Question

Найдите площадь фигуры ограничений
y=4x+x^2 и y=4+x

Answer 1

x²+4x=4+x
x²+3x-4=0
D=9+16=25
x1=-3+5/2=2/2=1
x2=-3-5/2=-8/2=-4

(-4;1) ∫(x²+3x-4) dx = x³/3+3x²/2-4x | (-4;1) = 1/3+3/2-4 - (-64/3+ 24+16) =

= 11/6-4+64/3-40 = -44+11/6+128/6= -44+139/6 = -43 6/6 + 23 1/6 =
=-20 5/6