주기적 함수로 정의된 구간 정적분의 최솟값

$g \left( a \right)$$\displaystyle = \int _{a} ^{a + 2} {f \left( x \right) dx}$라 하면 $g' \left( a \right) = f \left( a + 2 \right) - f \left( a \right)$ (ⅰ) $f \left( a + 2 \right)$의 값을 구하면 $5 \leq a \leq 6$에서 $7 \leq a + 2 \leq 8$ $f \left( a + 2 \right) = f \left( \left( a - 2 \right) + 4 \right) = f \left( a - 2 \right) + 1$ 이때 $3 \leq a - 2 \leq 4$이므로 조건 (가)에서 $\displaystyle f \left( a - 2 \right) = \frac{1}{2} \left( a - 2 \right) - 1 = \frac{1}{2} a - 2$ $\therefore$ $f \left( a + 2 \right)$$\displaystyle = \left( \frac{1}{2} a - 2 \right) + 1 = \frac{1}{2} a - 1$ (ⅱ) $f \left( a \right)$의 값을 구하면 $5 \leq a \leq 6$이므로 $f \left( a \right) = f \left( \left( a - 4 \right) + 4 \right) = f \left( a - 4 \right) + 1$ 이때 $1 \leq a - 4 \leq 2$이므로 조건 (가)에서 $f \left( a - 4 \right)$$= - \left( a - 4 \right) \left\{ \left( a - 4 \right) - 2 \right\}$ $= - \left( a - 4 \right) \left( a - 6 \right)$ $= - a ^{2} + 10 a - 24$ $\therefore$ $f \left( a \right)$$= - a ^{2} + 10 a - 23$ (ⅰ), (ⅱ)에 의해 $f \left( a + 2 \right)$$\displaystyle = \frac{1}{2} a - 1$, $f \left( a \right)$$= - a ^{2} + 10 a - 23$ 이므로 $g' \left( a \right)$$= f \left( a + 2 \right) - f \left( a \right)$ $\displaystyle = \left( \frac{1}{2} a - 1 \right) - \left( - a ^{2} + 10 a - 23 \right)$ $\displaystyle = a ^{2} - \frac{19}{2} a + 22$ $g' \left( a \right) = 0$에서 $2 a ^{2} - 19 a + 44 = 0$, $\left( 2 a - 11 \right) \left( a - 4 \right) = 0$ $\therefore$ $a = 4$ 또는 $\displaystyle a = \frac{11}{2}$ 이때 $5 \leq a \leq 6$이므로 $\displaystyle a = \frac{11}{2}$ 따라서 $g \left( a \right)$는 $\displaystyle a = \frac{11}{2}$에서 극솟값이면서 최솟값을 가지므로 $g \left( a \right)$의 최솟값은 $\displaystyle \int _{\frac{11}{2}} ^{\frac{15}{2}} f \left( x \right) dx$ 함수 $f \left( x \right)$를 $x$축의 방향으로 $- 4$만큼 평행이동하면 $\displaystyle \int _{\frac{11}{2}} ^{\frac{15}{2}} f \left( x \right) dx$$\displaystyle = \int _{\frac{3}{2}} ^{\frac{7}{2}} f \left( x + 4 \right) dx$ $\displaystyle = \int _{\frac{3}{2}} ^{\frac{7}{2}} \left\{ f \left( x \right) + 1 \right\} dx$ $\displaystyle = \int _{\frac{3}{2}} ^{\frac{7}{2}} {f \left( x \right) dx} + 2$ $\displaystyle = \int _{\frac{3}{2}} ^{2} {\left( - x ^{2} + 2 x \right) dx + \int _{2} ^{\frac{7}{2}} {\left( \frac{1}{2} x - 1 \right) dx}}$$+ 2$ $\displaystyle = \left[ - \frac{1}{3} x ^{3} + x ^{2} \right] _{\frac{3}{2}} ^{2} + \left[ \frac{1}{4} x ^{2} - x \right] _{2} ^{\frac{7}{2}}$$+ 2$ $\displaystyle = \frac{5}{24} + \frac{9}{16} + 2$ $\displaystyle = \frac{133}{48}$ 따라서 구하는 최솟값은 $\displaystyle \frac{133}{48}$이다.