найти площадь, ограниченную линиями

y=x^2-3x+3, y=-x^2+x+9

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