Question

Let $a$ and $b$ respectively be the points of local maximum and local minimum of the function $f\left(x\right)=2x^3-3x^2-12x.$
If $A$ is the total area of the region bounded by $y=f\left(x\right),$ the $x-$axis and the lines $x=a$ and $x=b,$ then $4A$ is equal to .

Answer 1

$f\left(x\right)=2x^3-3x^2-12x$
$\Rightarrow f^{\prime }\left(x\right)=6x^2-6x-12$
$=6(x-2)(x+1)$
$\Rightarrow f^{\prime }\left(x\right)=0\Rightarrow x=-1,2$
$x=-1$ is point of local maximum $\Rightarrow a=-1$
$x=2$ is point of local minimum $\Rightarrow b=2$
$f(-1)=8\text{and}f\left(2\right)=-20$
Required area is as shown in figure
Required area $=\underset{-1}{\overset{0}{\int }}\left(f\right(x)-0)dx+\underset{0}{\overset{2}{\int }}(0-f(x\left)\right)dx$
$={(\frac{1}{2}{x}^4-x^3-6x^2)}_{-1}^0-{(\frac{x^4}{2}-x^3-6x^2)}_0^2=\frac{57}{2}=A$
$\Rightarrow 4A=114$