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Малая33
@Малая33
July 2022
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Найти площадь фигуры, ограниченной линиями
1) Y=1/2cosx; y=0; x=0; y=x+2;
2) у=1/2х^2-4х+10; у=х+2
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nafanya2014
Verified answer
1) S= S(ΔABO)-S(криволинейного Δ СMDO)=
=(1/2)·AO·BO -∫⁰₍₋π/₂₎(1/2)cosxdx=
=2+(1/2)(sinx)|⁰₍₋π/₂₎=2+(1/2)sin0-(1/2)sin(-π/2)=2+(1/2)=2,5
2)S=S( трапеции АВСD) - S ( криволинейной ABMCD)=
=(4+10)·(8-2)/2-∫₂⁸((1/2)x²-4x+10)dx=
=42-((x³/6)-2x²+10x)|₂⁸=
=42-((8³/6)-2·8²+10·2)+((2³/6)-2·2²+10·2)=
=42-85+128-20-8+20=77
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Answers & Comments
Verified answer
1) S= S(ΔABO)-S(криволинейного Δ СMDO)==(1/2)·AO·BO -∫⁰₍₋π/₂₎(1/2)cosxdx=
=2+(1/2)(sinx)|⁰₍₋π/₂₎=2+(1/2)sin0-(1/2)sin(-π/2)=2+(1/2)=2,5
2)S=S( трапеции АВСD) - S ( криволинейной ABMCD)=
=(4+10)·(8-2)/2-∫₂⁸((1/2)x²-4x+10)dx=
=42-((x³/6)-2x²+10x)|₂⁸=
=42-((8³/6)-2·8²+10·2)+((2³/6)-2·2²+10·2)=
=42-85+128-20-8+20=77