/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 In Exercises \(35-38\) , find th... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercises \(35-38\) , find the function with the given derivative whose graph passes through the point \(P\) . $$f^{\prime}(x)=\frac{1}{x+2}, \quad x>-2, \quad P(-1,3)$$

Short Answer

Expert verified
The function with the given derivative that passes through the point \((-1,3)\) is \(f(x) = \ln(x+2) + 3\).

Step by step solution

01

Integral Calculation

We begin by finding the indefinite integral or antiderivative of the given derivative, \(f'(x) = \frac{1}{x + 2}\). This can be done by applying the rule of integrals. Integral of \(f'(x) = \frac{1}{x + 2}\) is \(\ln|x+2|\). But, as we're told that \(x > -2\), the absolute value isn't necessary, so the antiderivative function \(F(x)\) can be expressed as \(F(x) = \ln(x+2) + C\), where \(C\) is the constant of integration.
02

Calculation of Constant

We're given that the function passes through the point \((-1,3)\), which means when \(x = -1\), \(F(x) = 3\). We substitute these values into the function equation \(F(x) = \ln(x+2) + C\) to solve for C. Thus, we get \(3 = \ln((-1)+2) + C\), which simplifies to \(3 = \ln{1} + C\). Since \(\ln{1} = 0\), the constant \(C = 3\).
03

Formulate Complete Function

Now that we've determined that the constant of integration \(C = 3\), we can substitute this value back into our function equation to get the final function. Therefore, the function \(f(x)\) becomes \(f(x) = \ln(x+2) + 3\).

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Key Concepts

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

Constant of Integration
When we calculate an indefinite integral, it’s crucial to remember that there can be countless functions that differ only by a constant. This is where the constant of integration, often denoted as \( C \), comes into play.

In the context of functions, the constant of integration represents an entire family of functions that share the same derivative. It fills in the gap of having only part of the information by specifying a particular solution from this family. For example, when you integrate \( \frac{1}{x+2} \), you get \( \ln(x+2) + C \). Here, \( C \) could be any real number.

The constant becomes especially important when you have additional information or conditions such as a specific point that the function must pass through, as seen in problems like these. By using these conditions, we can solve for \( C \) and pinpoint the exact function from our family of functions.
Definite Integral
While this specific exercise focused on finding an indefinite integral, definite integrals play a different role in calculus. They are used to calculate the area under a curve between two bounds. Unlike indefinite integrals, definite integrals do not include a constant of integration since they result in a numerical value.

Definite integrals are written as \( \int_{a}^{b} f(x) \, dx \), which means finding the area under the curve of \( f(x) \) from \( x = a \) to \( x = b \). This process involves evaluating the antiderivative (indefinite integral) at these upper and lower bounds and subtracting the results.

Applications of definite integrals extend beyond finding areas; they can also determine volumes, calculate net changes, and even solve real-world problems that involve accumulation over an interval.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is a specific logarithm with a base of the mathematical constant \( e \) (approximately 2.71828). It is one of the most common functions you'll encounter in calculus, especially when dealing with integrals and derivatives.

The natural logarithm has key properties that make it particularly useful. For instance, \( \ln(1) = 0 \) and \( \ln(e) = 1 \). These properties can often simplify complex equations and are particularly handy when solving integration problems.

In the given problem, the use of the natural logarithm arises from integrating \( \frac{1}{x+2} \), resulting in \( \ln(x+2) \).
  • The derivative of \( \ln(x) \) is \( \frac{1}{x} \), which reverses the process of integration.
  • The inverse function of the natural logarithm is the exponential function, \( e^x \).
Understanding how to work with natural logarithms can significantly impact how efficiently you solve integrals and other calculus-related problems.

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Most popular questions from this chapter

Group Activity Cardiac Output In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Wurtzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 liters a minute. At rest it is likely to be a bit under 6 \(\mathrm{L} / \mathrm{min}\) . If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 \(\mathrm{L} / \mathrm{min.}\) Your cardiac output can be calculated with the formula $$$=\frac{Q}{D}$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{mL} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{mL} / \mathrm{min}\) and \(D=97-56=41 \mathrm{mL} / \mathrm{L}\) $$y=\frac{233 \mathrm{mL} / \mathrm{min}}{41 \mathrm{mL} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min}$$ fairly close to the 6 \(\mathrm{L} / \mathrm{min}\) that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D. Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

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