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Nikolaypetryk
@Nikolaypetryk
August 2022
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1) ∫с верху 2п с низу п/2 (cos8x-e^(-3x))dx; помогите пожалуйста очень нужно
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madikmucuy
1/8*sin8x+1/(3e^3x)|2π-π/2=1/8*sin16π+1/(3e^6π)-1/8*sin4π-1/(3e^1,5π)=
=1/8*0+1/(3e^6π)-1/8*0-1/(3e^1,5π)=(1-e^4π)/e^6π
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sedinalana
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Answers & Comments
=1/8*0+1/(3e^6π)-1/8*0-1/(3e^1,5π)=(1-e^4π)/e^6π
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Находим первообразные и подставляем границы