Draw a number line from –2 to 2. Put tick marks every one-tenth unit. Graph a point with the given coordinate.
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By inspecting the integrand, the "obvious" choice for substitution would be
<em>u</em> = <em>y</em> + <em>x</em>
<em>v</em> = <em>y</em> - <em>x</em>
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Solving for <em>x</em> and <em>y</em>, we would have
<em>x</em> = (<em>u</em> - <em>v</em>)/2
<em>y</em> = (<em>u</em> + <em>v</em>)/2
in which case the Jacobian and its determinant are
The trapezoid <em>R</em> has two of its edges on the lines <em>x</em> + <em>y</em> = 8 and <em>x</em> + <em>y</em> = 9, so right away, we have 8 ≤ <em>u</em> ≤ 9.
Then for <em>v</em>, we observe that when <em>x</em> = 0 (the lowest edge of <em>R</em>), <em>v</em> = <em>y</em> ; similarly, when <em>y</em> = 0 (the leftmost edge of <em>R</em>), <em>v</em> = -<em>x</em>. So
-<em>x</em> ≤ <em>v</em> ≤ <em>y</em>
-(<em>u</em> - <em>v</em>)/2 ≤ <em>v</em> ≤ (<em>u</em> + <em>v</em>)/2
-<em>u</em> + <em>v</em> ≤ 2<em>v</em> ≤ <em>u</em> + <em>v</em>
-<em>u</em> ≤ <em>v</em> ≤ <em>u</em>
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So, the integral becomes