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Ответ

Ответ дан
George Mitin
1)Первообразная от f(x)=x^2+4x
F(x)=1/3*x^3+2x^2+C
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2) График F(x) проходит через M(2;1), значит
F(2)=1
1/3*2^3+2*2^2+C=1
8/3+8+C=1
C=-7-2 -2/3
C= -9 -2/3
С= -(9 2/3)
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3) Ответ: первообразная от f(x), график которой проходит через точку M(2;1):
F(x)=1/3*x^3+2x^2-(9 2/3)
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Ответ

Ответ дан
Катя Лосева
f(x)=1/3x^3+2x^2+C; 1=8/3+8+C; C=-9 2/3; Ответ: f(x)=1/3x^3+2x^2-9 2/3.

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