Đáp án:
\[\int\limits_0^4 {{x^2}f'\left( x \right)dx} = - 16\]
Giải thích các bước giải:
\(\begin{array}{l}
t = 4x \Rightarrow \left\{ \begin{array}{l}
dt = 4dx\\
x = 0 \Rightarrow t = 4\\
x = 1 \Rightarrow t = 4
\end{array} \right.\\
\int\limits_0^1 {x.f\left( {4x} \right)dx = 1} \\
\Leftrightarrow \int\limits_0^4 {\frac{t}{4}.f\left( t \right).\frac{{dt}}{4}} = 1\\
\Leftrightarrow \frac{1}{{16}}\int\limits_0^4 {tf\left( t \right)dt} = 1\\
\Leftrightarrow \int\limits_0^4 {tf\left( t \right)dt} = 16\\
\Rightarrow \int\limits_0^4 {xf\left( x \right)dx} = 16\\
I = \int\limits_0^4 {{x^2}f'\left( x \right)dx} \\
\left\{ \begin{array}{l}
u = {x^2}\\
v' = f'\left( x \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 2x\\
v = f\left( x \right)
\end{array} \right.\\
\Rightarrow I = \mathop {\left. {{x^2}f\left( x \right)} \right|}\nolimits_0^4 - \int\limits_0^4 {2xf\left( x \right)dx} \\
= \left( {{4^2}.f\left( 4 \right) - {0^2}.f\left( 0 \right)} \right) - 2.\int\limits_0^4 {xf\left( x \right)dx} \\
= {4^2}.1 - 2.16\\
= - 16
\end{array}\)
