/*! 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 15 Use a table of integrals to eval... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 15

Use a table of integrals to evaluate the following indefinite integrals. Some of the integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table. $$\int \frac{x}{\sqrt{4 x+1}} d x$$

Short Answer

Expert verified
Question: Evaluate the indefinite integral: $$\int \frac{x}{\sqrt{4 x+1}} d x$$ Answer: $$\int \frac{x}{\sqrt{4 x+1}} d x = \frac{1}{16} \left(\frac{2}{3}(4x+1)^{\frac{3}{2}} - 2(4x+1)^{\frac{1}{2}}\right) + C$$

Step by step solution

01

Make a substitution

Let \(u = 4x + 1\). Then we have to find \(d u\) by differentiating \(u\) with respect to \(x\): $$\frac{d u}{d x} = 4$$ Now, solve for \(d x\): $$d x = \frac{d u}{4}$$ Now we can rewrite the integral in terms of \(u\): $$\int \frac{x}{\sqrt{4 x+1}} d x = \int \frac{x}{\sqrt{u}} \frac{d u}{4}$$ To rewrite the \(x\) term in terms of \(u\), solve for \(x\) from the substitution equation: $$x = \frac{u-1}{4}$$ Now the integral becomes: $$\int \frac{\frac{u-1}{4}}{\sqrt{u}} \frac{d u}{4}$$
02

Simplify the integral

Combine the constants and simplify the expression inside the integral: $$\int \frac{\frac{u-1}{4}}{\sqrt{u}} \frac{d u}{4} = \frac{1}{16} \int \frac{u-1}{\sqrt{u}} d u$$ Now, split the fraction inside the integral into two separate fractions: $$\frac{1}{16} \int \frac{u-1}{\sqrt{u}} d u = \frac{1}{16} \int \left(\frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}}\right) d u$$ Simplify the fractions further: $$\frac{1}{16} \int \left(\frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}}\right) d u = \frac{1}{16} \int \left(\sqrt{u} - u^{-\frac{1}{2}}\right) d u$$
03

Evaluate the integral

Integrate each term separately: $$\frac{1}{16} \int \left(\sqrt{u} - u^{-\frac{1}{2}}\right) d u = \frac{1}{16} \left(\int \sqrt{u} d u - \int u^{-\frac{1}{2}} d u\right)$$ Now we can use the power rule to integrate each term: $$\frac{1}{16} \left(\int \sqrt{u} d u - \int u^{-\frac{1}{2}} d u\right) = \frac{1}{16} \left(\frac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}}\right) + C$$ Where \(C\) is the constant of integration.
04

Substitute back the variable x

Now, we need to substitute back \(x\) into our solution. Recall that \(u = 4x + 1\). Replace \(u\) with this expression: $$\frac{1}{16} \left(\frac{2}{3}(4x+1)^{\frac{3}{2}} - 2(4x+1)^{\frac{1}{2}}\right) + C$$ And this is the final result of the indefinite integral evaluation: $$\int \frac{x}{\sqrt{4 x+1}} d x = \frac{1}{16} \left(\frac{2}{3}(4x+1)^{\frac{3}{2}} - 2(4x+1)^{\frac{1}{2}}\right) + C$$

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.

Integration Techniques
Integration is a fundamental tool in calculus, where it serves as the reverse operation of differentiation. Integration techniques involve a range of methods for finding the integral or antiderivative of functions. Some common techniques include the power rule, substitution method, integration by parts, and partial fractions decomposition.

When approaching an integral, one of the keys is recognizing patterns and applying the appropriate technique. Substitution is often useful when an integral contains a function and its derivative. Sometimes, manipulations such as completing the square are necessary before applying a technique. In the given exercise, the substitution method simplifies the integral into a form where the power rule can be applied.
Substitution Method
The substitution method, also known as u-substitution, is one of the most widely used techniques for integration. It is particularly handy when you notice a composite function that is the product of a function and its derivative. To apply this method, you choose a part of the integral to substitute with a new variable, usually denoted as u. This simplification can change a challenging integral into a much easier one.

In the example, the substitution u = 4x + 1 was selected because the derivative of 4x is a constant, making it a prime candidate for substitution. As the derivative of u with respect to x is 4, we can express dx in terms of du, simplifying the integral significantly. Remember, it's crucial to replace all instances of the original variable with the new one to perform the integration correctly.
Completing the Square
Completing the square is a technique used to convert a quadratic polynomial into a perfect square trinomial plus or minus a constant. This method is particularly useful when dealing with quadratics within integral expressions where the direct application of integration techniques is not possible.

To complete the square for a quadratic of the form ax^2 + bx + c, one rewrites it as a(x - h)^2 + k where h and k are constants derived from the original coefficients. This form exposes the core component of the quadratic as a square, which can significantly simplify the integration process. While not directly applied in this specific problem, completing the square can be very useful when integrating functions involving squareroots or when applying the substitution method to integrals of rational functions.
Power Rule for Integration
The power rule for integration is a direct counterpart to the power rule for differentiation. It allows us to integrate any power of x easily and is a fundamental tool in solving indefinite integrals. The rule states that the integral of x^n, where n is any real number except -1, is (x^{n+1})/(n+1), plus a constant of integration.

In the provided exercise, after the substitution is made and the integral is simplified, the power rule is used to integrate u^{1/2} and u^{-1/2}. These correspond to u raised to the power of 1/2 and -1/2, respectively. Following the power rule, the integrals become u^{3/2} times 2/3 and u^{1/2} times -2, plus the constant of integration. Recognizing when and how to apply the power rule is vital for any student learning integration.

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

The following integrals may require more than one table look-up. Evaluate the integrals using a table of integrals, and then check your answer with a computer algebra system. $$\int \frac{\sin ^{-1} a x}{x^{2}} d x, a>0$$

An important function in statistics is the Gaussian (or normal distribution, or bell-shaped curve), \(f(x)=e^{-a x^{2}}\) a. Graph the Gaussian for \(a=0.5,1,\) and 2. b. Given that \(\int_{-\infty}^{\infty} e^{-a x^{2}} d x=\sqrt{\frac{\pi}{a}},\) compute the area under the curves in part (a). c. Complete the square to evaluate \(\int_{-\infty}^{\infty} e^{-\left(a x^{2}+b x+c\right)} d x,\) where \(a>0, b,\) and \(c\) are real numbers.

A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ where we assume s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated using integration by parts: $$F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1}$$ Verify the following Laplace transforms, where a is a real number. $$f(t)=1 \rightarrow F(s)=\frac{1}{s}$$

Period of a pendulum A standard pendulum of length \(L\) that swings under the influence of gravity alone (no resistance) has a period of $$ T=\frac{4}{\omega} \int_{0}^{\pi / 2} \frac{d \varphi}{\sqrt{1-k^{2} \sin ^{2} \varphi}} $$ where \(\omega^{2}=g / L, k^{2}=\sin ^{2}\left(\theta_{0} / 2\right), g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(\theta_{0}\) is the initial angle from which the pendulum is released (in radians). Use numerical integration to approximate the period of a pendulum with \(L=1 \mathrm{m}\) that is released from an angle of \(\theta_{0}=\pi / 4\) rad.

Let \(R\) be the region between the curves \(y=e^{-c x}\) and \(y=-e^{-c x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c>0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves; see the Guided Project The exponential Eiffel Tower. ) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R .\) Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Geometry

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.