Cho hàm số $f(x)$ có đạo hàm trên $\mathbb{R}$ và thỏa mãn...

Đặt $I=\int\limits_{0}^{4}{{{x}^{2}}.{f}'(x)\text{d}x}$
Đặt $\left\{ \begin{aligned}
& u={{x}^{2}} \\
& \text{d}v={f}'(x)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=2x\text{d}x \\
& v=f(x) \\
\end{aligned} \right.$.
Khi đó $I=\left. {{x}^{2}}f(x) \right|_{0}^{4}-\int\limits_{0}^{4}{2f(x)\text{d}x=16f(4)-2}\int\limits_{0}^{4}{x.f(x)\text{d}x}$.
Xét $K=\int\limits_{0}^{4}{x.f(x)\text{d}x}=\int\limits_{0}^{4}{t.f(t)\text{d}t}$.
Đặt $t={{x}^{3}}+3x\Rightarrow \left\{ \begin{aligned}
& f(t)={{x}^{2}}+2 \\
& \text{d}t=(3{{x}^{2}}\text{+3)d}x \\
\end{aligned} \right.$.
$t=0\Rightarrow x=0; \text{ }t=4\Rightarrow x=1$.
Do đó $K=\int\limits_{0}^{1}{({{x}^{3}}+3x)({{x}^{2}}+2).(3{{x}^{2}}\text{+3)d}x=\dfrac{165}{8}}$.
$x=1\Rightarrow f(4)=3.$
Vậy $I=\int\limits_{0}^{4}{{{x}^{2}}.{f}'(x)\text{d}x}=16.3-2.\dfrac{165}{8}=\dfrac{27}{4}$.