Rewrite the boundary lines <em>y</em> = -1 - <em>x</em> and <em>y</em> = <em>x</em> - 1 as functions of <em>y </em>:
<em>y</em> = -1 - <em>x</em> ==> <em>x</em> = -1 - <em>y</em>
<em>y</em> = <em>x</em> - 1 ==> <em>x</em> = 1 + <em>y</em>
So if we let <em>x</em> range between these two lines, we need to let <em>y</em> vary between the point where these lines intersect, and the line <em>y</em> = 1.
This means the area is given by the integral,

The integral with respect to <em>x</em> is trivial:

For the remaining integral, integrate term-by-term to get

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line <em>y</em> = 1 and (0, -1)), so its area is 1/2*4*2 = 4.
Answer:
x = -3
Step-by-step explanation:
<span>I think your answer would be 270.04:)
Hope this helps:)
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