Ex 5. Functia f admite primitive daca are proprietatea Darboux.
Se pune problema continuitatii in x=2
Ls=limf(x) x→2 x<2 =lim(x+2)(x-2)/2·(x-2)=lim x+2)/2=2
Ld=lim f(x) x→2 x>2 (x²+3x+4)=2²+3·2+4=2
f(2)=(2+2)/2=4/2=2
Ls=Ld=f(s) functia e continua in 2 , si admite primitiva
Ex 6
∫x²/(x+1)dx= scazi si adui1 la numarator=∫(x²-1+1)dx/(x+1)=∫(x²-1)dx/(x+1)+∫dx/(x+1)=
∫(x+1)*(x-1)/(x+1)dx+ln(x+1)+C=∫(x-1)dx+ln(x+1)+C=∫xdx-∫dx+ln(x+1)+c=
x²/2-x+ln(x+1)+C relatia 1
∫xdx=x²/2+c relatia2
∫dx=x+c rel 3
Adui Relatiile 1 2, 3 si obtii
∫(f(x)-x+1)dx=x²/2-x+ln(x+1)-x²/2+x+C=ln(x+1)=C
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log[2](x+3)=2
(x+3)=2²
x+3=4
x=4-3=1
log[x[(2x-1)=2 Pui conditile de existenta x>0 x≠1
x²=2x-1=>x²-2x+1=0 (x-1)²=0 =?x=1 nu se accepta din cauza conditiilor de existenta a logaritmului
Adui
