Câu hỏi:
Cho hàm số \(f\left( x \right)\) thỏa mãn \( – xf’\left( x \right).\ln x + f\left( x \right) = 2{x^2}{f^2}\left( x \right),\,\,\forall x \in \left( {1; + \infty } \right)\), \(f\left( x \right) > 0,\forall x \in \left( {1; + \infty } \right)\)và \(f\left( {\rm{e}} \right) = \frac{1}{{{{\rm{e}}^2}}}\). Tính diện tích \(S\)hình phẳng giới hạn bởi đồ thị \(y = xf\left( x \right),y = 0,x = e,x = {e^2}\).
A. \(S = \frac{3}{2}\).
B. \(S = \frac{1}{2}\).
C. \(S = \frac{5}{3}\).
D. \(S = 2\).
LỜI GIẢI CHI TIẾT
Ta có: \( – xf’\left( x \right)\ln x + f\left( x \right) = 2{x^2}{f^2}\left( x \right) \Leftrightarrow – x\frac{{f’\left( x \right)}}{{{f^2}\left( x \right)}}\ln x + \frac{1}{{f\left( x \right)}} = 2{x^2}\), \(\forall x \in \left( {1; + \infty } \right)\).
\( \Leftrightarrow xg’\left( x \right).\ln x + g\left( x \right) = 2{x^2},\,\,\forall x \in \left( {1; + \infty } \right)\) với \(g\left( x \right) = \frac{1}{{f\left( x \right)}}\)
\( \Leftrightarrow g’\left( x \right)\ln x + \frac{{g\left( x \right)}}{x} = 2x\),\(\forall x \in \left( {1; + \infty } \right)\)\( \Rightarrow \int {g’\left( x \right)\ln x{\rm{d}}x + \int {\frac{{g\left( x \right)}}{x}{\rm{d}}x = \int {2x{\rm{d}}x} } } \)
\( \Leftrightarrow g\left( x \right)\ln x – \int {\frac{{g\left( x \right)}}{x}{\rm{d}}x + \int {\frac{{g\left( x \right)}}{x}{\rm{d}}x = {x^2} + C} } \)\( \Leftrightarrow g\left( x \right)\ln x = {x^2} + C\), \(\forall x \in \left( {1; + \infty } \right)\).
Do \(f\left( {\rm{e}} \right) = \frac{1}{{{{\rm{e}}^2}}} \Leftrightarrow g\left( e \right) = {e^2} \Leftrightarrow C = 0\).
Suy ra \(g\left( x \right)\ln x = {x^2}\),\(\forall x \in \left( {1; + \infty } \right)\)
\( \Rightarrow g\left( x \right) = \frac{{{x^2}}}{{\ln x}} > 0,\,\,\forall x \in \left( {1; + \infty } \right)\)
\( \Rightarrow y = xf\left( x \right) = \frac{x}{{g\left( x \right)}} = \frac{{\ln x}}{x}\), \(\forall x \in \left( {1; + \infty } \right)\).
Ta có\(S = \int_{\rm{e}}^{{{\rm{e}}^2}} {xf\left( x \right){\rm{d}}x} = \int_{\rm{e}}^{{{\rm{e}}^2}} {\frac{{\ln x}}{x}{\rm{d}}x} = \frac{1}{2}{\ln ^2}x\left| \begin{array}{l}{{\rm{e}}^2}\\{\rm{e}}\end{array} \right. = \frac{3}{2}\).
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