Đáp án:
\(\displaystyle\int\limits_0^4 {{x^2}f'\left( x \right)dx} = - 16\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
u = f\left( {4x} \right)\\
v' = x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 4f'\left( {4x} \right)\\
v = \dfrac{{{x^2}}}{2}
\end{array} \right.\\
\displaystyle\int\limits_0^1 {x.f\left( {4x} \right)dx} = 1\\
\Leftrightarrow \mathop {\left. {\dfrac{{{x^2}}}{2}.f\left( {4x} \right)} \right|}\nolimits_0^1 - \displaystyle\int\limits_0^1 {4.f'\left( {4x} \right).\dfrac{{{x^2}}}{2}dx} = 1\\
\Leftrightarrow \dfrac{1}{2}.f\left( 4 \right) - \displaystyle\int\limits_0^1 {2{x^2}.f'\left( {4x} \right).dx} = 1\\
\Leftrightarrow \dfrac{1}{2} - 2\displaystyle\int\limits_0^1 {{x^2}.f'\left( {4x} \right).dx} = 1\\
\Leftrightarrow \displaystyle\int\limits_0^1 {{x^2}.f'\left( {4x} \right).dx} = - \dfrac{1}{4}\,\,\,\,\,\,\left( 1 \right)\\
t = 4x \Rightarrow \left\{ \begin{array}{l}
x = 0 \Leftrightarrow t = 0\\
x = 1 \Leftrightarrow t = 4\\
dt = 4dx
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \displaystyle\int\limits_0^4 {{{\left( {\dfrac{t}{4}} \right)}^2}.f'\left( t \right).\dfrac{{dt}}{4}} = - \dfrac{1}{4}\\
\Leftrightarrow \dfrac{1}{{64}}\displaystyle\int\limits_0^4 {{t^2}f'\left( t \right)dt} = - \dfrac{1}{4}\\
\Leftrightarrow \displaystyle\int\limits_0^4 {{t^2}f'\left( t \right)dt} = - 16\\
\Leftrightarrow \displaystyle\int\limits_0^4 {{x^2}f'\left( x \right)dx} = - 16
\end{array}\)
