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msgula1970
@msgula1970
August 2022
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Для функции y=f(x) найдите первообразную, график которой проходит через начало координат:
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Gerren
3) f(x)=x^2/3+sin(x+pi/3)
F(x)=x^3/9-cos(x+pi/3)+C
точка (0;0)
0=0-cos(pi/3)+c
c=cospi/3
c=1/2
F(x)=x^3/9-cos(x+pi/3)+1/2
4)f(x)=-x^3/2+cos(x-pi/6)
F(x)=-x^4/8+sin(pi/6-x)+C
(0;0)
0=0-sinpi/6+c
c=1/2
F(x)=-x^4/8+sin(pi/6-x)+1/2
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Answers & Comments
F(x)=x^3/9-cos(x+pi/3)+C
точка (0;0)
0=0-cos(pi/3)+c
c=cospi/3
c=1/2
F(x)=x^3/9-cos(x+pi/3)+1/2
4)f(x)=-x^3/2+cos(x-pi/6)
F(x)=-x^4/8+sin(pi/6-x)+C
(0;0)
0=0-sinpi/6+c
c=1/2
F(x)=-x^4/8+sin(pi/6-x)+1/2