Răspuns
x≥0
F(x)=∫(2+3√x)dx=∫2dx+3∫√xdx=2x+3∫x^1/2dx=2x+3*x(1/2+1)/(1/2+1)+c=
2x+3x^(3/2)/(3/2)+c=2x+3*2x^(3/2)/3+c
2x+2x^(3/2)+c
f(0)=2*0+2*0+c=3
0+0+c=3
c=3
