/*! 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 70 Graph the function \(f(x)=\left(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 70

Graph the function \(f(x)=\left(16+x^{2}\right)^{-3 / 2}\) and find the area of the region bounded by the curve and the \(x\) -axis on the interval [0,3]

Short Answer

Expert verified
Answer: The area of the region is \(\frac{1}{60}\).

Step by step solution

01

Graph the function

To graph the function \(f(x)=\left(16+x^{2}\right)^{-\frac{3}{2}}\), we can plot a few points. The points should be chosen at reasonable intervals from the interval [0, 3]. For example, the values could be \(x=0,1,2,3\). Calculate the function values: - \(f(0)=\left(16+0\right)^{-\frac{3}{2}}=\frac{1}{4}\) - \(f(1)=\left(16+1\right)^{-\frac{3}{2}}=\frac{1}{27\sqrt{3}}\) - \(f(2)=\left(16+4\right)^{-\frac{3}{2}}=\frac{1}{216}\) - \(f(3)=\left(16+9\right)^{-\frac{3}{2}}=\frac{1}{125\sqrt{5}}\) By plotting the points \((0, \frac{1}{4})\), \((1, \frac{1}{27\sqrt{3}})\), \((2, \frac{1}{216})\), and \((3, \frac{1}{125\sqrt{5}})\), we obtain a decreasing curve that approaches the x-axis but never touches it.
02

Find the area under the curve

To find the area of the region bounded by the curve and the \(x\)-axis on the interval [0, 3], we need to compute the definite integral of the function over this interval. The integral is given by: \(\int_{0}^{3} \left(16+x^{2}\right)^{-\frac{3}{2}} dx\)
03

Perform a substitution (u-substitution)

To evaluate the integral, we can use a u-substitution method. Let \(u = 16 + x^2\). This means that \(du = 2x dx\). Now, we need to find the new bounds for the integral. When \(x=0\), \(u=16\). When \(x=3\), \(u=25\). So our integral becomes: \(\int_{16}^{25} u^{-\frac{3}{2}}\left(\frac{1}{2}\right) du\)
04

Evaluate the integral

Now we can evaluate the integral: \(\int_{16}^{25} u^{-\frac{3}{2}}\left(\frac{1}{2}\right) du = -\frac{1}{2}(\frac{2}{3})(\frac{1}{\sqrt{u}})\bigg\rvert_{16}^{25}\) Plug in the bounds: \(-\frac{1}{3}(\frac{1}{\sqrt{25}}-\frac{1}{\sqrt{16}})= -\frac{1}{3}( \frac{1}{5} - \frac{1}{4} )\)
05

Calculate the final result

Simplify the expression to find the area under the curve: \(-\frac{1}{3}( \frac{1}{5} - \frac{1}{4} ) = -\frac{1}{3}\times\frac{-1}{20}=\frac{1}{60}\) The area of the region bounded by the curve and the \(x\)-axis on the interval [0, 3] is \(\frac{1}{60}\).

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!

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

\(A\) total charge of \(Q\) is distributed uniformly on a line segment of length \(2 L\) along the \(y\) -axis (see figure). The \(x\) -component of the electric field at a point \((a, 0)\) on the \(x\) -axis is given by $$E_{x}(a)=\frac{k Q a}{2 L} \int_{-L}^{L} \frac{d y}{\left(a^{2}+y^{2}\right)^{3 / 2}}$$ where \(k\) is a physical constant and \(a>0\) a. Confirm that \(E_{x}(a)=\frac{k Q}{a \sqrt{a^{2}+L^{2}}}\) b. Letting \(\rho=Q / 2 L\) be the charge density on the line segment, show that if \(L \rightarrow \infty,\) then \(E_{x}(a)=2 k \rho / a\)

\(\pi < \frac{22}{7}\) One of the earliest approximations to \(\pi\) is \(\frac{22}{7} .\) Verify that \(0 < \int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x=\frac{22}{7}-\pi .\) Why can you conclude that \(\pi < \frac{22}{7} ?\)

Show that \(L=\lim _{n \rightarrow \infty}\left(\frac{1}{n} \ln n !-\ln n\right)=-1\) in the following steps. a. Note that \(n !=n(n-1)(n-2) \cdots 1\) and use \(\ln (a b)=\ln a+\ln b\) to show that $$ \begin{aligned} L &=\lim _{n \rightarrow \infty}\left[\left(\frac{1}{n} \sum_{k=1}^{n} \ln k\right)-\ln n\right] \\ &=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left(\frac{k}{n}\right) \end{aligned} $$ b. Identify the limit of this sum as a Riemann sum for \(\int_{0}^{1} \ln x d x\) Integrate this improper integral by parts and reach the desired conclusion.

The heights of U.S. men are normally distributed with a mean of 69 inches and a standard deviation of 3 inches. This means that the fraction of men with a height between \(a\) and \(b\) (with \(a

Refer to the summary box (Partial Fraction Decompositions) and evaluate the following integrals. $$\int \frac{x}{(x-1)\left(x^{2}+2 x+2\right)^{2}} d x$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

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.