/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 470 For the following exercises, fin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For the following exercises, find the antiderivative of the function. $$f(x)=\frac{1}{x^{2}}+x$$

Short Answer

Expert verified
Antiderivative: \( \int f(x) \, dx = -x^{-1} + \frac{x^2}{2} + C \).

Step by step solution

01

- Identify the Powers of Each Term

The function given is \( f(x) = \frac{1}{x^2} + x \). First, rewrite \( \frac{1}{x^2} \) as \( x^{-2} \) for easier integration. The expression becomes \( f(x) = x^{-2} + x \).
02

- Use the Power Rule for Antiderivatives

The power rule for antiderivatives states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \). Apply this rule to each term of the function:
03

- Antiderivative of the First Term

Find the antiderivative of the term \( x^{-2} \). According to the power rule, \( \int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = -x^{-1} \).
04

- Antiderivative of the Second Term

Find the antiderivative of the term \( x \). The power rule gives \( \int x \, dx = \frac{x^{1+1}}{1+1} = \frac{x^2}{2} \).
05

- Combine the Antiderivatives and Add Constant

Combine the results from Steps 3 and 4. The antiderivative is \( -x^{-1} + \frac{x^2}{2} + C \), where \( C \) is the constant of integration.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Power Rule for Integration
The Power Rule for Integration is an essential tool when finding antiderivatives of polynomial functions. It's similar to the Power Rule for differentiation, but works in the opposite direction. Given a function of the form \(x^n\), where \(n\) is not equal to \(-1\), the Power Rule states that the antiderivative is \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\). Here, \(C\) represents the constant of integration, which we'll discuss later.
To apply this rule:
  • Identify the power \(n\) of \(x\).
  • Add 1 to the exponent \((n+1)\).
  • Divide the term by the new exponent.
  • Don't forget to add the constant \(C\).
In our exercise, we first rewrote \(\frac{1}{x^2}\) as \(x^{-2}\), making it easier to apply the Power Rule. Each term in the function \(f(x) = x^{-2} + x\) was treated separately to find its antiderivative.
Integration Techniques
Integration is the process of finding an antiderivative, and efficient techniques can simplify this task. One basic technique is recognizing the structure of the expression so it conforms to a standard rule like the Power Rule. This technique was applied to both terms in our exercise's function.
For the first term, \(x^{-2}\), converting it from a reciprocal form to a power form \(x^n\) allowed us to use the Power Rule easily. For the second term, \(x\), the Power Rule straightforwardly applied, since \(x\) is essentially \(x^1\).
Knowing when and how to transform expressions:
  • Simplifies the integration process.
  • Makes the application of rules like the Power Rule more straightforward.
  • Reduces the potential for errors.
Ultimately, employing these techniques makes solving integrals a much smoother process.
Constant of Integration
In calculus, every time we integrate a function, we introduce a constant of integration \(C\). This is because integration is essentially reversing the differentiation process, and when differentiating a constant, you get zero, which means it doesn't appear in the derivative. Hence, its value is unknown when integrating.
This constant takes any real value, and its purpose is to represent an entire family of functions which have the same derivative. For example, functions like \(-x^{-1} + \frac{x^2}{2} + 3\) and \(-x^{-1} + \frac{x^2}{2} - 5\) would both have the same derivative \(\frac{1}{x^2} + x\).
Key characteristics of the constant of integration:
  • Is essential for the completeness of a general solution.
  • Accounts for any vertical shift in the antiderivative.
  • Symbolizes the infinite nature of antiderivative solutions.
Remembering to add \(C\) provides a complete and accurate solution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For the following exercises, set up, but do not evaluate, each optimization problem. You are the manager of an apartment complex with 50 units. When you set rent at \(\$ 800 / \mathrm{month}\) , all apartments are rented. As you increase rent by \(\$ 25 / \mathrm{month}\) , one fewer apartment is rented. Maintenance costs run \(\$ 50 / \mathrm{month}\) for each occupied unit. What is the rent that maximizes the total amount of profit?

True or False? Justify your answer with a proof or a counterexample. Assume that \(f(x)\) is continuous and differentiable unless stated otherwise. There is a function such that there is both an inflection point and a critical point for some value \(x=a\) .

Consider the roots of the equation. Show that the equation \(y=x^{3}+3 x^{2}+16\) has exactly one real root. What is it?

Find the critical points and the local and absolute extrema of the following functions on the given interval. $$f(x)=3 x^{4}-4 x^{3}-12 x^{2}+6 \text { over }[-3,3]$$

A rocket is launched into space; its kinetic energy is given by \(K(t)=\left(\frac{1}{2}\right) m(t) v(t)^{2},\) where \(K\) is the kinetic energy in joules, \(m\) is the mass of the rocket in kilograms, and \(v\) is the velocity of the rocket in meters/second. Assume the velocity is increasing at a rate of 15 \(\mathrm{m} / \mathrm{sec}^{2}\) and the mass is decreasing at a rate of 10 kg/sec because the fuel is being burned. At what rate is the rocket's kinetic energy changing when the mass is 2000 kg and the velocity is 5000 \(\mathrm{m} / \mathrm{sec}\) ? Give your answer in mega-Joules (MI), which is equivalent to \(10^{6} \mathrm{J} .\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.