration Constant
When we find an antiderivative, we are essentially reversing differentiation. This process is known as integration. One important aspect of integration is the integration constant, often denoted as \(C\). A function can have infinitely many antiderivatives because we can add any constant value to a particular antiderivative and still have the derivative of that entire expression equal to the original function. This \(C\) accounts for those constant differences.

For instance, if you consider the antiderivative of \( \frac{1}{x} \), it is \( \ln|x| + C \). Here, no specific value for \(C\) is essential because all these functions have the same derivative, \( \frac{1}{x} \).

The importance of the integration constant lies in ensuring that we encompass all possible functions in the family of antiderivatives. Without this constant, we would miss all the other potential solutions.
The Process of Differentiation
Differentiation is the process of finding the derivative of a function. It tells us how a function changes at any given point, essentially measuring the rate of change. When verifying our antiderivatives, differentiation plays a crucial role. After finding an antiderivative, we can differentiate it to see if we obtain the original function.

For example, if we differentiate \( \ln|x| + C \), we get \( \frac{1}{x} \). This confirms that our original function's derivative matches with the antiderivative we calculated. Similarly, differentiating \( 7\ln|x| + C \) should give \( \frac{7}{x} \), verifying correctness.

By using differentiation for verification, we ensure that our antiderivatives are indeed correct, thus reinforcing our understanding of the relationship between a function and its derivative.
Understanding the Constant Multiple Rule
The constant multiple rule is a simple yet powerful tool in calculus. It states that the derivative of a constant times a function is the constant times the derivative of the function. This rule is helpful when dealing with functions where constants are involved.

Consider the example of finding the antiderivative of \( \frac{7}{x} \). We can separate the constant 7 and focus solely on the function \( \frac{1}{x} \). By finding the antiderivative of \( \frac{1}{x} \) and then multiplying it by 7, we apply the constant multiple rule. Thus, the antiderivative is \(7\ln|x| + C\).

Understanding and applying the constant multiple rule allows us to handle more complex problems with ease, as it simplifies the process of integration by isolating constants from variables.

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Most popular questions from this chapter

Solve the initial value problems. $$\frac{d y}{d x}=9 x^{2}-4 x+5, \quad y(-1)=0$$

Solve the initial value problems. $$\frac{d v}{d t}=8 t+\csc ^{2} t, \quad v\left(\frac{\pi}{2}\right)=-7$$

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int \frac{\csc \theta}{\csc \theta-\sin \theta} d \theta$$

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a. When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2} \mathrm{cm} / \mathrm{sec}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ where \(r_{0}\) is the rest radius of the trachea in centimeters and \(c\) is a positive constant whose value depends in part on the length of the trachea. Show that \(v\) is greatest when \(r=(2 / 3) r_{0} ;\) that is, when the trachea is about \(33 \%\) contracted. The remarkable fact is that X-ray photographs confirm that the trachea contracts about this much during a cough. b. Take \(r_{0}\) to be 0.5 and \(c\) to be \(1,\) and graph \(v\) over the interval \(0 \leq r \leq 0.5 .\) Compare what you see with the claim that \(v\) is at a maximum when \(r=(2 / 3) r_{0}\).

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