Google
Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục, đồng biến trên $\mathbb{R}$, có đạo hàm trên $\mathbb{R}$. thỏa mãn ${{\left[ f'\left( x \right) \right]}^{2}}=4f\left( x \right){{e}^{2x}}$ với $\forall x\in \mathbb{R}$ và $f'\left( 0 \right)=2$. Khi đó $\int\limits_{0}^{\ln 2}{{{x}^{2}}f\left( x \right)dx}$ bằng
A. $2{{\ln }^{2}}2-\ln 2+\dfrac{3}{4}.$
B. $2{{\ln }^{2}}2-2\ln 2+\dfrac{1}{4}.$
C. ${{\ln }^{2}}2-2\ln 2+\dfrac{3}{4}.$
D. $2{{\ln }^{2}}2-2\ln 2+\dfrac{3}{4}.$
A. $2{{\ln }^{2}}2-\ln 2+\dfrac{3}{4}.$
B. $2{{\ln }^{2}}2-2\ln 2+\dfrac{1}{4}.$
C. ${{\ln }^{2}}2-2\ln 2+\dfrac{3}{4}.$
D. $2{{\ln }^{2}}2-2\ln 2+\dfrac{3}{4}.$
HD: Ta có: ${{\left[ f'\left( x \right) \right]}^{2}}=4f\left( x \right){{e}^{2x}}$, do $f\left( x \right)$ liên tục, đồng biến trên $\mathbb{R}$ nên $f'\left( x \right)>0$
Mà ${{e}^{2x}}>0$ suy ra $f\left( x \right)>0$, ta có: $f'\left( x \right)=2\sqrt{f\left( x \right)}.{{e}^{x}}\Leftrightarrow \dfrac{f'\left( x \right)}{2\sqrt{f\left( x \right)}}={{e}^{x}}$
Lấy nguyên hàm hai vế ta được $\begin{aligned}
& \int{\dfrac{f'\left( x \right)dx}{2\sqrt{f\left( x \right)}}=\int{{{e}^{x}}dx\Leftrightarrow \int{\dfrac{d\left[ f\left( x \right) \right]}{2\sqrt{f\left( x \right)}}={{e}^{x}}+C\Leftrightarrow \sqrt{f\left( x \right)}={{e}^{x}}+C}}} \\
& \\
\end{aligned}$
$\Rightarrow f\left( x \right)={{\left( {{e}^{x}}+C \right)}^{2}}\Rightarrow f'\left( x \right)=2\left( {{e}^{x}}+C \right).{{e}^{x}}$
Thay $x=0\Rightarrow f'\left( 0 \right)=2\left( 1+C \right).1\Leftrightarrow C=0\Rightarrow f\left( x \right)={{e}^{2x}}$
Khi đó $\int\limits_{0}^{\ln 2}{{{x}^{2}}f\left( x \right)dx=}\int\limits_{0}^{\ln 2}{{{x}^{2}}{{e}^{2x}}dx\approx 0,3246={{\ln }^{2}}2-2\ln 2+\dfrac{3}{4}.}$
Mà ${{e}^{2x}}>0$ suy ra $f\left( x \right)>0$, ta có: $f'\left( x \right)=2\sqrt{f\left( x \right)}.{{e}^{x}}\Leftrightarrow \dfrac{f'\left( x \right)}{2\sqrt{f\left( x \right)}}={{e}^{x}}$
Lấy nguyên hàm hai vế ta được $\begin{aligned}
& \int{\dfrac{f'\left( x \right)dx}{2\sqrt{f\left( x \right)}}=\int{{{e}^{x}}dx\Leftrightarrow \int{\dfrac{d\left[ f\left( x \right) \right]}{2\sqrt{f\left( x \right)}}={{e}^{x}}+C\Leftrightarrow \sqrt{f\left( x \right)}={{e}^{x}}+C}}} \\
& \\
\end{aligned}$
$\Rightarrow f\left( x \right)={{\left( {{e}^{x}}+C \right)}^{2}}\Rightarrow f'\left( x \right)=2\left( {{e}^{x}}+C \right).{{e}^{x}}$
Thay $x=0\Rightarrow f'\left( 0 \right)=2\left( 1+C \right).1\Leftrightarrow C=0\Rightarrow f\left( x \right)={{e}^{2x}}$
Khi đó $\int\limits_{0}^{\ln 2}{{{x}^{2}}f\left( x \right)dx=}\int\limits_{0}^{\ln 2}{{{x}^{2}}{{e}^{2x}}dx\approx 0,3246={{\ln }^{2}}2-2\ln 2+\dfrac{3}{4}.}$
Đáp án C.