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Use the method of cylindrical shells to find the …

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Problem 63 Hard Difficulty

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the curves about the given axis.

$ y = e^{-x} $ , $ y = 0 $ , $ x = -1 $ , $ x = 0 $ ; about $ x = 1 $


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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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Anna Marie Vagnozzi

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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
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Problem 71
Problem 72
Problem 73
Problem 74

Video Transcript

The problem is use the method of cylindrical shelves to find the volume generated by rating the region bounded by the curves of the given axis here y is equal to 2 negative x y. Equal to 0 x is negative. 1 x equals to 0 at x, equals to 1, so we can do. The aphis is x y and it's a graph of y equal to e to negative, x and x, equal to negative 1 and about its axis. This is x. If 1 is the region. Is this part and rotate this region about axis x, equal to 1 by using the miser riddle of celindrical shells we can see the volume is equal to 2 pi integral from negative 1 to 0. Here x is a distant distance from the point to the x x to the axis of the line. So here it is, is 1 minus x times function. F x is 2 negative x, so notice that this is the distance was a point to the axis between the point and the axis distance. So this is 1 minus x. Now we can use integration by parts to solve this integral. The formula is integral from a to b: u times, vix is equal to u times v from a to b minus integral from integral u prime v d x from a to b now for our problem, we can write. U is equal to 1 minus x and prime is equal to 2 negative x, then prime is equal to negative 1 and v is equal to negative to negative x. Then this integral is equal to 2 pi. By using this formula we can say this is 2 pi times u times v, so this is negative 1 minus x to negative x from negative 1 to 0 and minus integral from negative 1 to 0 prime times, so this is to negative x dx. So this is equal to 2 pi, plain 0 and negative 1 to these functions. We have. This is negative: 1 minus negative, 1 minus minus 1 is 2 and e to 1 is negative and minus integral of e to negative x is negative to negative x. This is past negative x from negative 1 to 0 point, so this is equal to 2 pi 2 e minus 1 plus this is 1 minus e. So i answer is 2 pi ti.

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Related Topics

Integration Techniques

Top Calculus 2 / BC Educators
Grace He

Numerade Educator

Anna Marie Vagnozzi

Campbell University

Kayleah Tsai

Harvey Mudd College

Joseph Lentino

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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