Giải thích các bước giải:
Ta có :
$f(x).f'(x)=\sqrt{4-x^2}$
$\to \int f(x).f'(x)dx=\int\sqrt{4-x^2}dx$
$\to \int f(x)d(f(x))=2\arcsin \left(\dfrac{1}{2}x\right)+\sin \left(2\arcsin \left(\dfrac{1}{2}x\right)\right)+c$
$\to \dfrac{f^2(x)}{2}=2\arcsin \left(\dfrac{1}{2}x\right)+\sin \left(2\arcsin \left(\dfrac{1}{2}x\right)\right)+c$
Vì $f(0)=3$
$\to \dfrac{f^2(0)}{2}=2\arcsin \left(\dfrac{1}{2}.0\right)+\sin \left(2\arcsin \left(\dfrac{1}{2}.0\right)\right)+c=c$
$\to c=3$
$\to \dfrac{f^2(x)}{2}=2\arcsin \left(\dfrac{1}{2}x\right)+\sin \left(2\arcsin \left(\dfrac{1}{2}x\right)\right)+3$
$\to \dfrac{f^2(2)}{2}=\pi+3$
$\to f^2(2)=2\pi+6$
