/*! 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 1 Find the indefinite integrels in... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 1

Find the indefinite integrels in \(1-20\). \(\int \sqrt{2+x} d x \quad(a d d+C)\)

Short Answer

Expert verified
The indefinite integral is \( \frac{2(2+x)^{3/2}}{3} + C \).

Step by step solution

01

Identify the Type of Integral

This integral involves the square root of an expression in the form of \( \sqrt{2+x} \). The goal is to integrate this with respect to \( x \) and add a constant of integration, \( C \).
02

Substitution

Let's use the substitution method to simplify the integral. Let \( u = 2 + x \). Then, \( du = dx \). Substitute these into the integral to get \( \int \sqrt{u} \, du \).
03

Applying Power Rule for Integration

Using the power rule for integration, which states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \) for \( n eq -1 \), we integrate \( u^{1/2} \) to get \( \frac{u^{3/2}}{3/2} + C \).
04

Simplify the Result

Simplify \( \frac{u^{3/2}}{3/2} + C \) into \( \frac{2u^{3/2}}{3} + C \).
05

Substitute Back the Original Variable

Return to the original variable by substituting \( u = 2 + x \) back into the expression to get \( \frac{2(2+x)^{3/2}}{3} + C \).

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 91影视!

Key Concepts

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

Substitution Method
The substitution method is a handy tool when dealing with functions that seem complicated at first glance. It transforms an integral into a simpler form, making the integration process more manageable. In this specific exercise, we applied the substitution method to the integral \( \int \sqrt{2+x} \, dx \). Here's how it works step-by-step:
  • First, identify a substitution that will simplify the expression. Often, this "substitute" is the inside part of a composite function. For instance, if your function contains \( \sqrt{2+x} \), a useful substitution would be \( u = 2 + x \).
  • Next, compute the derivative of the substitution, \( du \). For this example, since \( u = 2 + x \), \( du = dx \).
  • Now, replace all instances of \( x \) in the integral with \( u \) and \( du \). This changes our integral from \( \int \sqrt{2+x} \, dx \) to \( \int \sqrt{u} \, du \).
By simplifying the integrand, the integral becomes easier to solve. After integration, always remember to substitute back the original variable.
Power Rule for Integration
The power rule for integration is a fundamental technique in calculus used to integrate power functions. It states that for any function of the form \( u^n \), where \( n eq -1 \), the integral is given by:\[ \int u^n \, du = \frac{u^{n+1}}{n+1} + C \]In our exercise, after substitution, we were left with the integral \( \int \sqrt{u} \, du \). To apply the power rule here:
  • Notice that \( \sqrt{u} \) can be rewritten as \( u^{1/2} \).
  • Using the power rule, integrate \( u^{1/2} \), giving \( \frac{u^{3/2}}{3/2} + C \).
  • Simplify this result to \( \frac{2u^{3/2}}{3} + C \).
Applying the power rule correctly is crucial in determining the antiderivative of power functions, allowing for more complex integral calculations.
Constant of Integration
The concept of a constant of integration, represented as \( C \), is central to finding indefinite integrals. Whenever we integrate a function without given limits, we introduce this constant.An indefinite integral represents a family of functions, all of which differ by a constant value. Why include it?
  • Integration essentially reverses differentiation; however, differentiating different constants (like \( C \)) results in zero. This implies that many different functions can have the same derivative, i.e., the initial function before difference.
  • By including \( C \), we ensure that we account for any possible constant that could have been there originally before differentiation.
In our solution for \( \int \sqrt{2+x} \, dx \), \( C \) was added at the end, yielding the expression \( \frac{2(2+x)^{3/2}}{3} + C \). Always remember: the constant of integration is necessary to represent all potential antiderivatives of a function.

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

Suppose \(d f / d x=2 x\). We know that \(d\left(x^{2}\right) / d x=2 x\). How do we prove that \(f(x)=x^{2}+C ?\)

The hypervolume of a four-dimensional sphere is \(H=\) \(\frac{1}{2} \pi^{2} r^{4}\). Therefore the area (volume?) of its three-dimensional surface \(x^{2}+y^{2}+z^{2}+t^{2}=r^{2}\) is _______.

Find the indefinite integrels in \(1-20\). \(\int t^{3} d t / \sqrt{1+t^{2}}\)

Find the indefinite integrels in \(1-20\). \(\int t \sqrt{1-t^{2}} d t\)

Find an antiderivative \(f(x)\) for \(v(x)\). Then compute the definite integral \(\int_{0}^{1} r(x) d x=f(1)-f(0)\). $$ 1 \sqrt{x}\left(\text { or } x^{-1 / 2}\right) $$
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation
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.