Найти площадь y = – x2 + 4x – 3 y = 0
–x2+4x–3=0 x2–4x+3=0 D=16–12=4 x1=1; x2=3 S= ∫ 31(–x2+4x–3)dx=((–x3/3)+4·(x2/2)–3x)|31= =(–33/3)+4·(32/2)–3·3–((–13/3)+4·(12/2)–3·1)= =–9+18–9–((–1/3)+2–3)=0–(–4/3)=0+(4/3)=4/3=1 целая 1/3