Đáp án

Giải thích

Ta có: \(\left( {\sqrt x  + \sqrt {x + 2} } \right)f'\left( x \right) = 1 \Leftrightarrow f'\left( x \right) = \frac{1}{{\sqrt x  + \sqrt {x + 2} }} \Leftrightarrow \mathop \smallint \nolimits^ f'\left( x \right)dx = \mathop \smallint \nolimits^ \frac{{dx}}{{\sqrt x  + \sqrt {x + 2} }}\)

\( \Leftrightarrow f\left( x \right) = \frac{1}{2}\mathop \smallint \nolimits^ \left( {\sqrt {x + 2}  - \sqrt x } \right)dx = \frac{1}{2}.\frac{2}{3}\left( {\sqrt {{{(x + 2)}^3}}  - \sqrt {{x^3}} } \right) + C = \frac{1}{3}\left( {\sqrt {{{(x + 2)}^3}}  - \sqrt {{x^3}} } \right) + C\)

\(f\left( 0 \right) = \frac{1}{3} \Leftrightarrow \frac{1}{3}.2\sqrt 2  + C = \frac{1}{3} \Leftrightarrow C = \frac{1}{3}\left( {1 - 2\sqrt 2 } \right)\)

\( \Rightarrow f\left( x \right) = \frac{1}{3}\left( {\sqrt {{{(x + 2)}^3}}  - \sqrt {{x^3}}  + 1 - 2\sqrt 2 } \right)\)

\[ \Rightarrow \mathop \smallint \nolimits^ _0^2f\left( x \right)dx = \frac{1}{3}\mathop \smallint \nolimits^ _0^2\left( {\sqrt {{{(x + 2)}^3}}  - \sqrt {{x^3}}  + 1 - 2\sqrt 2 } \right)dx = \left. {\frac{1}{3}.\left( {\frac{2}{5}\sqrt {{{(x + 2)}^5}}  - \frac{2}{5}\sqrt {{x^5}}  + (1 - 2\sqrt 2 )x} \right)} \right|_0^2 = \frac{{74 - 36\sqrt 2 }}{{15}}\]