A function $f ( x )$ that is continuous on the set of all real numbers satisfies the following condition. When $n - 1 \leq x < n$, $| f ( x ) | = | 6 ( x - n + 1 ) ( x - n ) |$. (Here, $n$ is a natural number.) For the function $$g ( x ) = \int _ { 0 } ^ { x } f ( t ) d t - \int _ { x } ^ { 4 } f ( t ) d t$$ defined on the open interval $(0, 4)$, when $g ( x )$ has a minimum value of 0 at $x = 2$, what is the value of $\int _ { \frac { 1 } { 2 } } ^ { 4 } f ( x ) d x$? [4 points] (1) $- \frac { 3 } { 2 }$ (2) $- \frac { 1 } { 2 }$ (3) $\frac { 1 } { 2 }$ (4) $\frac { 3 } { 2 }$ (5) $\frac { 5 } { 2 }$
A function $f ( x )$ that is continuous on the set of all real numbers satisfies the following condition.\\
When $n - 1 \leq x < n$, $| f ( x ) | = | 6 ( x - n + 1 ) ( x - n ) |$.\\
(Here, $n$ is a natural number.)
For the function
$$g ( x ) = \int _ { 0 } ^ { x } f ( t ) d t - \int _ { x } ^ { 4 } f ( t ) d t$$
defined on the open interval $(0, 4)$, when $g ( x )$ has a minimum value of 0 at $x = 2$, what is the value of $\int _ { \frac { 1 } { 2 } } ^ { 4 } f ( x ) d x$? [4 points]\\
(1) $- \frac { 3 } { 2 }$\\
(2) $- \frac { 1 } { 2 }$\\
(3) $\frac { 1 } { 2 }$\\
(4) $\frac { 3 } { 2 }$\\
(5) $\frac { 5 } { 2 }$