一道比较复杂的定积分,求大神教一下怎么做..谢谢

2025-05-10 21:31:48
推荐回答(1个)
回答1:

分为2个部分
∫[0,2]2ydy = y^2[0,2] = 4

∫[0,2]y√(2y-y^2)dy
= ∫[0,2]y√[1-(y-1)^2]dy
y-1=sinu y=0 u=-π/2 y=2,u=π/2
=∫[-π/2,π/2] (1+sinu)cosudsinu
=∫[-π/2,π/2] (1+sinu)cosu^2du
= ∫[-π/2,π/2](1+sinu)(1-sinu^2)du
= ∫[-π/2,π/2](1+sinu-sinu^2-sinu^3)du
=[1-cosu+(1/2)(cos2u-1)+cosu-cosu^3/3] |[-π/2,π/2]
=π/2