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Evaluate the integral. $ \displaystyle \int \s…

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Problem 30 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int_{-1}^2 \mid e^x - 1 \mid dx $


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Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 5

Strategy for Integration

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

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
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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
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Problem 42
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Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
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Problem 68
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Problem 70
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Problem 75
Problem 76
Problem 77
Problem 78
Problem 79
Problem 80
Problem 81
Problem 82
Problem 83
Problem 84

Video Transcript

Let's evaluate this inaugural by first evaluating into grand so recall by definition of absolute value, that's t If the input is positive and it's negative t otherwise so Over here we have e to the X minus one. So here recall that either the ex is bigger than one. If X is bigger than zero and then either the X is less than one if X is less than zero. So here's the sea. This You could just recall the graph at zero we get one so access to the right of zero eat of exes above one. Otherwise access the left of zero than you've X is less than one. So this tells us that either that X minus one equals well, if either the ax is bigger than one. If either that X is bigger than one, then e x minus one is positive. So this will be true if X is between zero and two. Still here and then otherwise we have one minus either the ex if X is between negative one and zero. So I did here is break up the interval Negative wants it too into two parts because the absolute value is either the X minus one between zero and two. No, otherwise, between zero negative one. If we're in this region here, X is less than zero. It means even the X is less than one. So either X minus one is negative. And that's why we multiply this by negative one to get one minus either the ex. So with that being said, we have to animals here the first from negative one zero. And then between zero and two. We have either the X minus one. So now we don't have to deal with the absolute value anymore. So let's just go ahead and integrate X minus. Either the ex negative one zero. And then we have either the x minus, x Sierra two. So couldn't plug those in. You get zero minus one minus negative, one minus each of the negative one. That's when you us for the first expression. And then for the second one. This is what we have. It's good and simplify that. We get one over he Plus he squared minus three and that's your final answer

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Related Topics

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Top Calculus 2 / BC Educators
Catherine Ross

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Baylor University

Kristen Karbon

University of Michigan - Ann Arbor

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Boston College

Calculus 2 / BC Courses

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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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