Google
Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( 0 \right)=\sqrt{3}$ và $f\left( x \right).{f}'\left( x \right)=2{{\cos }^{2}}x-3,\forall x\in \mathbb{R}$.
Khi đó $\int\limits_{0}^{\dfrac{\pi }{4}}{{{f}^{2}}\left( x \right)dx}$ bằng
A. $\dfrac{4+6\pi -{{\pi }^{2}}}{8}$.
B. $\dfrac{-4+6\pi -{{\pi }^{2}}}{8}$.
C. $\dfrac{4-4\pi -{{\pi }^{2}}}{8}$.
D. $\dfrac{4+4\pi -{{\pi }^{2}}}{8}$.
Khi đó $\int\limits_{0}^{\dfrac{\pi }{4}}{{{f}^{2}}\left( x \right)dx}$ bằng
A. $\dfrac{4+6\pi -{{\pi }^{2}}}{8}$.
B. $\dfrac{-4+6\pi -{{\pi }^{2}}}{8}$.
C. $\dfrac{4-4\pi -{{\pi }^{2}}}{8}$.
D. $\dfrac{4+4\pi -{{\pi }^{2}}}{8}$.
Ta có: $f\left( x \right).{f}'\left( x \right)=2{{\cos }^{2}}x-3=1+\cos 2x-3=\cos 2x-2$
Lấy nguyên hàm 2 vế ta được
$\int{f\left( x \right).{f}'\left( x \right)dx}=\int{\left( \cos 2x-2 \right)dx}\Leftrightarrow \int{f\left( x \right)d\left[ f\left( x \right) \right]}=\dfrac{\sin 2x}{2}-2x+C$
$\Leftrightarrow \dfrac{{{f}^{2}}\left( x \right)}{2}=\dfrac{\sin 2x}{2}-2x+C\Leftrightarrow {{f}^{2}}\left( x \right)=\sin 2x-4x+C$
Mặt khác $f\left( 0 \right)=\sqrt{3}$ nên $C=3\Rightarrow {{f}^{2}}\left( x \right)=\sin 2x-4x+3$
Suy ra $\int\limits_{0}^{\dfrac{\pi }{4}}{{{f}^{2}}\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \sin 2x-4x+3 \right)dx}=\left. \left( -\dfrac{\cos 2x}{2}-2{{x}^{2}}+3x \right) \right|_{0}^{\dfrac{\pi }{4}}=-\dfrac{{{\pi }^{2}}}{8}+\dfrac{3\pi }{4}+\dfrac{1}{2}$.
Lấy nguyên hàm 2 vế ta được
$\int{f\left( x \right).{f}'\left( x \right)dx}=\int{\left( \cos 2x-2 \right)dx}\Leftrightarrow \int{f\left( x \right)d\left[ f\left( x \right) \right]}=\dfrac{\sin 2x}{2}-2x+C$
$\Leftrightarrow \dfrac{{{f}^{2}}\left( x \right)}{2}=\dfrac{\sin 2x}{2}-2x+C\Leftrightarrow {{f}^{2}}\left( x \right)=\sin 2x-4x+C$
Mặt khác $f\left( 0 \right)=\sqrt{3}$ nên $C=3\Rightarrow {{f}^{2}}\left( x \right)=\sin 2x-4x+3$
Suy ra $\int\limits_{0}^{\dfrac{\pi }{4}}{{{f}^{2}}\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \sin 2x-4x+3 \right)dx}=\left. \left( -\dfrac{\cos 2x}{2}-2{{x}^{2}}+3x \right) \right|_{0}^{\dfrac{\pi }{4}}=-\dfrac{{{\pi }^{2}}}{8}+\dfrac{3\pi }{4}+\dfrac{1}{2}$.
Đáp án A.