Đề thi THPT QG 2019 – mã đề 104

Cho hàm số \(f\left( x \right)\) có đạo hàm liên tục trên R. Biết \(f\left( 3 \right) = 1\) và \(\int\limits_0^1 {xf\left( {3x} \right)dx = 1} \), khi đó \(\int\limits_0^3 {{x^2}f'\left( x \right)dx} \) bằng:

Xét \(\int\limits_0^1 {xf\left( {3x} \right)dx = 1} \).

Đặt \(t = 3x \Leftrightarrow dt = 3dx\). Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = 1 \Rightarrow t = 3\end{array} \right.\).

\( \Rightarrow \int\limits_0^1 {xf\left( {3x} \right)dx}  = 1 \Leftrightarrow \int\limits_0^3 {\dfrac{t}{3}f\left( t \right)\dfrac{{dt}}{3}}  = 1\)\( \Leftrightarrow \dfrac{1}{9}\int\limits_0^3 {tf\left( t \right)dt}  = 1\)

\( \Leftrightarrow \int\limits_0^3 {tf\left( t \right)dt}  = 9 \Leftrightarrow 2\int\limits_0^3 {xf\left( x \right)dx}  = 18\)\( \Rightarrow \int\limits_0^3 {2xf\left( x \right)dx}  = 18\)

Ta có: \(\int\limits_0^3 {\left[ {{x^2}f'\left( x \right) + 2xf\left( x \right)} \right]dx} \)\( = \int\limits_0^3 {{x^2}f'\left( x \right)dx}  + \int\limits_0^3 {2xf\left( x \right)dx} \)

\(\begin{array}{l} \Leftrightarrow \int\limits_0^3 {\left( {{x^2}f\left( x \right)} \right)'dx}  = \int\limits_0^3 {{x^2}f'\left( x \right)dx}  + 18\\ \Leftrightarrow \left. {{x^2}f\left( x \right)} \right|_0^3 = \int\limits_0^3 {{x^2}f'\left( x \right)dx}  + 18\\ \Leftrightarrow 9f\left( 3 \right) = \int\limits_0^3 {{x^2}f'\left( x \right)dx}  + 18\\ \Leftrightarrow 9.1 = \int\limits_0^3 {{x^2}f'\left( x \right)dx}  + 18\\ \Leftrightarrow \int\limits_0^3 {{x^2}f'\left( x \right)dx}  = 9 - 18 =  - 9\end{array}\)