Integral of ∫ ∫ x√(x^2+y) dxdy?
∫ ∫ x√(x^2+y) dxdy
R=[0,1]x[0,1]
I just need to figure out how to derive the "dx" portion...I don't remember how to use substitution
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∫ ∫ x√(x^2+y) dxdy
R=[0,1]x[0,1]
I just need to figure out how to derive the "dx" portion...I don't remember how to use substitution
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I'm not totally sure of this but since no-one else has answered.
INT [0, 1] (INT [0, 1] x*(x^2 + y)^(1/2) dx ) dy
= INT [0, 1] (2x/3)*(x^2 + y)^(3/2) dx evaluated over y = [0, 1]
= INT [0, 1] (2x/3)*(x^2 + 1)^(3/2) - (2x/3)*(x^2)^(3/2) dx
= INT [0, 1] (2x/3)*(x^2 + 1)^(3/2) - (2/3)*x^4 dx
= (2/15)*(x^2 + 1)^(5/2) - (2/15)*x^5 evaluated over x = [0, 1]
= [(2/15)*2^(5/2) - (2/15)] - [(2/15)*1^(5/2) - 0]
= (8/15)*sqrt(2) - 4/15