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Evaluate the integral.

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

$\frac{1}{e}+e^{2}-3$

Integration Techniques

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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