합성 조건 정적분

조건 (가)에 의하여 $x < 1$일 때 $f \left( x \right) = - x ^{2} + 4 x + C \left( \right.$단, $C$는 적분상수$)$ 조건 (나)에 의하여 $x > 0$일 때 $2 xf' \left( x ^{2} + 1 \right) = 2 ae ^{2 x} + b$ $\displaystyle f' \left( x ^{2} + 1 \right) = \frac{2 ae ^{2 x} + b}{2 x}$ $\displaystyle f' \left( 1 \right) = \lim\limits _{x \rightarrow 0 +} {} f' \left( x ^{2} + 1 \right) = \lim\limits _{x \rightarrow 0 +} {} \frac{2 ae ^{2 x} + b}{2 x}$ $\displaystyle \lim\limits _{x \rightarrow 0 +} 2 x = 0$이므로 $\displaystyle \lim\limits _{x \rightarrow 0 +} {} \left( 2 ae ^{2 x} + b \right) = 0$ $2 a + b = 0$, $b = - 2 a$ $\displaystyle f' \left( x ^{2} + 1 \right) = \frac{2 ae ^{2 x} + b}{2 x} = \frac{2 ae ^{2 x} - 2 a}{2 x}$ 함수 $f' ( x )$가 $x = 1$에서 연속이므로 $\displaystyle \lim\limits _{x \rightarrow 1 -} f' ( x )$$\displaystyle = \lim\limits _{x \rightarrow 1 +} {} f' \left( x \right) = f' \left( 1 \right)$ $\displaystyle \lim\limits _{x \rightarrow 1 -} {} f' \left( x \right) = - 2 + 4 = 2$ $\displaystyle \lim\limits _{x \rightarrow 1 +} {} f' \left( x \right)$$\displaystyle = \lim\limits _{s \rightarrow 0 +} {} f' \left( s ^{2} + 1 \right)$$\displaystyle = \lim\limits _{s \rightarrow 0 +} {} \frac{2 a \left( e ^{2 s} - 1 \right)}{2 s} = 2 a$ $f' ( 1 ) = 2$ $2 = 2 a$, $a = 1$, $b = - 2$ 함수 $f ( x )$가 $x = 1$에서 연속이므로 $\displaystyle \lim\limits _{x \rightarrow 1 -} {} f \left( x \right) = - 1 ^{2} + 4 \times 1 + C = C + 3$ $\displaystyle \lim\limits _{x \rightarrow 1 +} {} f \left( x \right)$$\displaystyle = \lim\limits _{s \rightarrow 0 +} {} f \left( s ^{2} + 1 \right)$$\displaystyle = \lim\limits _{s \rightarrow 0 +} {} \left( e ^{2 s} - 2 s \right) = 1$ $f \left( 1 \right) = 1$ $C + 3 = 1$이므로 $C = - 2$ 그러므로 $x < 1$일 때, $f \left( x \right) = - x ^{2} + 4 x - 2$ $x \geq 0$일 때, $f \left( x ^{2} + 1 \right) = e ^{2 x} - 2 x$ $\displaystyle \int _{0} ^{5} {} f \left( x \right) dx = \int _{0} ^{1} {} f \left( x \right) dx + \int _{1} ^{5} {} f \left( x \right) dx$ $\displaystyle \int _{0} ^{1} {} f \left( x \right) dx = \int _{0} ^{1} {} \left( - x ^{2} + 4 x - 2 \right) dx = - \frac{1}{3}$ $\displaystyle \int _{1} ^{5} {} f \left( x \right) dx$에서 $x = t ^{2} + 1 \left( t \geq 0 \right)$이라 하면 $\displaystyle \frac{d x}{d t} = 2 t$ $\displaystyle \int _{1} ^{5} {} f \left( x \right) dx$$\displaystyle = \int _{0} ^{2} {} f \left( t ^{2} + 1 \right) 2 tdt$ $\displaystyle = \int _{0} ^{2} {} 2 t \left( e ^{2 t} - 2 t \right) dt$ $\displaystyle = \int _{0} ^{2} {} \left( 2 te ^{2 t} - 4 t ^{2} \right) dt$ $\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} te ^{2 t} \end{array} \right] _{0} ^{2} - \int _{0} ^{2} {} e ^{2 t} dt - \int _{0} ^{2} {} 4 t ^{2} dt$ $\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} te ^{2 t} - \frac{1}{2} e ^{2 t} - \frac{4}{3} t ^{3} \end{array} \right] _{0} ^{2}$ $\displaystyle = \frac{3}{2} e ^{4} - \frac{61}{6}$ $\displaystyle \int _{0} ^{5} {} f \left( x \right) dx = \left( - \frac{1}{3} \right) + \left( \frac{3}{2} e ^{4} - \frac{61}{6} \right) = \frac{3}{2} e ^{4} - \frac{21}{2}$ 에서 $\displaystyle p = \frac{3}{2}$, $\displaystyle q = \frac{21}{2}$ 따라서 $p + q = 12$