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Chapter 4: Problem 65

Right, or wrong? Say which for each formula and give a brief reason for each answer. \(\begin{array}{l}\text { a. }{\int(2 x+1)^{2} d x=\frac{(2 x+1)^{3}}{3}+C} \\\ \text { b. }{\int 3(2 x+1)^{2} d x=(2 x+1)^{3}+C} \\ {\text { c. } \int 6(2 x+1)^{2} d x=(2 x+1)^{3}+C}\end{array}\)

Short Answer

Expert verified
Formula a is incorrect; b is incorrect; c is correct.

Step by step solution

01

Analyzing Formula a

The integral given is \( \int (2x+1)^2 \, dx \). To integrate this, we need to use substitution because it is in the form of \((ax+b)^n\). Let \( u = 2x+1 \), then \( du = 2 \, dx \), thus \( dx = \frac{du}{2} \). The integral becomes \( \int \left(u^2 \cdot \frac{1}{2}\right) \, du = \frac{1}{2} \int u^2 \, du \). Integrating, we get \( \frac{1}{2} \cdot \frac{u^3}{3} + C = \frac{(2x+1)^3}{6} + C \), not \( \frac{(2x+1)^3}{3} + C \). Therefore, Formula a is incorrect.
02

Analyzing Formula b

For Formula b, \( \int 3(2x+1)^2 \, dx \), we again use substitution: \( u = 2x+1 \), \( du = 2 \, dx \), then \( dx = \frac{du}{2} \). The integral becomes \( \int 3 \cdot u^2 \cdot \frac{1}{2} \, du = \frac{3}{2} \int u^2 \, du \). Integrating, \( \frac{3}{2} \cdot \frac{u^3}{3} + C = \frac{(2x+1)^3}{2} + C \). Thus, Formula b is incorrect because it claims the result is \( (2x+1)^3 + C \).
03

Analyzing Formula c

In Formula c, \( \int 6(2x+1)^2 \, dx \), let \( u = 2x+1 \) and \( du = 2 \, dx \), so \( dx = \frac{du}{2} \). The integral becomes \( \int 6 \cdot u^2 \cdot \frac{1}{2} \, du = 3 \int u^2 \, du \). Integrating yields \( 3 \cdot \frac{u^3}{3} + C = (2x+1)^3 + C \). This matches the result given, so Formula c is correct.

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Key Concepts

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

Substitution Method
The substitution method is a powerful tool in calculus integration. It is used when you have an integral that can be simplified by changing the variable.
This works well when the integrand (the function being integrated) contains a composite function, like \((ax+b)^n\).
Here's how it works:
  • Identify a part of the integrand that can be substituted with a single variable, usually denoted as \(u\).
  • Write down your substitution, for instance, let \(u = 2x + 1\). This makes the inside of our function simpler.
  • Find the derivative of your substitution with respect to \(x\), which we call \(du\). For our example, \(du = 2\,dx\), which implies \(dx = \frac{du}{2}\).
  • Rewrite the integral in terms of \(u\) and \(du\) and integrate. For instance, the integral \(\int (2x+1)^2 \, dx\) turns into \(\frac{1}{2} \int u^2 \, du\).
  • Finally, substitute back the original variable to express the answer in terms of \(x\).
Substitution simplifies the problem, often turning a difficult integral into something easier to manage.
Integration Techniques
Integration techniques are various methods applied to solve integrals. There are numerous techniques, but some of the most common include:
  • Substitution Method: Covered in detail above.
  • Integration by Parts: Useful for products of different types of functions. Based on the product rule for differentiation.
  • Partial Fraction Decomposition: Used when integrating rational functions by breaking them down into simpler parts.
  • Trigonometric Substitution: Suitable for integrals involving square roots of quadratic expressions. Each technique has its specific applications and advantages, depending on the form of the integral you are working with. Familiarity with these techniques allows for versatility and efficiency in solving various integral problems.
Definite and Indefinite Integrals
Understanding the difference between definite and indefinite integrals is crucial in calculus.

**Indefinite Integrals**:
  • Represent the antiderivative of a function, which means reversing the process of differentiation.
  • Expressed with the integration symbol and usually contain a constant of integration \(C\), such as \(\int f(x) \, dx = F(x) + C\).
  • The constant \(C\) accounts for all possible antiderivatives of a given function, since differentiation of such constants results in zero.
**Definite Integrals**:
  • Involve computing the area under a curve over a specific interval \([a, b]\).
  • Expressed as \(\int_a^b f(x) \, dx\) and result in a number representing the net area.
  • No constant \(C\) appears because the limits \(a\) and \(b\) define a specific area, eliminating ambiguity.
These integrals not only deal with theoretical solutions but also have practical applications, such as calculating areas, volumes, and other quantities in physics and engineering.

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Most popular questions from this chapter

The hammer and the feather When Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonstrate that in a vacuum all bodies fall with the same (constant) acceleration, he dropped them from about 4 ft above the ground. The television footage of the event shows the hammer and the feather falling more slowly than on Earth, where, in a vacuum, they would have taken only half a second to fall the 4 ft. How long did it take the hammer and feather to fall 4 ft on the moon? To find out, solve the following initial value problem for \(s\) as a function of \(t\) . Then find the value of \(t\) that makes \(s\) equal to 0 . \begin{equation} \begin{array}{ll}{\text { Differential equation: }} & {\frac{d^{2} s}{d t^{2}}=-5.2 \mathrm{ft} / \mathrm{sec}^{2}} \\ {\text { Initial conditions: }} & {\frac{d s}{d t}=0 \text { and } s=4 \text { when } t=0}\end{array} \end{equation}

Motion with constant acceleration The standard equation for the position \(s\) of a body moving with a constant acceleration \(a\) along a coordinate line is \begin{equation} s=\frac{a}{2} t^{2}+v_{0} t+s_{0} \quad\quad\quad\quad\quad\quad\text {(1)} \end{equation} where \(v_{0}\) and \(s_{0}\) are the body's velocity and position at time \(t=0 .\) Derive this equation by solving the initial value problem \begin{equation} \begin{array}{ll}{\text { Differential equation: }} & {\frac{d^{2} s}{d t^{2}}=a} \\ {\text { Initial conditions: }} & {\frac{d s}{d t}=v_{0} \text { and } s=s_{0} \text { when } t=0}\end{array} \end{equation}

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