Cho hàm số bậc hai $y=f\left( x \right)$ có đồ thị $\left( P...

Giả sử $d:y=mx+n$, $\left( P \right):f\left( x \right)=ax_{{}}^{2}+bx+c$
Tư đồ thị ta có:
Đường thẳng $d$ đi qua $A\left( 3;2 \right),B\left( 6;5 \right)$ nên có $\left\{ \begin{aligned}
& 3m+n=2 \\
& 6m+n=5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& m=1 \\
& n=-1 \\
\end{aligned} \right. $ $ \Rightarrow d:y=x-1$.
Đồ thị $\left( P \right)$ đi qua $A\left( 3;2 \right),B\left( 6;5 \right)$ nên có $\left\{ \begin{aligned}
& 9a+3b+c=2 \\
& 36a+6b+c=5 \\
\end{aligned} \right.$
Và $S=\int\limits_{3}^{6}{\left( x-1-a{{x}^{2}}-bx-c \right)\text{d}x}=\dfrac{9}{2}$ $\Leftrightarrow \left. \left( \dfrac{{{x}^{2}}}{2}-x-\dfrac{a{{x}^{3}}}{3}-\dfrac{b{{x}^{2}}}{2}-cx \right) \right|_{3}^{6}=\dfrac{9}{2}$
$\Leftrightarrow 63a+\dfrac{27}{2}b+3c=6$
Do đó ta có hệ phương trình $\left\{ \begin{aligned}
& 9a+3b+c=2 \\
& 36a+6b+c=5 \\
& 63a+\dfrac{27}{2}b+3c=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=1 \\
& b=-8 \\
& c=17 \\
\end{aligned} \right.$
$\Rightarrow f\left( x \right)={{x}^{2}}-8x+17\Rightarrow {f}'\left( x \right)=2x-8$.
Suy ra $\int\limits_{3}^{6}{\left( 2x-3 \right){f}'\left( x \right)\text{d}x}=\int\limits_{3}^{6}{\left( 2x-3 \right)\left( 2x-8 \right)\text{d}x}$ $=\int\limits_{3}^{6}{\left( 4{{x}^{2}}-22x+24 \right)\text{d}x}$
$=\left. \left( \dfrac{4{{x}^{3}}}{3}-11{{x}^{2}}+24x \right) \right|_{3}^{6}=27$