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91Ó°ÊÓ

Choose the correct formula: a. \(\int u d v=u v+\int v d u \quad\) or b. \(\int u d v=u v-\int v d u\)

Short Answer

Expert verified
The correct formula is b: \( \int u \, dv = uv - \int v \, du \).

Step by step solution

01

Identify the Context

The problem is asking us to identify the correct formula for integration by parts, which is a strategy used to integrate products of functions.
02

Recall the Integration by Parts Formula

The integration by parts formula is derived from the product rule for differentiation and is stated as: \( \int u \, dv = uv - \int v \, du \).
03

Compare Given Options

Option a states: \( \int u \, dv = uv + \int v \, du \). Option b states: \( \int u \, dv = uv - \int v \, du \).
04

Select the Correct Formula

The correct formula for integration by parts, considering the minus sign derived from the product rule, is Option b: \( \int u \, dv = uv - \int v \, du \).

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Key Concepts

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

Calculus Integration
Calculus integration is a fundamental concept in calculus dealing with finding the antiderivative or integral of a function. This process is essential because it helps in calculating areas under curves and solving problems related to accumulation. When integrating simple functions, we often use basic integration rules, such as power rules or trigonometric integrals. However, when dealing with more complex functions, particularly products of different functions, simple rules are not sufficient. Here is where advanced techniques like integration by parts, partial fraction decomposition, or substitution come into play.

Integration by parts itself is a powerful technique derived from the product rule of differentiation, specifically tailored to handle the integration of products of two functions efficiently. Understanding this concept thoroughly allows students to tackle a wide range of integration problems with confidence.

For practical applications, integration is used in physics for computing quantities like displacement or areas, and in economics for consumer and producer surplus calculations. Thus, mastering integration processes not only supports academic success but also provides tools for real-world problem solving.
Product Rule for Differentiation
The product rule for differentiation is a critical rule in calculus, used to differentiate functions that are products of two or more functions. Expressed mathematically, if you have two functions, say \( u(x) \) and \( v(x) \), their derivative according to the product rule is stated as:

\[ (uv)' = u'v + uv' \]

This formula ensures that when we take the derivative of a product, we consider both changes in each function and how they contribute to the change in the whole product. The product rule forms the basis for understanding the integration by parts technique because integration by parts can be viewed as an inverse application of the product rule.

Furthermore, the product rule underscores the importance of understanding how functions interact and change relative to each other. It is a foundational tool used across many fields of study. When applied meticulously, it allows for the differentiation of complex expressions encountered in fields as varied as physics, engineering, and economics.
Integration Techniques
In calculus, integration techniques are essential tools that provide various methods for finding antiderivatives. Each technique is tailored for different types of functions, allowing us to integrate more complex or compound functions that simple antiderivative rules can't handle.

Some common integration techniques include:
  • **Integration by Substitution**: This involves changing variables to make an integral easier to solve, and is especially useful when dealing with composite functions.
  • **Integration by Parts**: As outlined earlier, this technique is ideal for integrating products of functions. Using the formula \( \int u \, dv = uv - \int v \, du \), it simplifies integrals that involve polynomial and exponential, logarithmic, or trigonometric functions.
  • **Partial Fraction Decomposition**: It is used when a rational function is to be integrated. By expressing the function as a sum of simpler fractions, integration becomes more straightforward.
By learning these techniques, you become equipped with a mathematics toolkit necessary for solving a broad array of integral problems. Mastery of these methods enhances understanding and provides the groundwork for more advanced studies or applications in real-life scenarios.

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Most popular questions from this chapter

$$ \text { Use integration by parts to find each integral. } $$ $$ \int \sqrt[3]{x} \ln x d x $$

Integration by parts often involves finding integrals like the following when integrating \(d v\) to find \(v\). Find the following integrals without using integration by parts (using formulas 1 through 7 on the inside back cover). Be ready to find similar integrals during the integration by parts procedure. $$ \int(x-1) d x $$

Fick's Law governs the diffusion of a solute across a cell membrane. According to Fick's Law, the concentration \(y(t)\) of the solute inside the cell at time \(t\) satisfies \(\frac{d y}{d t}=\frac{k A}{V}\left(C_{0}-y\right)\), where \(k\) is the diffusion constant, \(A\) is the area of the cell membrane, \(V\) is the volume of the cell, and \(C_{0}\) is the concentration outside the cell. a. Find the general solution of this differential equation. (Your solution will involve the constants \(k, A, V\), and \(\left.C_{0} .\right)\) b. Find the particular solution that satisfies the initial condition \(y(0)=y_{0}\), where \(y_{0}\) is the initial concentration inside the cell.

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 $$ R=\frac{p_{0}}{I_{0}} \frac{1-e^{-K T / R}}{1-e^{-K t_{0} / R}} $$

$$ \text { Use integration by parts to find each integral. } $$ $$ \int(x-1) e^{x} d x \quad[\text { Hint: Take } u=x-1 .] $$

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