JEE Main 2021 (Online) 26th August Evening ShiftQuestion 1
Let $a$ and $b$ respectively be the points of local maximum and local minimum of the function $f(x)=2x^{3}-3x^{2}-12x$. If $A$ is the total area of the region bounded by $y=f(x)$, the $x$-axis and the lines $x=a$ and $x=b$, then $4A$ is equal to $\_\_\_\_\_\_$ .
Application of Integrals
Answer$114$
Solution
$f'(x)=6x^2-6x-12=0\Rightarrow x=-1$ (max), $x=2$ (min), so $a=-1,b=2$. $f(x)=x(2x^2-3x-12)$ has no zero in $(-1,2)$ except $0$; $f>0$ on $[-1,0]$, $f<0$ on $[0,2]$. $\int_{-1}^0 f=\left[\frac{x^4}{2}-x^3-6x^2\right]_{-1}^0=\frac92$, $\int_0^2 f=8-8-24=-24$. $A=\frac92+24=\frac{57}{2}\Rightarrow 4A=114$.