Let f(x)=|x+1|,|x+2|,,|x+5|.

Question

Let $f(x)=\max\{|x+1|,|x+2|,\ldots,|x+5|\}$. Then $\displaystyle\int_{-6}^0 f(x)\,dx$ is equal to\_\_\_

Answer: 21

Accepted Answer

The upper envelope of $|x+1|,\ldots,|x+5|$ on $[-6,0]$ is determined by the two extreme terms $|x+1|$ and $|x+5|$, which cross at $x=-3$ (both equal $2$). So $f(x)=|x+1|$ on $[-6,-3]$ and $f(x)=|x+5|$ on $[-3,0]$:

$$\int_{-6}^{-3}-(x+1)\,dx + \int_{-3}^0(x+5)\,dx = 10.5+10.5 = 21$$

Solutions · 1