A calculus problem by Jose Sacramento

Calculus Level 4

∫ 0 1 ∫ 0 1 x − y ( x + y ) 3 d x d y + ∫ 0 1 ∫ 0 1 x − y ( x + y ) 3 d y d x = ? \int_0^1 \int_0^1 \dfrac{ x-y} {(x+y)^3} \, dx dy + \int_0^1 \int_0^1 \dfrac{ x-y} {(x+y)^3} \, dy dx = \, ?

1 0

Jose Sacramento by Jose Sacramento , 66, Portugal

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1 solution

Abhishek Sinha
Feb 24, 2016

∫ 0 1 ∫ 0 1 x − y ( x + y ) 3 d x d y = ∫ 0 1 ∫ 0 1 x ( x + y ) 3 d x d y − ∫ 0 1 ∫ 0 1 y ( x + y ) 3 d x d y = ( a ) ∫ 0 1 ∫ 0 1 x ( x + y ) 3 d x d y − ∫ 0 1 ∫ 0 1 x ( x + y ) 3 d x d y = 0 \int_{0}^{1} \int_{0}^{1} \frac{x-y}{(x+y)^3} dxdy = \int_{0}^{1} \int_{0}^{1} \frac{x}{(x+y)^3} dxdy- \int_{0}^{1} \int_{0}^{1} \frac{y}{(x+y)^3} dxdy\stackrel{(a)}{=}\int_{0}^{1} \int_{0}^{1} \frac{x}{(x+y)^3} dxdy- \int_{0}^{1} \int_{0}^{1} \frac{x}{(x+y)^3} dxdy=0 Where the equality (a) is obtained by interchanging x x and y y .

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