Question

O primitiva cu F(0)=3 plz

Answer 1

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