.f′(x)=6x2−6x−12=6(x2−x−2)
=6(x−2)(x+1)

Clearly, {f}^{'}(-{1}^{-})>0&{f}^{'}(-{1}^{+})<0
Hence, x=−1 is the point of local maximum.
Also, {f}^{'}({2}^{-})<0&{f}^{'}({2}^{+})>0
Hence, x=2 is the point of local minimum.
\therefore a=-1&b=2
Points=(2,-20)&(-1,7)

A=∫−10(2x3−3x2−12x)dx−∫02(2x3−3x2−12x)dx
=(2x4−x3−6x2)−10−(2x4−x3−6x2)02
=29+24=257⇒4A=114