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In Exercises \(1-16,\) find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. \begin{equation} \text { a. } \frac{2}{3} x^{-1 / 3} \quad \text { b. } \frac{1}{3} x^{-2 / 3} \quad \text { c. }-\frac{1}{3} x^{-4 / 3} \end{equation}

Short Answer

Expert verified
Antiderivatives are: a) \(x^{2/3} + C\), b) \(x^{1/3} + C\), c) \(x^{-1/3} + C\).

Step by step solution

01

Understand the Task

We are asked to find the antiderivatives of three different functions. Remember that finding an antiderivative is essentially integrating the function.
02

Review the Power Rule of Integration

Recall the power rule for integration: \( \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \), where \(neq -1\), and \(C\) is the constant of integration.
03

Antiderivative of Function a

Given function: \( \frac{2}{3} x^{-1/3} \).Using the power rule: \[ \int \frac{2}{3} x^{-1/3} \, dx = \frac{2}{3} \times \frac{x^{(-1/3) + 1}}{(-1/3) + 1} + C = \frac{2}{3} \times \frac{x^{2/3}}{2/3} + C = x^{2/3} + C. \]
04

Antiderivative of Function b

Given function: \( \frac{1}{3} x^{-2/3} \).Using the power rule:\[ \int \frac{1}{3} x^{-2/3} \, dx = \frac{1}{3} \times \frac{x^{(-2/3) + 1}}{(-2/3) + 1} + C = \frac{1}{3} \times \frac{x^{1/3}}{1/3} + C = x^{1/3} + C. \]
05

Antiderivative of Function c

Given function: \(-\frac{1}{3} x^{-4/3} \).Using the power rule:\[ \int -\frac{1}{3} x^{-4/3} \, dx = -\frac{1}{3} \times \frac{x^{(-4/3) + 1}}{(-4/3) + 1} + C = -\frac{1}{3} \times \frac{x^{-1/3}}{-1/3} + C = x^{-1/3} + C. \]
06

Verification by Differentiation

Differentiate the antiderivatives found to ensure they match the original functions:- Differentiate \( x^{2/3} + C \): gives \( \frac{2}{3} x^{-1/3} \).- Differentiate \( x^{1/3} + C \): gives \( \frac{1}{3} x^{-2/3} \).- Differentiate \( x^{-1/3} + C \): gives \( -\frac{1}{3} x^{-4/3} \).Each matches the original function, confirming our answers.

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Key Concepts

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

Integration
Integration is a fundamental concept in calculus that allows us to find antiderivatives. It is essentially the process of finding a function whose derivative matches a given function. In simpler terms, integration is used to find the original function when only the rate of change is known.
Antiderivatives are found using integration, and they include a constant "+C" because there are infinitely many functions with the same derivative differing only by a constant. This constant exists because when you differentiate, the constant disappears, so it must be considered when integrating.
  • Antiderivative is another term for the integral of a function.
  • The definite integral calculates the area under a curve over a specific interval, while the indefinite integral determines the original function up to a constant.
The key to finding the antiderivative is using rules, such as the Power Rule, which guides how calculations are performed during integration.
Power Rule
The Power Rule is a powerful tool for both differentiation and integration. For integration, the Power Rule helps find the antiderivative of functions that are a power of \(x\).
The Power Rule for integration states that to integrate a function of the form \(x^n\), you add one to the exponent and then divide by the new exponent. Mathematically, it’s written as:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]Where \(n\) is not equal to \(-1\). This rule simplifies the process of integration significantly:
  • Add 1 to the exponent: \(x^n ightarrow x^{n+1}\).
  • Divide by the new exponent: \((n+1)\).
  • Add the constant "+C" at the end to account for any constant that was lost during differentiation.
This Power Rule is a shortcut that makes it easy to find antiderivatives without having to perform more complex calculations.
Differentiation
Differentiation is the process used to determine the derivative of a function, which represents the rate of change. It is the opposite operation of integration. While integration provides the original function from its derivative, differentiation takes you from the original function to the derivative.
In checking your work in problems involving integration, differentiation serves as a verification tool. After finding an antiderivative, you can differentiate it to ensure that it yields the original function.
  • To differentiate a power of \(x\), use the rule: bring down the exponent as a coefficient, then subtract one from the exponent.
  • For example, differentiating \(x^{n+1}\) results in \((n+1)x^n\).
This process supports the accuracy of integration and reassures that the antiderivatives calculated are correct. Differentiation and integration are two sides of the same coin, each essential for verifying and solving calculus problems effectively.

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Most popular questions from this chapter

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

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