Ta có: $-x^2+6x-5=-x^2+4x-3 \Leftrightarrow x=1$
$-x^2+6x-5=3x-15 \Leftrightarrow \left[\begin{array}{l}x=5\\x=-2\end{array}\right.$
$-x^2+4x-3=3x-15 \Leftrightarrow \left[\begin{array}{l}x=4\\x=-3\end{array}\right.$
Diện tích hình phẳng cần tìm là:
$S=\int\limits_1^4|(-x^2+6x-5)-(-x^2+4x-3)|dx+\int\limits_4^5|(-x^2+6x-5)-(3x-15)|dx$
$=\int\limits_1^4(2x-2)dx+\int\limits_4^5(-x^2+3x+10)dx$
$=(x^2-2x)\left|\begin{array}{l}4\\1\end{array}\right.+(\dfrac{-x^3}{3}+\dfrac{3x^2}{2}+10x)\left|\begin{array}{l}5\\4\end{array}\right.$
$=\dfrac{73}{6}$