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  4. If f(x)= min (|x-1|,|x|,|x+1|), then the value of 6(∫ limits-11 f(x) d x) is

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Q. If $f(x)=\min (|x-1|,|x|,|x+1|)$, then the value of $6\left(\int\limits_{-1}^1 f(x) d x\right)$ is

NTA AbhyasNTA Abhyas 2022

Solution:

Solution
$\int\limits _{- 1}^{1} f \left(x\right) d x=$ shaded area
$=2(\frac{1}{2}$ base⋅ height)
$=2\left(\frac{1}{2} \cdot 1 \cdot \frac{1}{2}\right)=\frac{1}{2}$