정적분으로 정의된 함수의 최솟값

$g ( x )$$\displaystyle = \int _{- 1} ^{x} {f ( t ) \left\{ 2 x - f ( t ) \right\}} dt$ $\displaystyle = 2 x \int _{- 1} ^{x} {f ( t ) dt} - \int _{- 1} ^{x} \left\{ f ( t ) \right\} ^{2} dt$ 양변을 $x$에 대하여 미분하면 $g' ( x )$$\displaystyle = 2 \int _{- 1} ^{x} {f ( t ) dt + 2 xf ( x ) - \left\{ f ( x ) \right\} ^{2}}$ (ⅰ) $x < - 1$일 때, $g' ( x )$$\displaystyle = 2 \int _{- 1} ^{x} {0 dt + 2 x \cdot 0 - \left\{ 0 \right\} ^{2}}$$= 0$ (ⅱ) $- 1 \leq x < 0$일 때, $g' ( x )$$\displaystyle = 2 \int _{- 1} ^{x} {( 1 + t ) dt + 2 x ( 1 + x ) - \left( 1 + x \right) ^{2}}$ $= \left[ ( 1 + t ) ^{2} \right] _{- 1} ^{x} + x ^{2} - 1$ $= ( 1 + x ) ^{2} + x ^{2} - 1$ $= 2 x ^{2} + 2 x$ $= 2 x ( x + 1 )$ (ⅲ) $0 \leq x < 1$일 때, $g' ( x )$$\displaystyle = 2 \left\{ \int _{- 1} ^{0} {( 1 + t ) dt + \int _{0} ^{x} ( 1 - t ) dt \}} \right.$ $+ 2 x ( 1 - x ) - \left( 1 - x \right) ^{2}$ $\displaystyle = 1 - \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} ( 1 - t ) ^{2} \end{array} \right] _{0} ^{x} - 3 x ^{2} + 4 x - 1$ $= - \left( 1 - x \right) ^{2} + 1 - 3 x ^{2} + 4 x$ $= - 4 x ^{2} + 6 x$ $= - 2 x \left( 2 x - 3 \right)$ (ⅳ) $x \geq 1$일 때, $g' ( x )$$\displaystyle = 2 \left\{ \int _{- 1} ^{0} {( 1 + t ) dt + \int _{0} ^{1} ( 1 - t ) dt + \int _{1} ^{x} {0 dt} \}} \right.$$+ 2 x \cdot 0 - \left( 0 \right) ^{2}$ $\displaystyle = 2 \left( \frac{1}{2} + \frac{1}{2} \right)$$= 2$ 따라서 $y = g' ( x )$의 그래프를 그리면 다음과 같다.

따라서 $x = 0$일 때, 극소이면서 최소가 되므로 최솟값은 $g ( 0 )$이다. $\therefore$$g ( 0 )$$\displaystyle = 2 \cdot 0 \cdot \int _{- 1} ^{0} {( 1 + t ) dt} - \int _{- 1} ^{0} ( 1 + t ) ^{2} dt$ $\displaystyle = 0 - \left[ \frac{1}{3} ( 1 + t ) ^{3} \right] _{- 1} ^{0}$ $\displaystyle = - \frac{1}{3}$