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Use a graph to find approximate x-coordinates of …

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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 $


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Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Related Topics

Integration Techniques

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Top Calculus 2 / BC Educators
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Missouri State University

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Lectures

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01:53

Integration Techniques - Intro

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

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27:53

Basic Techniques

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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Watch More Solved Questions in Chapter 7

Problem 1
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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
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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.

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Related Topics

Integration Techniques

Top Calculus 2 / BC Educators
Catherine Ross

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Samuel Hannah

University of Nottingham

Michael Jacobsen

Idaho State University

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Boston College

Calculus 2 / BC Courses

Lectures

Video Thumbnail

01:53

Integration Techniques - Intro

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

Video Thumbnail

27:53

Basic Techniques

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

Join Course
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