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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( 4 \right)=\dfrac{4}{3}$ và $f\left( x \right)=x\left( 1+\dfrac{1}{\sqrt{x}}-{f}'\left( x \right) \right)$, $\forall x>0$. Khi đó $\int\limits_{1}^{4}{xf\left( x \right)\text{d}x}$ bằng
A. $\dfrac{1283}{30}$.
B. $-\dfrac{157}{30}$.
C. $\dfrac{157}{30}$.
D. $-\dfrac{1283}{30}$.
A. $\dfrac{1283}{30}$.
B. $-\dfrac{157}{30}$.
C. $\dfrac{157}{30}$.
D. $-\dfrac{1283}{30}$.
Với mọi $ x>0$ ta có: $f\left( x \right)+x{f}'\left( x \right)=x+\sqrt{x}$ $\Leftrightarrow {{\left( x.f\left( x \right) \right)}^{\prime }}=x+\sqrt{x}$.
Lấy nguyên hàm hai vế ta được: $x.f\left( x \right)=\dfrac{{{x}^{2}}}{2}+\dfrac{2}{3}x\sqrt{x}+C$.
Mà $f\left( 4 \right)=\dfrac{4}{3}\Rightarrow C=-8$ $\Rightarrow xf\left( x \right)=\dfrac{{{x}^{2}}}{2}+\dfrac{2}{3}x\sqrt{x}-8$.
Vậy $\int\limits_{1}^{4}{xf\left( x \right)\text{d}x}=\int\limits_{1}^{4}{\left( \dfrac{{{x}^{2}}}{2}+\dfrac{2}{3}x\sqrt{x}-8 \right)}\text{d}x=-\dfrac{157}{30}.$
Lấy nguyên hàm hai vế ta được: $x.f\left( x \right)=\dfrac{{{x}^{2}}}{2}+\dfrac{2}{3}x\sqrt{x}+C$.
Mà $f\left( 4 \right)=\dfrac{4}{3}\Rightarrow C=-8$ $\Rightarrow xf\left( x \right)=\dfrac{{{x}^{2}}}{2}+\dfrac{2}{3}x\sqrt{x}-8$.
Vậy $\int\limits_{1}^{4}{xf\left( x \right)\text{d}x}=\int\limits_{1}^{4}{\left( \dfrac{{{x}^{2}}}{2}+\dfrac{2}{3}x\sqrt{x}-8 \right)}\text{d}x=-\dfrac{157}{30}.$
Đáp án B.