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Evaluate the integral.

$ \displaystyle \int^{2}_{-1} \bigl( x - 2 \mid x \mid \bigr) \,dx $

$$-3.5$$

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Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

We are looking at the integral from negative 1-2 of X -2 Absolute Value of X. Uh huh. Dx So if I were trying to do this I would look for the area under the curve. Um And I would actually start evaluating points along the curve like when X is negative one because if X is negative one this is now positive one. This X so again negative one minus two which would be negative three. I would do the same thing with 00 minus zero would still give me zero and I would figure out when X is one will be again one minus uh two times one is, so one minus two would give me one. Um And then same thing with X equals two because there'd be two minus four which would be negative too. So if I I'm understanding this problem correctly uh it's gonna be like a V. Shape like that And I can create two triangles. Um and that you can evaluate it strangles one half times the base times the height. Um And then the other triangle would be right here. Hopefully all of this is correct. Otherwise I'm gonna be Off a little bit one and we're going to add to it one half Times the width of two times the height of two. Um And the only thing is because they're both under the X. Axis, they both need to be negative. So I'm looking at -3/2. That would be -4 has because that's how you multiply fractions just enumerated with numerator with numerator over denominator. So that's how I got those. So that comes out to be negative seven halves Which is the same thing as negative 3.5. So I did this correctly. I would leave it like that. Yeah.