Let $n$ be a positive integer. Define
$$f ( x ) = \min \{ | x - 1 | , | x - 2 | , \ldots , | x - n | \}$$
Then $\int _ { 0 } ^ { n + 1 } f ( x ) d x$ equals\\
(A) $\frac { ( n + 4 ) } { 4 }$\\
(B) $\frac { ( n + 3 ) } { 4 }$\\
(C) $\frac { ( n + 2 ) } { 2 }$\\
(D) $\frac { ( n + 2 ) } { 4 }$