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Инессо4ка
@Инессо4ка
June 2022
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решить определенный интеграл (верхний предел 1, нижний - 1)
sin2*x-cos*3x+1 dx
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11qqaa
[-1;1] ∫ (sin2x-cos3x+1) dx = = [-1;1] (-1/2 cos2x - 1/3 sin3x +x) = = -1/2 ( cos (2*1) - cos(2*(-1) ) - 1/3 ( sin(3*1) - sin(3*(-1) ) + 1 -(-1) == -1/2 ( cos (2) - cos(-2) ) - 1/3 ( sin(3) - sin(-3) ) + 2 == -1/2 ( cos (2) - cos(2) ) - 1/3 ( sin(3) + sin(3) ) + 2 == -1/2 * 0 - 1/3 * 2sin(3) + 2 == 2/3 * ( 3 - sin(3) )
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