Pt ca F sa fie primitiva a lui f, trebuie ca F' = f;
F'(x) = [tex] (\frac{2x+1}{3}* \sqrt{2x+1} )'[/tex]
se aplica (fg)' = f' * g + f * g'
si [tex](\sqrt{u})' =u'* \frac{1}{2 \sqrt{u}} [/tex]
[tex] \frac{1}{3}(2* \sqrt{2x+1}+(2x+1)* \frac{2}{2* \sqrt{2x+1}}) [/tex]
[tex] \frac{1}{3}(\frac{2*\sqrt{2x+1}* \sqrt{2x+1}+2x+1}{\sqrt{2x+1}})[/tex]
[tex]\frac{1}{3}(\frac{2*(2x+1)+2x+1}{\sqrt{2x+1}}) = \frac{1}{3}(\frac{3*(2x+1)}{\sqrt{2x+1}})[/tex]
[tex]\frac{(2x+1)}{\sqrt{2x+1}}= \sqrt{2x+1} [/tex]
=> F' = f, F primitiva a lui f
b) aria suprafetei e integrala definita a lui f intre 4 si 12, iar primitiva F o cunoastem deja:
[tex] \int\limits^a_b {x} \, dx = F(b) - F(a)[/tex]
deci aria = F(12) - F(4) =
[tex]\frac{2*12+1}{3}* \sqrt{2*12+1} - \frac{2*4+1}{3}* \sqrt{2*4+1} [/tex]
[tex]\frac{25}{3}* \sqrt{25} - \frac{9}{3}* \sqrt{9}= \frac{125}{3} - \frac{27}{3} = \frac{98}{3} [/tex]
