Select the locations on the number line to plot the points 11 /4 and 19 /4 .
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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>
<em />
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>
<em />
So, the integral becomes
First, isolate y on one side to find some information about the graph.
3x+4y=-10
4y=-10-3x
y=(-10-3x)/4
Now find out the x and y intercepts of this equation;
(0)=(-10-3x)/4
4x0=-10-3x
0=-10-3x
3x=-10
x=-10/3 or -3.3333 (x-intercept)
y=(-10-3(0))/4
4y=-10
y=-10/4 or -5/2 = -2.5 (y-intercept)
Plot these points onto your graph and connect the lines.
I have attached a graph below, I'm sorry I couldn't do it on paper.
Hope I helped :)
3x+4y=-10
4y=-10-3x
y=(-10-3x)/4
Now find out the x and y intercepts of this equation;
(0)=(-10-3x)/4
4x0=-10-3x
0=-10-3x
3x=-10
x=-10/3 or -3.3333 (x-intercept)
y=(-10-3(0))/4
4y=-10
y=-10/4 or -5/2 = -2.5 (y-intercept)
Plot these points onto your graph and connect the lines.
I have attached a graph below, I'm sorry I couldn't do it on paper.
Hope I helped :)