Ответ:
1) 8/3 ; 2) 8
Пошаговое объяснение:
1)
∫ (x^2+x-1) dx [0;2] = (1/3 x^3 + 1/2 x^2 -x) [0;2] =
= (1/3 *2^3 + 1/2 *2^2 -2) - (1/3 *0^3 + 1/2 *0^2 -0) =
= 8/3 + 2 -2 - 0 = 8/3
2)
x^3 - 4x = x (x^2-4) = x (x-2)(x+2)
x = -2; x =0; x =2
∫ (x^3 - 4x) dx [-2;0] + ∫ -(x^3 - 4x) dx [0;2] = (x^4/4 - 2x^2) [-2;0] - - (x^4/4 - 2x^2) [0;2] = 0 - ((-2)^4/4 - 2*(-2)^2)) - (2^4/4 - 2*2^2)+0 =
= 0 + 4 +4 +0 = 8