Evaluate the given integral by making an appropriate change of variables. ∬_(R)7(x-6y)/(3x-y)dA, where R is the parallelogram enclosed by the lines x-6y=0,x-6y=9,3x-y=2, and 3x-y=3
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$$ \iint_R \frac{x-6y}{3x-y} dA $$ where R is the parallelogram enclosed by the lines $$x-6y=0, x-6y=9, 3x-y=2, \text{ and } 3x-y=3$$ Show more…
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