/*! 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 39 Evaluate the integrals. Some int... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Evaluate the integrals. Some integrals do not require integration by parts. $$ \int x^{3} \sqrt{x^{2}+1} d x $$

Short Answer

Expert verified
The integral is \( \frac{1}{5} (x^{2}+1)^{5/2} - \frac{1}{3} (x^{2}+1)^{3/2} + C \).

Step by step solution

01

Determine Technique

The integral \( \int x^{3} \sqrt{x^{2}+1} \, dx \) involves a combination of polynomial and radical expressions. Integration by substitution is appropriate here.
02

Choose a Substitution

Set \( u = x^{2} + 1 \). Then, \( du = 2x \, dx \), or equivalently, \( dx = \frac{du}{2x} \). This substitution is chosen to simplify the expression under the square root.
03

Rewrite the Integral

Substitute \( u \) and \( dx \) into the integral: \[ \int x^{3} \sqrt{x^{2}+1} \, dx = \int x^{3} \sqrt{u} \cdot \frac{du}{2x} \]Simplify to get: \[ \int \frac{x^{2} \sqrt{u}}{2} \, du \]
04

Express \( x^{2} \) in Terms of \( u \)

From the substitution \( u = x^2 + 1 \), we find \( x^2 = u - 1 \). Substitute this back in: \[ \int \frac{(u-1)\sqrt{u}}{2} \, du \]
05

Simplify and Integrate

Distribute and simplify the integral: \[ \int \frac{u^{3/2} - u^{1/2}}{2} \, du = \frac{1}{2} \left( \int u^{3/2} \, du - \int u^{1/2} \, du \right) \] Integrate term-by-term: \[ \frac{1}{2} \left( \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} \right) + C \]Simplifier further: \[ \frac{1}{5} u^{5/2} - \frac{1}{3} u^{3/2} + C \]
06

Substitute Back

Replace \( u \) with \( x^{2} + 1 \): \[ \frac{1}{5} (x^{2}+1)^{5/2} - \frac{1}{3} (x^{2}+1)^{3/2} + C \] This is the result of the integral with respect to \( x \).

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.

Understanding Integrals
Integrals are a fundamental concept in calculus. They represent the area under a curve in a given interval and are essential for calculating things like displacement, area, and total accumulated quantities. Integrals come in two main types: definite and indefinite. A definite integral has limits and calculates a specific numerical area, while an indefinite integral finds a general form of antiderivatives and adds the constant of integration, usually denoted by +C. When dealing with integrals, it's crucial to identify the structure of the integrand to choose an appropriate integration technique. This could be basic integration, substitution, or integration by parts, among others. In this problem, the integrand involves polynomial functions and radical expressions, making it suitable for substitution.
Polynomial Functions
Polynomial functions are expressions that consist of variables and coefficients combined using only addition, subtraction, multiplication, and non-negative integer exponentiation. The function in the exercise, \( x^3 \), is a polynomial function and can be integrated using standard rules or through substitution when combined with other types of expressions.Polynomials are essential in integration because their derivatives and antiderivatives are quite predictable. The power rule, which states that \( \int x^n\, dx = \frac{x^{n+1}}{n+1} + C \) for any real number \( n eq -1 \), is often used for integrating simple polynomial terms. For more complex polynomials, especially when combined with other functions, substitution might be needed to simplify the process.
Handling Radical Expressions
Radical expressions involve roots, like the square root \( \sqrt{x^2 + 1} \) in our exercise. These can complicate integrals because they are not straightforward to integrate by standard means. Radicals often appear together with polynomial terms, requiring careful manipulation. In our problem, the expression under the square root \( \sqrt{x^2 + 1} \) is simplified using substitution, wherein \( u = x^2 + 1 \). Substitution transforms the radical into a function of \( u \), making the integral easier to manage. Whenever radicals are involved, look for substitution opportunities to rewrite the integral in terms of a more straightforward expression.
Exploring Integration Techniques
Integration techniques are strategies used to solve complex integrals. The most common methods include substitution, integration by parts, partial fraction decomposition, and trigonometric substitutions. The goal of these techniques is to simplify integrands into forms that are easier to integrate. In our scenario, integration by substitution is utilized. This technique works by changing the variable of integration to simplify the integral. Identify a part of the integrand that can be replaced with a new variable. In this problem, \( u = x^2 + 1 \) serves as a substitute variable that simplifies the integral's complexity. The process involves replacing \( x \, dx \) with the corresponding \( du \) expression, streamlining the integration process. Successfully applying substitution often relies on recognizing patterns and transforming messy integrands into manageable forms.

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

Effects of an antihistamine The concentration of an antihistamine in the bloodstream of a healthy adult is modeled by $$C=12.5-4 \ln \left(t^{2}-3 t+4\right)$$ where \(C\) is measured in grams per liter and \(t\) is the time in hours since the medication was taken. What is the average level of concentration in the bloodstream over a 6 -hr period?

Find the value of the constant \(c\) so that the given function is a probability density function for a random variable over the specified interval. \(f(x)=c x \sqrt{25-x^{2}}\) over \([0,5]\)

Usable values of the sine-integral function The sine-integral function, $$\operatorname{Si}(x)=\int_{0}^{x} \frac{\sin t}{t} d t$$ is one of the many functions in engineering whose formulas cannot be simplified. There is no elementary formula for the antiderivative of \((\sin t) / t .\) The values of \(S \mathrm{Si}(x),\) however, are readily estimated by numerical integration. Although the notation does not show it explicitly, the function being integrated is $$f(t)=\left\\{\begin{array}{cl}{\frac{\sin t}{t},} & {t \neq 0} \\ {1,} & {t=0}\end{array}\right.$$ the continuous extension of \((\sin t) / t\) to the interval \([0, x]\) . The function has derivatives of all orders at every point of its domain. Its graph is smooth, and you can expect good results from Simpson's Rule. a. Use the fact that \(\left|f^{(4)}\right| \leq 1\) on \([0, \pi / 2]\) to give an upper bound for the error that will occur if $$\mathrm{Si}\left(\frac{\pi}{2}\right)=\int_{0}^{\pi / 2} \frac{\sin t}{t} d t$$ is estimated by Simpson's Rule with \(n=4\) b. Estimate \(\operatorname{Si}(\pi / 2)\) by Simpson's Rule with \(n=4\) . c. Express the error bound you found in part (a) as a percentage of the value you found in part (b).

Moment about \(y\) -axis A thin plate of constant density \(\delta=1\) occupies the region enclosed by the curve \(y=36 /(2 x+3)\) and the line \(x=3\) in the first quadrant. Find the moment of the plate about the \(y\) -axis.

Use numerical integration to estimate the value of $$\sin ^{-1} 0.6=\int_{0}^{0.6} \frac{d x}{\sqrt{1-x^{2}}}$$ For reference, \(\sin ^{-1} 0.6=0.64350\) to five decimal places.

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.