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Chapter 4: Problem 4

Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(2 x^{-3}\) b. \(\frac{x^{-3}}{2}+x^{2}\) c. \(-x^{-3}+x-1\)

Short Answer

Expert verified
a) \(-x^{-2} + C\) b) \(-\frac{1}{4}x^{-2} + \frac{x^3}{3} + C\) c) \(-\frac{x^{-2}}{2} + \frac{x^2}{2} - x + C\)

Step by step solution

01

Understand the Problem

We need to find an antiderivative for each given function. An antiderivative is a function whose derivative equals the given function. We will use basic integration rules to solve each part.
02

Solve Part (a)

The given function is \(2 x^{-3}\). We use the power rule for antiderivatives: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\). Here, \(n = -3\), so the antiderivative is \(2 \cdot \frac{x^{-3+1}}{-3+1} = 2 \cdot \frac{x^{-2}}{-2} = -x^{-2} + C\), where \(C\) is the constant of integration.
03

Confirm Part (a) by Differentiation

Differentiate \(-x^{-2} + C\). We find \(\frac{d}{dx}(-x^{-2}) = 2x^{-3}\), confirming that \(-x^{-2} + C\) is correct.
04

Solve Part (b)

The function is \(\frac{x^{-3}}{2} + x^2\). Integrate each term separately: \(\int \frac{x^{-3}}{2} \, dx = \frac{1}{2} \cdot \int x^{-3} \, dx = \frac{1}{2} \cdot \frac{x^{-2}}{-2} = -\frac{1}{4}x^{-2}\) and \(\int x^2 \, dx = \frac{x^3}{3}\). So the antiderivative is \(-\frac{1}{4}x^{-2} + \frac{x^3}{3} + C\).
05

Confirm Part (b) by Differentiation

Differentiate \(-\frac{1}{4}x^{-2} + \frac{x^3}{3} + C\). The derivative \(\frac{d}{dx}(-\frac{1}{4}x^{-2}) = \frac{x^{-3}}{2}\) and \(\frac{d}{dx}(\frac{x^3}{3}) = x^2\), match the original function \(\frac{x^{-3}}{2} + x^2\).
06

Solve Part (c)

The function is \(-x^{-3}+x-1\). Apply the power rule: \(\int -x^{-3} \, dx = -\frac{x^{-2}}{2}\), \(\int x \, dx = \frac{x^2}{2}\), and \(\int -1 \, dx = -x\). Combine the terms: \(-\frac{x^{-2}}{2} + \frac{x^2}{2} - x + C\).
07

Confirm Part (c) by Differentiation

Differentiate \(-\frac{x^{-2}}{2} + \frac{x^2}{2} - x + C\). The derivatives are \(\frac{d}{dx}(-\frac{x^{-2}}{2}) = x^{-3}\), \(\frac{d}{dx}(\frac{x^2}{2}) = x\), and \(\frac{d}{dx}(-x) = -1\), which combine to the original function \(-x^{-3} + x - 1\).

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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. It is essentially the reverse process of differentiation.
While differentiation breaks down a function to find its rate of change, integration assembles these rates to recover the original function or determine an area under a curve.
When you integrate a function, you're looking for an antiderivative, which is a function whose derivative matches the given function. In our exercise, we used integration to find antiderivatives for functions like \(2x^{-3}\) and \(-x^{-3} + x - 1\).

Key steps in integration include:
  • Identify the function to integrate.
  • Apply integration rules, such as the power rule.
  • Add a constant of integration, \(C\), since differentiation removes constants during the process.
Using these steps for each term of the given functions allowed us to find their antiderivatives easily.
Differentiation
Differentiation is the process of finding the derivative of a function. A derivative represents the rate at which a function is changing at any given point.
In our context, differentiation is used to verify that our integration results in the correct antiderivative by ensuring that the derivative of the antiderivative returns the original function.

Here's a breakdown of the differentiation process in our exercise:
  • Differentiate each term of the antiderivative obtained from integration.
  • Check each result to ensure it matches the original function given in the exercise.
For example, after finding that the antiderivative of \(-x^{-3} + x - 1\) was \(-\frac{x^{-2}}{2} + \frac{x^2}{2} - x + C\), we differentiated it to ensure we got back \(-x^{-3} + x - 1\). This consistency proves the correctness of the integration.
Power Rule
The power rule is an essential tool in both differentiation and integration. It simplifies processes by providing a straightforward formula to follow.
When differentiating, the power rule states:
  • If \(y = x^n\), then \(\frac{dy}{dx} = nx^{n-1}\).
In integration, it's similarly simple but in reverse:
  • If \(f(x) = x^n\), then \(\int f(x)\,dx = \frac{x^{n+1}}{n+1} + C\), where \(n eq -1\).
The power rule is particularly useful for polynomial functions, like those in our exercise. It allowed us to calculate antiderivatives of terms like \(2x^{-3}\) by easily adjusting the exponent and coefficient, and always remembering to add \(C\), the constant of integration.
Mastering the power rule enables students to tackle most basic integration and differentiation problems confidently and correctly.

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Most popular questions from this chapter

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

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