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Chapter 6: Problem 3

1-6. For each integral, state the number of the integration formula (from the inside back cover) and the values of the constants \(a\) and \(b\) so that the formula fits the integral. (Do not evaluate the integral.) $$ \int \frac{1}{x \sqrt{-x+7}} d x $$

Short Answer

Expert verified
Use a formula with \( a = -1 \) and \( b = 7 \).

Step by step solution

01

Identify Integral Type

The integral \( \int \frac{1}{x \sqrt{-x+7}} \, dx \) resembles the form involving a square root of a linear function. We can recognize it as associated with formulas that deal with square roots in the denominator.
02

Express Integral in Standard Form

The integral has the form \( \int \frac{1}{x \sqrt{-x+7}} \, dx \). Let's rearrange it: \( \int \frac{1}{x \sqrt{-1(x - 7)}} \, dx \). This suggests a transformation of the form \( x = 7 - u \) might be useful with the square root part \( \sqrt{-x+7} = \sqrt{7-x} \).
03

Set Constants a and b

From the last step, we see that the linear term inside the square root \( -x + 7 \) can be rewritten as \( \sqrt{-1(x - 7)} \), giving us \( a = -1 \) and \( b = 7 \). Therefore, formula involving \( \sqrt{a(x-b)} \) is used with our constants identified.
04

Match Formula to Integral

The formula that fits this integral involves expressing \( \sqrt{a(x - b)} \) in the form \( \sqrt{-1(x - 7)} \), meaning our integral belongs to the group where the quadratic term inside the square root is linearly substituted with the representation \( \sqrt{a(x - b)} \).

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Key Concepts

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

Integral Transformation
Integral transformation is a powerful technique used to simplify and solve complex integrals by changing the variables. This method helps us convert the integral into a more manageable form. In the given exercise, the integral transformation is applied by identifying that the given integral can be rewritten using a different variable substitution. For the integral \( \int \frac{1}{x \sqrt{-x+7}} \, dx \), we perform a transformation by rewriting it as \( \int \frac{1}{x \sqrt{-1(x - 7)}} \, dx \).
Substituting the linear function within the square root, we transform the integral to better fit a standard formula. The key idea is to use a transformation like \( x = 7 - u \), hence altering the bounds and simplifying the problem.
This process not only helps in recognizing the patterns in the integral but also identifies constants such as \( a \) and \( b \), which are crucial for applying specific integral formulae effectively.
Square Root in Integrals
Integrals containing square roots can often be tricky to deal with due to their non-linear nature. However, recognizing patterns can help solve them effectively. The given exercise involves the square root \( \sqrt{-x + 7} \), rewriting it as \( \sqrt{7-x} \).
This recognition fits the integral into a form related to standard formulas involving square roots. When you see a term like \( \sqrt{-x+7} \), you should think about transformations or substitutions that can make the integral easier to manage.
Often, rewriting the integral helps classify it under a known formula, allowing you to then apply well-established techniques to find solutions, even if you do not evaluate the integral entirely but identify its type and structure.
Constants in Integration Formulas
Understanding the role of constants in integration formulas is central to solving integrals effectively. Constants such as \( a \) and \( b \) are used to match an integral to a known formula, simplifying the solution process. In our problem, constants emerge from rewriting\( -x + 7 \) as \( \sqrt{-1(x - 7)} \) which leads to setting \( a = -1 \) and \( b = 7 \).
These constants reflect the linear taking place inside the square root and align the given integral with standard integrals you might find in a formula reference.
Identifying these constants properly allows us to utilize specific formulas tied to each integral structure, streamlining our problem-solving process without directly solving the integral.

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Most popular questions from this chapter

Each equation follows from the integration by parts formula by replacing \(u\) by \(f(x)\) and \(v\) by a particular function. What is the function \(v\) ? $$ \int f(x) \frac{1}{x} d x=f(x) \ln x-\int \ln x f^{\prime}(x) d x $$

Find in two different ways and check that your answers agree. \(\int x(x-2)^{5} d x\) a. Use integration by parts. b. Use the substitution \(u=x-2 \quad\) (so \(x\) is replaced by \(u+2\) ) and then multiply out the integrand.

In the reservoir model, the heart is viewed as a balloon that swells as it fills with blood (during a period called the systole), and then at time \(t_{0}\) it shuts a valve and contracts to force the blood out (the diastole). Let \(p(t)\) represent the pressure in the heart at time \(t\) a. During the diastole, which lasts from \(t_{0}\) to time \(T, p(t)\) satisfies the differential equation $$\frac{d p}{d t}=-\frac{K}{R} p$$ Find the general solution \(p(t)\) of this differential equation. ( \(K\) and \(R\) are positive constants determined, respectively, by the strength of the heart and the resistance of the arteries. The differential equation states that as the heart contracts, the pressure decreases \((d p / d t\) is negative) in proportion to itself.) b. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0} . \quad\left(p_{0}\right.\) is a constant representing the pressure at the transition time \(t_{0}\).) c. During the systole, which lasts from time 0 to time \(t_{0}\), the pressure \(p(t)\) satisfies the differential equation $$\frac{d p}{d t}=K I_{0}-\frac{K}{R} p$$ Find the general solution of this differential equation. \(\left(I_{0}\right.\) is a positive constant representing the constant rate of blood flow into the heart while it is expanding.) [Hint: Use the same \(u\) -substitution technique that was used in Example 7.] d. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0}\). e. In parts (b) and (d) you found the formulas for the pressure \(p(t)\) during the diastole \(\left(t_{0} \leq t \leq T\right)\) and the systole \(\left(0 \leq t \leq t_{0}\right)\). Since the heart behaves in a cyclic fashion, these functions must satisfy \(p(T)=p(0)\). Equate the solutions at these times (use the correct formula for each time) to derive the important relationship $$ R=\frac{p_{0}}{I_{0}} \frac{1-e^{-K T / R}}{1-e^{-K t_{0} / R}} $$

$$ \text { Use integration by parts to find each integral. } $$ $$ \int \frac{\ln t}{\sqrt{t}} d t $$

$$ \text { Use integration by parts to find each integral. } $$ $$ \int(x+b) e^{a x} d x \quad(a \neq 0) $$
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