Problem

Use a graph to find approximate x-coordinates of …

05:05

Question

Answered step-by-step

Problem 59 Hard Difficulty

Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find (approximately) the area of the region bounded by the curves.

$ y = \arcsin \left(\frac{1}{2} x \right) $, $ y = 2 - x^2 $

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Wen Zheng

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Watch More Solved Questions in Chapter 7

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Video Transcript

The problem is use a graph to find approx. Approximate x, coordinates of the points of intersection of given curves then find it area of the region bounded by the curves curve is y, is equal to sine x. Over 2 and y is equal to 2 minus x square. The first lesson: look at the graph of these 2 curves. The right 1 is y is equal to oxide x over 2 and the blue 1 is equal to 2 minus x square. We can see its intersection point is negative 1.7551 and 1.1 sen 2, and between these 2 points the blue curve is greater than the red curve. So we have a is equal to negative 1.751 and b is equal to 1.12, and i ride area of the region. Bonded by the curves, as integral from a to b 2 minus x, squared minus oxine, half x dx. So this is equal to so. The integral of 2 minus x squared is equal to 2 x, minus 1 third x to 3 power from a to b minus the integral from a to b oxide 1 half x x. For this integral. We can use integration by parts that, u is equal to oxide x over 2 and v. Prim is equal to 1. It then is equal to 1 over rotie 1 minus x, squared over 4 times by half. So this is equal to 1 over root of 4 minus x, squared and v is equal to x. So this integral is equal to x times oxide 1, over 2 times x, from a to b minus integral x, over relative 4 minus x, squared dx, and for this integral we can use. U substitution, not y is equal to 4 minus x. Squared and y is equal to negative 2 x dx, so we can write this integral is equal to negative, integral from a to b 1 half du over relative. U so this is a to negative root of? U so. This is root of 4 minus x square. From a to b, then we can write. Area of this region is equal to area of this region is 2 minus x square minus sine 1 over 2 x x, so this is equal to 2 x, minus 1 third x 2 cube minus oxide x over 2 times x and plus retto 4 minus x Square from a to b by using calculator we can find this is equal to 3.9993.

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem