Cho hàm số $y=f(x)$ có đạo hàm trên đoạn $\left[ 1;2...

$\int\limits_{1}^{2}{{f}'(x)dx=f\left( 2 \right)-f\left( 1 \right)}=-1$ $\Rightarrow 2\int\limits_{1}^{2}{{f}'(x)dx=-2}$
$\int\limits_{1}^{2}{{{x}^{2}}.{{({f}'(x))}^{2}}dx=2}\Leftrightarrow \int\limits_{1}^{2}{{{\left[ x.{f}'(x) \right]}^{2}}dx-2}=0\Leftrightarrow \int\limits_{1}^{2}{{{\left[ x.{f}'(x) \right]}^{2}}dx+}2\int\limits_{1}^{2}{{f}'(x)dx=0}$.
$\Leftrightarrow \int\limits_{1}^{2}{\left[ {{x}^{2}}.{{\left[ {f}'(x) \right]}^{2}}+2{f}'(x) \right]dx=0}$
$\Rightarrow {{x}^{2}}.{{\left[ {f}'(x) \right]}^{2}}+2{f}'(x)=0\Leftrightarrow {f}'(x)\left[ {{x}^{2}}.{f}'(x)+2 \right]=0$ $\Leftrightarrow \left[ \begin{aligned}
& {f}'(x)=0\quad \left( Loai \right) \\
& {{x}^{2}}.{f}'(x)+2=0 \\
\end{aligned} \right.$
$\Rightarrow {{x}^{2}}.{f}'(x)+2=0\Rightarrow {f}'(x)=\dfrac{-2}{{{x}^{2}}}$
$\Rightarrow f\left( x \right)=\int{{f}'(x)dx}=\dfrac{2}{x}+C$
$f(1)=2\Leftrightarrow \dfrac{2}{1}+C=2\Leftrightarrow C=0$
Diện tích: $\int\limits_{1}^{2}{g(x)dx}=\int\limits_{1}^{2}{{{x}^{4}}.f(x)}dx=\int\limits_{1}^{2}{{{x}^{4}}.\dfrac{2}{x}}dx=\int\limits_{1}^{2}{2{{x}^{3}}}dx=\dfrac{15}{2}$