JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Area Under Curves

The five expressions $|x+1|,|x+2|,|x+3|,|x+4|,|x+5|$ measure the distance of $x$ from the points $-1,-2,-3,-4,-5$, so their maximum is always the distance to whichever of those points lies farthest from $x$. The extreme points are $-1$ and $-5$, with midpoint $-3$: for $x>-3$ the farther point is $-5$, giving $f(x)=|x+5|$, while for $x<-3$ the farther point is $-1$, giving $f(x)=|x+1|$. This splits the integral at $x=-3$. On $[-6,-3]$ the variable stays below $-1$, so $f(x)=-(x+1)$ and ${\int }_{-6}^{-3}-(x+1)dx={\left[-\frac{x^2}{2}-x\right]}_{-6}^{-3}=\left(-\frac{9}{2}+3\right)-\left(-18+6\right)=\frac{21}{2}$. On $[-3,0]$ the variable stays above $-5$, so $f(x)=x+5$ and ${\int }_{-3}^0(x+5)dx={\left[\frac{x^2}{2}+5x\right]}_{-3}^0=0-\left(\frac{9}{2}-15\right)=\frac{21}{2}$. The two equal pieces add to $\frac{21}{2}+\frac{21}{2}=21$.