Question

Let $a$ and $b$ be the points of local maximum and local minimum respectively of the function $f\left(x\right)=2x^3-3x^2-12x.$ If $A$ is the area of region bounded by $y=f\left(x\right),$ $x-$axis, $y-$axis and $x=b,$ then the value of $2A$(in sq. units) is equal to

Answer 1

Given: $f\left(x\right)=2x^3-3x^2-12x.$
$\Rightarrow f^{\prime }\left(x\right)=6x^2-6x-12$
For max./min, $f^{\prime }\left(x\right)=0$
$\therefore 6(x^2-x-2)=0$
$\Rightarrow x=2,-1$
and $f^{\prime \prime }\left(x\right)=12x-6$
Since $f^{\prime \prime }(-1)=-18<0,$ therefore $x=-1$ is point of local maximum.
$f^{\prime \prime }\left(2\right)=18<0,$ therefore $x=2$ is point of local minimum.
$\therefore a=-1$ and $b=2$
Figure:


Required area
$A=-\underset{0}{\overset{2}{\int }}(2x^3-3x^2-12x)dx$
$\Rightarrow A=-{[\frac{x^4}{2}-x^3-6x^2]}_0^2$
$\Rightarrow A=-[8-8-24]=24$ sq. units
$\therefore 2A=48$ sq. units