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Question
Evaluate ∫(INT 8 to 2) ∫(INT 16x to 9) (3xy)^.5 dy dx
I'm having a bit of trouble working with the 16x interval. I'm not sure if I'm setting it up wrong or if I'm getting tangled up in the algebra towards the end of the problem. Please help.
Thanks
2 9
∫ ∫ √(3xy) dy dx =
8 16x
2 9
∫ dx ∫ √(3xy) dy =
8 16x
2 9
∫ dx √(3x) ∫ √y dy =
8 16x
2
∫ dx { √(3x)*(2/3)*(y)^(3/2) } [from 16x to 9] =
8
2
∫ √(3x)*(2/3)*[(9)^(3/2)-(16x)^(3/2)] dx =
8
2
∫ √x*(2/√3)*(9)^(3/2) dx -
8
2
√3*16*∫ (x)^(3/2) dx =
8
This integral is immediate, I'll leave it to you from here.
Alon.
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Alon Mandes
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