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Use the method of cylindrical shells to find the …

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Problem 60 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 = x \ln (x + 1) $ , $ y = 3x - 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
Catherine Ross

Missouri State University

Anna Marie Vagnozzi

Campbell University

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Oregon State University

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Harvey Mudd 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.

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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 approximate x coordinates of the points of intersection with given curves then find the area of the region bondage curves here y is equal to x times. L n, x, plus 1 and y is equal to 3 x minus x. Squared by using some graphic device we can find, the graph is as follows. So y 1 is y, is equal to x, times, x, plus 1 and a blue 1, as is equal to 3 x minus x square, and we can see its intersection. Points are 0 and 1.926, so we can let a is equal to 0 and b is equal to 1.926. Then, and we can see between these 2 points. The blue curve is on top of the right 1. So we have every bonds. 2 curves is equal to integral, from a to e 3 x, minus x, squared minus x times n x, plus 1 dx. So this is equal to 3 over 2 x, squared minus 1, half 11 third to 3 power from a to b minus integral from a to b x times, l n x, plus 1 dx. But then we compute the integral of x times x, plus 1. I see we kind o use integration by parts, but first we cannot. Y is equal to x, plus 1 used substitution and we have. This is equal to the integral of y minus 1 times n x, o n y d y, and then we can use? U substitution, that? U is equal to l, n y and prime, is equal to y minus 1. Then u, prime, is equal to 1. Over y and v is equal to 1 half y squared and minus y. Then this integral is equal to u times v. So this is 1 half y squared minus y times n y minus integral of prim times v. So this is 1 half y minus 1. This is equal to 1 half y squared minus y n y minus. This is upon force y square, minus y and as cos, the number c now and we convert y back to x. So this is 1 half x, plus 1 square minus x, plus 1 times l n x, plus 1 minus a force x, plus 1 square minus x, plus 1 and for our problem is integral definitely integral from a to b of this value. So this is from a to b, then this is equal to this 1 and from a to b now, by using a calculator we can find. The value of this 1 is about 1.69 twoix.

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Related Topics

Integration Techniques

Top Calculus 2 / BC Educators
Catherine Ross

Missouri State University

Anna Marie Vagnozzi

Campbell University

Heather Zimmers

Oregon State University

Kayleah Tsai

Harvey Mudd 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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