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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục, có đạo hàm trên $\mathbb{R}$ thỏa mãn $x-{{f}^{3}}\left( x \right)-f\left( x \right)+3=0$. Tính tích phân $I=\int\limits_{-1}^{7}{x{f}'\left( x \right)\text{d}x}$
A. $\dfrac{5}{4}$.
B. $\dfrac{3}{4}$.
C. $\dfrac{9}{4}$.
D. $\dfrac{51}{4}$.
A. $\dfrac{5}{4}$.
B. $\dfrac{3}{4}$.
C. $\dfrac{9}{4}$.
D. $\dfrac{51}{4}$.
Từ giả thiết ta có: $x-{{f}^{3}}\left( x \right)-f\left( x \right)+3=0$
$x=7\Rightarrow 7-{{f}^{3}}\left( 7 \right)-f\left( 7 \right)+3=0\Leftrightarrow f\left( 7 \right)=2$
$x=-1\Rightarrow -1-{{f}^{3}}\left( -1 \right)-f\left( -1 \right)+3=0\Leftrightarrow f\left( -1 \right)=1$.
$x-{{f}^{3}}\left( x \right)-f\left( x \right)+3=0\Rightarrow x{f}'\left( x \right)=\left[ {{f}^{3}}\left( x \right)+f\left( x \right)-3 \right]{f}'\left( x \right)$
$\Rightarrow \int\limits_{-1}^{7}{x{f}'\left( x \right)\text{d}x}=\int\limits_{-1}^{7}{\left[ {{f}^{3}}\left( x \right)+f\left( x \right)-3 \right]{f}'\left( x \right)\text{d}x}=\left. \left[ \dfrac{1}{4}{{f}^{4}}\left( x \right)+\dfrac{1}{2}{{f}^{2}}\left( x \right)-3f\left( x \right) \right] \right|_{-1}^{7}$
$=\dfrac{1}{4}{{f}^{4}}\left( 7 \right)+\dfrac{1}{2}{{f}^{2}}\left( 7 \right)-3f\left( 7 \right)-\left[ \dfrac{1}{4}{{f}^{4}}\left( -1 \right)+\dfrac{1}{2}{{f}^{2}}\left( -1 \right)-3f\left( -1 \right) \right]=0-\left( -\dfrac{9}{4} \right)=\dfrac{9}{4}$
$x=7\Rightarrow 7-{{f}^{3}}\left( 7 \right)-f\left( 7 \right)+3=0\Leftrightarrow f\left( 7 \right)=2$
$x=-1\Rightarrow -1-{{f}^{3}}\left( -1 \right)-f\left( -1 \right)+3=0\Leftrightarrow f\left( -1 \right)=1$.
$x-{{f}^{3}}\left( x \right)-f\left( x \right)+3=0\Rightarrow x{f}'\left( x \right)=\left[ {{f}^{3}}\left( x \right)+f\left( x \right)-3 \right]{f}'\left( x \right)$
$\Rightarrow \int\limits_{-1}^{7}{x{f}'\left( x \right)\text{d}x}=\int\limits_{-1}^{7}{\left[ {{f}^{3}}\left( x \right)+f\left( x \right)-3 \right]{f}'\left( x \right)\text{d}x}=\left. \left[ \dfrac{1}{4}{{f}^{4}}\left( x \right)+\dfrac{1}{2}{{f}^{2}}\left( x \right)-3f\left( x \right) \right] \right|_{-1}^{7}$
$=\dfrac{1}{4}{{f}^{4}}\left( 7 \right)+\dfrac{1}{2}{{f}^{2}}\left( 7 \right)-3f\left( 7 \right)-\left[ \dfrac{1}{4}{{f}^{4}}\left( -1 \right)+\dfrac{1}{2}{{f}^{2}}\left( -1 \right)-3f\left( -1 \right) \right]=0-\left( -\dfrac{9}{4} \right)=\dfrac{9}{4}$
Đáp án C.