Hướng dẫn giải

Đặt \(\left\{ \begin{array}{l}u = f\left( x \right)\\dv = {x^2}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = f'\left( x \right)dx\\v = \frac{{{x^3}}}{3}\end{array} \right.\)

Ta có \(\int\limits_0^1 {{x^2}f\left( x \right)dx} = \frac{{{x^3}f\left( x \right)}}{3}\left| {_{\scriptstyle\atop\scriptstyle0}^{\scriptstyle1\atop\scriptstyle}} \right. - \frac{1}{3}\int\limits_0^1 {{x^3}f'\left( x \right)dx} \)

\( \Rightarrow - \frac{1}{3}\int\limits_0^1 {{x^3}.f'\left( x \right)} dx = \frac{1}{3} \Rightarrow \int\limits_0^1 {{x^3}.f'\left( x \right)d{\rm{x}} = - 1.} \)

Cách 1: Ta có \(\int\limits_0^1 {{{\left[ {f'\left( x \right)} \right]}^2}dx = 7} \) (1).

\(\int\limits_0^1 {{x^6}dx = \frac{{{x^7}}}{7}\left| {_{\scriptstyle\atop\scriptstyle0}^{\scriptstyle1\atop\scriptstyle}} \right. = \frac{1}{7} \Rightarrow \int\limits_0^1 {49{x^6}dx} = \frac{1}{7}.49 = 7} \) (2).

\(\int\limits_0^1 {{x^3}.f'\left( x \right)dx = - 1 \Rightarrow \int\limits_0^1 {14{x^3}.f'\left( x \right)dx} = - 14} \) (3).

Cộng hai vế (1), (2) và (3) suy ra

\(\int\limits_0^1 {{{\left[ {f'\left( x \right)} \right]}^2}dx + \int\limits_0^1 {49{x^6}dx} + \int\limits_0^1 {14{x^3}.f'\left( x \right)dx = 0} } \)

\( \Rightarrow \int\limits_0^1 {{{\left[ {f\left( x \right) + 7{x^3}} \right]}^2}dx = 0.} \)

Do \({\left[ {f'\left( x \right) + 7{x^3}} \right]^2} \ge 0 \Rightarrow \int\limits_0^1 {{{\left[ {f'\left( x \right) + 7{x^3}} \right]}^2}} dx \ge 0\). Mà \(\int\limits_0^1 {{{\left[ {f'\left( x \right) + 7{x^3}} \right]}^2}} dx = 0 \Rightarrow f'\left( x \right) = - 7{x^3}.\)

\(f\left( x \right) = - \frac{{7{x^4}}}{4} + C.\)

Mà \(f\left( 1 \right) = 0 \Rightarrow - \frac{7}{4} + C = 0 \Rightarrow C = \frac{7}{4}.\)

Do đó \(f\left( x \right) = - \frac{{7{x^4}}}{4} + \frac{7}{4}.\)

Vậy \(\int\limits_0^1 {f\left( x \right)dx} = \int\limits_0^1 {\left( { - \frac{{7{x^4}}}{4} + \frac{7}{4}} \right)dx = \frac{7}{5}.} \)

Cách 2: Tương tự như trên ta có \(\int\limits_0^1 {{x^3}.f'\left( x \right)dx = - 1} \)

Áp dụng BĐT Cauchy-Schwarz, ta có

\(7 = 7{\left( {\int\limits_0^1 {{x^3}f'\left( x \right)dx} } \right)^2} \le 7\left( {\int\limits_0^1 {{{\left( {{x^3}} \right)}^2}dx} } \right).\left( {\int\limits_0^1 {{{\left[ {f'\left( x \right)} \right]}^2}dx} } \right)\)

\( \Leftrightarrow 7 \le 7.\frac{1}{7}.\int\limits_0^1 {{{\left[ {f'\left( x \right)} \right]}^2}dx = \int\limits_0^1 {{{\left[ {f'\left( x \right)} \right]}^2}dx.} } \)

Dấu bằng xảy ra khi và chỉ khi \(f'\left( x \right) = a{x^3},\) với \(a \in \mathbb{R}\).

Ta có \(\int\limits_0^1 {{x^3}.f'\left( x \right)dx = - 1} \Rightarrow \int\limits_0^1 {{x^3}.a{x^3}dx} = - 1 \Rightarrow a = - 7.\)

Suy ra \(f'\left( x \right) = - 7{x^3} \Rightarrow f\left( x \right) = - \frac{{7{x^4}}}{4} + C\), mà \(f\left( 1 \right) = 0\) nên \(C = \frac{7}{4}.\)

Do đó \(f\left( x \right) = \frac{7}{4}\left( {1 - {x^4}} \right),\forall x \in \left[ {0;1} \right]\).

Vậy \(\int\limits_0^1 {f\left( x \right)dx} = \int\limits_0^1 {\left( { - \frac{{7{x^4}}}{4} + \frac{7}{4}} \right)dx = \frac{7}{5}.} \)