Vei rezolva inegalitatea x²>x+2 <=>
x²-x-2>0
rezolvi ecuatia atsata x1= -1, x2=2, Intre radacini functia e negativa deci
x²-x-2≤0 adica x²<x-2 Pt x∈[-1 ,2]∩[-3,2]=[-1,2]
si x²-x-2>0 in intervalul (-∞ ,-1)U[2,∞) ∩[-3,2]=[-3,-1)
Explicitezi functia
f(x)={ x+2 pt x∈[-1,2]
{x² pt x∈[-3, -1)
Problema continuitatii se pune in x=-1.Calculezi limitele laterale
Ls : x→-1; x≤ -1 lim (x+2)=-1+2=1
Ld: x→ -1 x> -1 lim (x²=1
f(-1)=-1+2=1
Ls=Ld=f(-1)=1 .Functia e continua in 1 si este integrabila pe domeniul de definitie
I=∫x²dx x∈[-3 ,-1]dx +∫(x+2)dx x∈[-1 ,2]
pt x∈[-3, -1] ∫x²dx=x³/3=conf formulei Leibniz Newton [(-1/3)]³-(-3/3)³]=-1/27+1=26/27
x∈[-1 ,2] ∫(x+2)dx=(x²/2+2x) l-1↑2=4/4+4-1/2-2=5/2
I=26/27+5/2=187/54
