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Evaluate the integral.
$ \displaystyle \int_{-1}^2 \mid e^x - 1 \mid dx $
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Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 5
Strategy for Integration
Integration Techniques
Missouri State University
Baylor University
University of Michigan - Ann Arbor
Boston College
Lectures
01:53
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.
27:53
In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.
03:51
Evaluate the integral.…
00:54
Evaluate the definite inte…
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03:21
Evaluate the given integra…
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04:03
01:13
Evaluate the integrals.
01:19
01:51
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evaluate Integral
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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