Evaluate the integral below.

Question

Evaluate the integral below.

 

∫_0^1¦∫_y^(2−y)¦ ∫_0^(2−x−y)¦ xy dz dx dy

 

 

Answer 1

∫₀¹ ∫_y^2-y ∫₀^2-x-y xy dz dx dy

 

= ∫₀¹ ∫_y^2-y [xyz]₀^2-x-y dx dy

 

= ∫₀¹ ∫_y^2-y (2xy - x²y - xy²) dx dy

 

= ∫₀¹ [x²y - x³y/3 - x²y²]_y^2-y

 

= ∫₀¹ (4y/3 - 2y² + 2y⁴/3) dy

 

= [2y²/3 - 2y³/3 + 2y⁵/15]₀¹

 

= 2/3 - 2/3  + 2/15

 

= 2/15