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14091991
@14091991
August 2022
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вычислите площадь фигуры ограниченной линиями
y=3sinx,y=-2sinx, 0<=x<=2пи/3
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sangers1959
Verified answer
Y=3sinx y=-2sinx 0≤x≤2π/3
S=∫(3sinx-(-2sinx))dx(2π/3;0)=∫(5sinx)dx(2π/3;0)=-5cosx(2π/3;0)=-5*((-0,5)-1)=7,5.
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Thanks 5
14091991
можешь построить график
sangers1959
Я на работе, здесь нет условий чертить и фотографировать.
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Answers & Comments
Verified answer
Y=3sinx y=-2sinx 0≤x≤2π/3S=∫(3sinx-(-2sinx))dx(2π/3;0)=∫(5sinx)dx(2π/3;0)=-5cosx(2π/3;0)=-5*((-0,5)-1)=7,5.