определим границы площади
y=x^2-3x+3,
y=-x^2+x+9
приравняем по у
x^2-3x+3 = -x^2+x+9
2x^2-4x-6=0
x^2-2x-3=0
x1=-1 ; x2=3
интервал [-1;3]
S = S2 - S1 =
=∫(-x^2+x+9)dx - ∫(x^2-3x+3)dx =
= -x^3/3+x^2/2+9x |[-1;3] - x^3/3-3x^2/2+3x |[-1;3] =
=-3^3/3+3^2/2+9*3-[-(-1)^3/3+(-1)^2/2+9(-1)]-(3^3/3-3*3^2/2+3*3-[(-1)^3/3-3*(-1)^2/2+3*(-1)])=
=64/3
ответ 64/3
y=x^2-3x+3,
y=-x^2+x+9
x^2-3x+3 = -x^2+x+9
2x^2-4x-6=0
x^2-2x-3=0
x1=-1 ; x2=3
S = S2 - S1 =
=∫(-x^2+x+9)dx - ∫(x^2-3x+3)dx =
= -x^3/3+x^2/2+9x |[-1;3] - x^3/3-3x^2/2+3x |[-1;3] =
=-3^3/3+3^2/2+9*3-[-(-1)^3/3+(-1)^2/2+9(-1)]-(3^3/3-3*3^2/2+3*3-[(-1)^3/3-3*(-1)^2/2+3*(-1)])=
=64/3
