Cho hàm số $f\left( x...

Đặt $m=\int_{0}^{1}|f(x)| \mathrm{d} x,(m \geq 0) \Rightarrow f(x)=x^{3}-4 m x$

Do $f(1)>0 \Rightarrow 1-4 m>0 \Leftrightarrow m<\dfrac{1}{4} \Rightarrow m \in\left[0 ; \dfrac{1}{4}\right)$.
Khi đó $m=\int_{0}^{1}|f(x)| \mathrm{d} x=\int_{0}^{1}\left|x^{3}-4 m x\right| \mathrm{d} x=\int_{0}^{2 \sqrt{m}}\left|x^{3}-4 m x\right| \mathrm{d} x+\int_{2 \sqrt{m}}^{1}\left|x^{3}-4 m x\right| \mathrm{d} x$
$
\begin{aligned}
&\Leftrightarrow m=-\int_{0}^{2 \sqrt{m}}\left(x^{3}-4 m x\right) \mathrm{d} x+\int_{2 \sqrt{m}}^{1}\left(x^{3}-4 m x\right) \mathrm{d} x \\
&\Leftrightarrow m=-\left.\left(\dfrac{1}{4} x^{4}-2 m x^{2}\right)\right|_{0} ^{2 \sqrt{m}}+\left.\left(\dfrac{1}{4} x^{4}-2 m x^{2}\right)\right|_{2 \sqrt{m}} ^{1}
\end{aligned}
$
$
\Leftrightarrow m=-\left.\left(\dfrac{1}{4} x^{4}-2 m x^{2}\right)\right|_{0} ^{2 \sqrt{m}}+\left.\left(\dfrac{1}{4} x^{4}-2 m x^{2}\right)\right|_{2 \sqrt{m}} ^{1} \Leftrightarrow 8 m^{2}-3 m+\dfrac{1}{4}=0 \Leftrightarrow\left[\begin{array}{l}
m=\dfrac{1}{4} \\
m=\dfrac{1}{8}
\end{array}\right.
$
Vì $m \in\left[0 ; \dfrac{1}{4}\right)$ nên $m=\dfrac{1}{8}$.
Khi đó $f(x)=x^{3}-\dfrac{1}{2} x \Rightarrow f(4)=62$.