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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục, nhận giá trị dương trên $\left( 0; +\infty \right)$, $f\left( 1 \right)=1$ và thỏa mãn ${{x}^{3}}f\left( x \right)+2{{f}^{3}}\left( x \right)=2{{x}^{4}}{f}'\left( x \right), \forall x\in \left( 0; +\infty \right)$. Tính diện tích hình phẳng giới hạn bởi đồ thị hàm số $y=f\left( x \right)$, trục hoành và hai đường thẳng $x=1; x=4$.
A. $\dfrac{15}{2}$.
B. $\dfrac{14}{3}$.
C. $\dfrac{255}{4}$.
D. $\dfrac{62}{5}$.
A. $\dfrac{15}{2}$.
B. $\dfrac{14}{3}$.
C. $\dfrac{255}{4}$.
D. $\dfrac{62}{5}$.
Ta có ${{x}^{3}}f\left( x \right)+2{{f}^{3}}\left( x \right)=2{{x}^{4}}{f}'\left( x \right)\Leftrightarrow {{x}^{3}}\left[ f\left( x \right)-2x{f}'\left( x \right) \right]=-2{{f}^{3}}\left( x \right)\Leftrightarrow \dfrac{{{x}^{3}}\left[ f\left( x \right)-2x{f}'\left( x \right) \right]}{{{f}^{3}}\left( x \right)}=-2$
$\Leftrightarrow {{x}^{3}}.\dfrac{{{f}^{2}}\left( x \right)-2xf\left( x \right).{f}'\left( x \right)}{{{f}^{4}}\left( x \right)}=-2\Leftrightarrow {{x}^{3}}.\dfrac{{{\left( x \right)}^{\prime }}.{{f}^{2}}\left( x \right)-x.{{\left[ {{f}^{2}}\left( x \right) \right]}^{\prime }}}{{{f}^{4}}\left( x \right)}=-2$ $\Leftrightarrow {{\left[ \dfrac{x}{{{f}^{2}}\left( x \right)} \right]}^{\prime }}=\dfrac{-2}{{{x}^{3}}}={{\left( \dfrac{1}{{{x}^{2}}} \right)}^{\prime }}$ $\Rightarrow \dfrac{x}{{{f}^{2}}\left( x \right)}=\dfrac{1}{{{x}^{2}}}+C$. Cho $x=1\Rightarrow 1=1+C\Leftrightarrow C=0\Rightarrow {{f}^{2}}\left( x \right)={{x}^{3}}\Leftrightarrow f\left( x \right)=x\sqrt{x}$.
Vậy $S=\int\limits_{1}^{4}{x\sqrt{x} \text{d}x}=\dfrac{62}{5}$.
$\Leftrightarrow {{x}^{3}}.\dfrac{{{f}^{2}}\left( x \right)-2xf\left( x \right).{f}'\left( x \right)}{{{f}^{4}}\left( x \right)}=-2\Leftrightarrow {{x}^{3}}.\dfrac{{{\left( x \right)}^{\prime }}.{{f}^{2}}\left( x \right)-x.{{\left[ {{f}^{2}}\left( x \right) \right]}^{\prime }}}{{{f}^{4}}\left( x \right)}=-2$ $\Leftrightarrow {{\left[ \dfrac{x}{{{f}^{2}}\left( x \right)} \right]}^{\prime }}=\dfrac{-2}{{{x}^{3}}}={{\left( \dfrac{1}{{{x}^{2}}} \right)}^{\prime }}$ $\Rightarrow \dfrac{x}{{{f}^{2}}\left( x \right)}=\dfrac{1}{{{x}^{2}}}+C$. Cho $x=1\Rightarrow 1=1+C\Leftrightarrow C=0\Rightarrow {{f}^{2}}\left( x \right)={{x}^{3}}\Leftrightarrow f\left( x \right)=x\sqrt{x}$.
Vậy $S=\int\limits_{1}^{4}{x\sqrt{x} \text{d}x}=\dfrac{62}{5}$.
Đáp án D.