Q. 78 Area under a Curve; The area und... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 78 (page 452)

Area under a Curve; The area under the graph of $y = \frac{1}{\sqrt{1 鈭� x^{2}}}$ and above the $x$ -axis between $x = a$ and $x = b$ is given by $\sin^{鈭� 1} 鈦� b 鈭� \sin^{鈭� 1} 鈦� a$ .
See the figure,
(a) Find the exact area under the graph of $y = \frac{1}{\sqrt{1 鈭� x^{2}}}$
and above the $x$ -axis between $x = 0$ and $x = \frac{\sqrt{3}}{2}$ .
(b) Find the exact area under the graph of $y = \frac{1}{\sqrt{1 鈭� x^{2}}}$ and above the $x$ -axis between $x = - \frac{1}{2}$ and $x = \frac{1}{2}$

Short Answer

Expert verified

(a) The area under the graph of $y = \frac{1}{\sqrt{1 鈭� x^{2}}}$ and above the $x$ -axis between $x = 0$ and $x = \frac{\sqrt{3}}{2}$ is $\frac{蟺}{3}$ .

(b) The area under the graph of $y = \frac{1}{\sqrt{1 鈭� x^{2}}}$ and above the $x$ -axis between $x = - \frac{1}{2}$ and $x = \frac{1}{2}$ is $\frac{蟺}{3}$ .

Step by step solution

Step 1. Given data

The area under the graph of $y = \frac{1}{\sqrt{1 鈭� x^{2}}}$ and above the $x$ -axis between $x = a$ and $x = b$ is $\sin^{鈭� 1} 鈦� b 鈭� \sin^{鈭� 1} 鈦� a$

We have to find the exact area of $y = \frac{1}{\sqrt{1 鈭� x^{2}}}$ and above the axis between $x = 0$ and $x = \frac{\sqrt{3}}{2}$ and when $x = - \frac{1}{2}$ and $x = \frac{1}{2}$ .

Step 2. The exact area under the graph and above the x-axis between x = 0 and x=32.

Let $a = 0$ and $b = \frac{\sqrt{3}}{2}$

Therefore,

The exact area $(A)$ under the graph and above the $x$ -axis

$A = \sin^{鈭� 1} 鈦� \frac{\sqrt{3}}{2} 鈭� \sin^{鈭� 1} 鈦� 0 = \frac{蟺}{3} 鈭� 0 = \frac{蟺}{3} 鈭� 0$

Step 3. The exact area under the graph and above the x-axis between x=-12 and x=12.

Let $a = - \frac{1}{2}$ and $b = \frac{1}{2}$

Therefore,

The exact area $(A)$ under the graph and above the $x$ -axis

$\begin{aligned} A & = \sin^{鈭� 1} 鈦� (\frac{鈭� 1}{2}) 鈭� \sin^{鈭� 1} 鈦� (\frac{1}{2}) \\ = \frac{蟺}{6} 鈭� \frac{鈭� 蟺}{6} \\ = \frac{蟺}{6} + \frac{蟺}{6} \\ = \frac{蟺}{3} \end{aligned}$

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

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Short Answer

Step by step solution

Step 1. Given data

Step 2. The exact area under the graph and above the x-axis between x = 0 and x=32.

Step 3. The exact area under the graph and above the x-axis between x=-12 and x=12.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Decision Maths

Logic and Functions

Theoretical and Mathematical Physics

Pure Maths

Statistics

Study anywhere. Anytime. Across all devices.