Cho hàm số \(f\left( x \right)\) có đạo hàm liên tục trên đoạn \(\left[ {0;1} \right]\) thỏa mãn \(f\left( 1 \right) = 0\), \(\int\limits_0^1 {{{\left[ {f'\left( x \right)} \right]}^2}dx}  = 7\) và \(\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}x.\cos xf\left( {\sin x} \right)dx}  = \dfrac{1}{3}\). Tính tích phân \(\int\limits_0^1 {f\left( x \right)dx} \) bằng:

Xét tích phân \({I_1} = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}x\cos xf\left( {\sin x} \right)dx} \).

Đặt \(t = \sin x \Rightarrow dt = \cos xdx\).

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = \dfrac{\pi }{2} \Rightarrow t = 1\end{array} \right.\).

\( \Rightarrow {I_1} = \int\limits_0^1 {{t^2}f\left( t \right)dt}  = \int\limits_0^1 {{x^2}f\left( x \right)dx}  = \dfrac{1}{3}\) (tính chất không phụ thuộc biến số).

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

\(\begin{array}{l} \Rightarrow {I_1} = \left. {\dfrac{1}{3}{x^3}f\left( x \right)} \right|_0^1 - \dfrac{1}{3}\int\limits_0^1 {{x^3}f'\left( x \right)dx}  = \dfrac{1}{3}\\ \Rightarrow \dfrac{1}{3}f\left( 1 \right) - \dfrac{1}{3}\int\limits_0^1 {{x^3}f'\left( x \right)dx}  = \dfrac{1}{3}\\ \Leftrightarrow \dfrac{1}{3}.0 - \dfrac{1}{3}\int\limits_0^1 {{x^3}f'\left( x \right)dx}  = \dfrac{1}{3}\\ \Leftrightarrow \int\limits_0^1 {{x^3}f'\left( x \right)dx}  =  - 1\end{array}\)

Ta có:

\(\begin{array}{l}\int\limits_0^1 {{{\left[ {f'\left( x \right) + 7{x^3}} \right]}^2}dx}  = \int\limits_0^1 {{{\left[ {f'\left( x \right)} \right]}^2}dx}  + 14\int\limits_0^1 {{x^3}f'\left( x \right)dx}  + 49\int\limits_0^1 {{x^6}dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7 - 14 + 7 = 0\\ \Rightarrow \int\limits_0^1 {{{\left[ {f'\left( x \right) + 7{x^3}} \right]}^2}dx}  = 0\\ \Leftrightarrow f'\left( x \right) + 7{x^3} = 0 \Leftrightarrow f'\left( x \right) =  - 7{x^3}\\ \Rightarrow f\left( x \right) = \int {f'\left( x \right)dx}  = \int { - 7{x^3}dx}  = \dfrac{{ - 7}}{4}{x^4} + C\end{array}\)

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

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

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