Help with simplifying integral

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SUMMARY

The discussion centers on simplifying a double integral involving the expression 6 - (2/3)x + 4 over the limits of 0 to 0 for both variables. The integral to solve is ∫∫ [12 - 2x - (3y/4)] dy dx. The user attempts to simplify the integral but encounters confusion regarding the coefficients and terms, specifically the factor of (1/4) and the correct representation of (3y²/2) as (3y²/8). Clarifications on these points are essential for progressing with the integration.

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Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for examples of common pitfalls in double integral simplification.

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Homework Statement


We need to solve the double integral:

6 -2/3x+4
∫ ∫ [12-2x-(3y/4)] dy dx
0 0

Homework Equations


(a+b)2 = a2 + 2ab + b2


The Attempt at a Solution


I've got to:

6
(1/4) ∫ [12y-2xy-(3y2/2)] evaluated at -2/3x + 4 and 0.
0
6
(1/4) ∫ [12(-2/3x + 4) - 2x(-2/3x + 4) - (3(-2/3x + 4)2/2)]
0

This is where I'm stuck. I could solve the rest of the integral if I could simplify this right. Sometimes, I think simplifying the integral is harder than the integration itself.
 
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6
(1/4) ∫ [12y-2xy-(3y2/2)] evaluated at -2/3x + 4 and 0.
0
6

This isn't right.Where did the 1/4 come from? And it's not (3y2/2), it's 3y2/8.
Which, (1/4)(3/2)=(3/8), but then the other terms are wrong, because (1/4)12= 3, and it's not 3y.
 

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