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Câu hỏi: Cho hàm số $y=f\left( x \right)>0$ xác định, có đạo hàm trên đoạn $\left[ 0;1 \right]$ và thỏa mãn $g\left( x \right)=1+2018\int\limits_{0}^{x}{f\left( t \right)dt},g\left( x \right)={{f}^{2}}\left( x \right)$. Tính $\int\limits_{0}^{1}{\sqrt{g\left( x \right)}dx}$.
A. $\dfrac{1011}{2}$.
B. $\dfrac{1009}{2}$.
C. $\dfrac{2019}{2}$.
D. 505.
A. $\dfrac{1011}{2}$.
B. $\dfrac{1009}{2}$.
C. $\dfrac{2019}{2}$.
D. 505.
Ta có $g\left( x \right)=1+2018\int\limits_{0}^{x}{f\left( t \right)dt}\Rightarrow {g}'\left( x \right)=2018f\left( x \right)=2018\sqrt{g\left( x \right)}$
Suy ra $\dfrac{{g}'\left( x \right)}{\sqrt{g\left( x \right)}}=2018\Leftrightarrow \int\limits_{0}^{t}{\dfrac{{g}'\left( x \right)}{\sqrt{g\left( x \right)}}}dx=2018\int\limits_{0}^{t}{dx}\Rightarrow 2\left( \sqrt{g\left( t \right)}-1 \right)=2018t$
$\Rightarrow \sqrt{g\left( t \right)}=1009t+1\Rightarrow \int\limits_{0}^{1}{\sqrt{g\left( t \right)}dt}=\dfrac{1011}{2}$.
Suy ra $\dfrac{{g}'\left( x \right)}{\sqrt{g\left( x \right)}}=2018\Leftrightarrow \int\limits_{0}^{t}{\dfrac{{g}'\left( x \right)}{\sqrt{g\left( x \right)}}}dx=2018\int\limits_{0}^{t}{dx}\Rightarrow 2\left( \sqrt{g\left( t \right)}-1 \right)=2018t$
$\Rightarrow \sqrt{g\left( t \right)}=1009t+1\Rightarrow \int\limits_{0}^{1}{\sqrt{g\left( t \right)}dt}=\dfrac{1011}{2}$.
Đáp án A.