\(0 \leqq \alpha \leqq \beta\) をみたす実数 \(\alpha , \beta\) と, \(2\) 次式 \(f(x) = x^2 -( \alpha +\beta )x + \alpha \beta\) について, \[ \displaystyle\int _ {-1}^1 f(x) \, dx = 1 \] が成立しているとする. このとき定積分 \[ S = \displaystyle\int _ {0}^{\alpha} f(x) \, dx \] を \(\alpha\) の式で表し, \(S\) がとりうる値の最大値を求めよ.

【 解 答 】

\[\begin{align} \displaystyle\int _ {-1}^1 f(x) \, dx & = 2 \displaystyle\int _ {0}^1 ( x^2 +\alpha \beta ) \, dx \\ & = 2 \left[ \dfrac{x^3}{3} +\alpha \beta x \right] _ 0^1 \\ & = 2 \left( \dfrac{1}{3} +\alpha \beta \right) \end{align}\] なので \[\begin{gather} 2 \left( \dfrac{1}{3} +\alpha \beta \right) = 1 \\ \text{∴} \quad \alpha \beta = \dfrac{1}{6} \end{gather}\] したがって, \(\alpha \neq 0\) の場合を考えればよく \[ \beta = \dfrac{1}{6 \alpha} \] \(0 \lt \alpha \leqq \beta\) なので \[\begin{align} \alpha & \leqq \dfrac{1}{6 \alpha} \\ \text{∴} \quad 0 & \lt \alpha \leqq \dfrac{1}{\sqrt{6}} \quad ... [1] \end{align}\] このとき \[\begin{align} S & = \left[ \dfrac{x^3}{3} -\dfrac{\alpha +\beta}{2} x^2 +\alpha \beta x \right] _ 0^{\alpha} \\ & = \dfrac{\alpha^3}{3} -\dfrac{\alpha^2}{2} \left( \alpha +\dfrac{1}{6 \alpha} \right) +\dfrac{\alpha}{6} \\ & = \underline{\dfrac{1}{12} ( \alpha -2 \alpha^3 )} \end{align}\] したがって \[\begin{align} \dfrac{d S}{d \alpha} & = \dfrac{1}{12} ( 1 -6 \alpha^2 ) \\ & = -\dfrac{1}{12} \left( \alpha +\dfrac{1}{\sqrt{6}} \right) \left( \alpha -\dfrac{1}{\sqrt{6}} \right) \geqq 0 \end{align}\] よって, \(\alpha = \dfrac{1}{\sqrt{6}}\) のとき, \(S\) は最大値 \[\begin{align} S & = \dfrac{1}{12} \left( \dfrac{1}{\sqrt{6}} -\dfrac{2}{6 \sqrt{6}} \right) \\ & = \dfrac{1}{12} \cdot \dfrac{2}{3 \sqrt{6}} \\ & = \underline{\dfrac{\sqrt{6}}{108}} \end{align}\]