如题所述
第1个回答 2017-07-10
∫(0→1) dx ∫(0→(1-x)/2) dy ∫(0→1-x-2y) xdz
=∫(0→1) dx ∫(0→(1-x)/2) dy xz |(0→1-x-2y)
=∫(0→1) dx ∫(0→(1-x)/2) dy x(1-x-2y - 0)
=∫(0→1) dx ∫(0→(1-x)/2) x(1-x-2y)dy
=∫(0→1) dx x(y-xy-y²) |(0→(1-x)/2)
=∫(0→1) x[(1-x)/2 - x(1-x)/2 -(1-x)²/4]dx
=(1/4) ∫(0→1) (x-2x²+x³) dx
=(1/4) (x²/2 - 2x³/3 + x^4/4) |(0→1)
=(1/4) (1/2 - 2/3 +1/4)
=1/48本回答被网友采纳
=∫(0→1) dx ∫(0→(1-x)/2) dy xz |(0→1-x-2y)
=∫(0→1) dx ∫(0→(1-x)/2) dy x(1-x-2y - 0)
=∫(0→1) dx ∫(0→(1-x)/2) x(1-x-2y)dy
=∫(0→1) dx x(y-xy-y²) |(0→(1-x)/2)
=∫(0→1) x[(1-x)/2 - x(1-x)/2 -(1-x)²/4]dx
=(1/4) ∫(0→1) (x-2x²+x³) dx
=(1/4) (x²/2 - 2x³/3 + x^4/4) |(0→1)
=(1/4) (1/2 - 2/3 +1/4)
=1/48本回答被网友采纳
