Let n be a positive integer, then $\int_{0}^{n + 1} \min {| x - 1 |, | x - 2 |, | x - 3 |, ....... | x - n |} d x$ equals

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$\frac{(n + 1)}{4}$

$\frac{(n + 3)}{4}$

$\frac{(n + 2)}{4}$

$\frac{(n + 4)}{4}$

answer is C.

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Detailed Solution

$\int_{0}^{n + 1} \min {| x - 1 |, | x - 2 |, | x - 3 |, ....... | x - n |} d x = \frac{1}{2} (1) + \frac{1}{2} \times \frac{1}{2} \times (n - 1) + \frac{1}{2} \times (1) = \frac{n + 3}{4}$

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