MCQ
Let $f(x)=\max \{|x+1|,|x+2|, \ldots,|x+5|\}$. Then $\int_{-6}^{0} f(x) d x$ is equal to
- A$20$
- B$40$
- ✓$21$
- D$41$
✓
Answer
Correct option: C.
$21$
c
$f(x)=\max \{|x+1|,|x+2|,|x+3|,|x+4|,|x+5|\}$
$f(x)=\max \{|x+1|,|x+2|,|x+3|,|x+4|,|x+5|\}$
$\int_{-6}^{0} f(x) d x=\int_{-6}^{-3}|x+1| d x+\int_{-3}^{0}|x+5| d x$
$=-\int_{-6}^{-3}(x+1) d x+\int_{-3}^{0}(x+5) d x$
$=-\left[\frac{x^{2}}{2}+x\right]_{-6}^{-3}+\left[\frac{x^{2}}{2}+5 x\right]_{-3}^{0}$
$=-\left[\left(\frac{9}{2}-3\right)-(18-6)\right]+\left[0-\left(\frac{9}{2}-15\right)\right]$
$=-\left[\frac{3}{2}-12\right]+\frac{21}{2}=\frac{21}{2}+\frac{21}{2}=21$
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