Cho hàm số \(y = f\left( x \right) >0\) xác định, có đạo hàm trên đoạn \(\left[ {0;1} \right]\) và thỏa mãn: \(g\left( x \right) = 1 + 2020\int\limits_0^x {f\left( t \right){\rm{dt}}} \), \(g\left( x \right) = {f^2}\left( x \right)\). Tính \(\int\limits_0^1 {\sqrt {g\left( x \right)} {\rm{d}}x} \).

Ta có \(g\left( x \right) = 1 + 2020\int\limits_0^x {f\left( t \right){\rm{dt}}} \)\( \Rightarrow g'\left( x \right) = 2020f\left( x \right) = 2020\sqrt {g\left( x \right)} \)

\( \Rightarrow \frac{{g'\left( x \right)}}{{\sqrt {g\left( x \right)} }} = 2020\)\( \Rightarrow \int\limits_0^t {\frac{{g'\left( x \right)}}{{\sqrt {g\left( x \right)} }}{\rm{d}}x = 2020} \int\limits_0^t {{\rm{d}}x} \)\( \Rightarrow \left. {2\left( {\sqrt {g\left( x \right)} } \right)} \right|_0^t = \left. {2020x} \right|_0^t\)

\( \Rightarrow 2\left( {\sqrt {g\left( t \right)} - 1} \right) = 2020t\) (do \(g\left( 0 \right) = 1\))

\( \Rightarrow \sqrt {g\left( t \right)} = 1010t + 1\)

\( \Rightarrow \int\limits_0^1 {\sqrt {g\left( t \right)} {\rm{dt}}} = \left. {\left( {505{t^2} + t} \right)} \right|_0^1 = 506\).