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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\left[ 0;1 \right]$ đồng thời thỏa mãn các điều kiện $f\left( 1 \right)=2$ và ${{f}^{2}}\left( x \right)-xf\left( x \right)f'\left( x \right)=2x+6,\forall x\in \left[ 0;1 \right]$. Tích phân $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}$ bằng
A. 3.
B. 4.
C. 5.
D. 6.
A. 3.
B. 4.
C. 5.
D. 6.
Lấy tích phân hai vế đẳng thức đã cho $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}-\int\limits_{0}^{1}{xf\left( x \right)f'\left( x \right)dx}=\int\limits_{0}^{1}{\left( 2x+6 \right)dx}=7$
Mặt khác tích phân từng phần có:
$\int\limits_{0}^{1}{xf\left( x \right)f'\left( x \right)dx}=\int\limits_{0}^{1}{xd\left( \dfrac{1}{2}{{f}^{2}}\left( x \right) \right)}=\left. \dfrac{x}{2}{{f}^{2}}\left( x \right) \right|_{0}^{1}-\dfrac{1}{2}\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=\dfrac{1}{2}{{f}^{2}}\left( 1 \right)-\dfrac{1}{2}\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=2-\dfrac{1}{2}\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}$
Vậy $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}-\left( 2-\dfrac{1}{2}\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx} \right)=7\Leftrightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=6$
Mặt khác tích phân từng phần có:
$\int\limits_{0}^{1}{xf\left( x \right)f'\left( x \right)dx}=\int\limits_{0}^{1}{xd\left( \dfrac{1}{2}{{f}^{2}}\left( x \right) \right)}=\left. \dfrac{x}{2}{{f}^{2}}\left( x \right) \right|_{0}^{1}-\dfrac{1}{2}\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=\dfrac{1}{2}{{f}^{2}}\left( 1 \right)-\dfrac{1}{2}\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=2-\dfrac{1}{2}\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}$
Vậy $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}-\left( 2-\dfrac{1}{2}\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx} \right)=7\Leftrightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=6$
Đáp án D.