実数 \(x\) に対して, \(f(x) = \displaystyle\int _ 0^{\frac{\pi}{2}} \left| \cos t -x \sin 2t \right| \, dt\) とおく.
(1) 関数 \(f(x)\) の最小値を求めよ.
(2) 定積分 \(\displaystyle\int _ 0^1 f(x) \, dx\) を求めよ.
【 解 答 】
(1)
\(g(x) = \cos t -x \sin 2t\) , \(G(x) = \displaystyle\int g(x) \, dx\) とおく. \[\begin{align} g(x) & = \cos t -2x \sin t \cos t = \cos t \left( 1 -2x \sin t \right) \\ G(x) & = \sin t +\dfrac{x}{2} \cos 2t +C \quad ( C \text{は積分定数} ) \end{align}\]
1* \(2x \leqq 1\) すなわち \(x \leqq \dfrac{1}{2}\) のとき
\(0 \leqq t \leqq \dfrac{\pi}{2}\) において, \(g(t) \geqq 0\) なので \[\begin{align} f(x) & = G\left( \dfrac{\pi}{2} \right) -G(0) \\ & = \left( 1-\dfrac{x}{2} \right) -\dfrac{x}{2} \\ & = 1-x \geqq \dfrac{1}{2} \end{align}\]2* \(2x \gt 1\) すなわち \(x \gt \dfrac{1}{2}\) のとき
\(\sin \alpha = \dfrac{1}{2x} \ \left( 0 \leqq \alpha \leqq \dfrac{\pi}{2} \right) \quad ... [1]\) をみたす. \(\alpha\) をおくと, \(g(t)\) の符号は下表のようになる. \[ \begin{array}{c|ccccc} t & 0 & \cdots & \alpha & \cdots & \dfrac{\pi}{2} \\ \hline g(t) & & + & 0 & - & \end{array} \] したがって \[\begin{align} f(x) & = \left\{ G( \alpha ) -G(0) \right\} -\left\{ G \left( \dfrac{\pi}{2} \right) -G( \alpha ) \right\} \\ & = -\dfrac{x}{2} -\left( 1-\dfrac{x}{2} \right) +2 \left( \sin \alpha +\dfrac{x}{2} \cos 2\alpha \right) \\ & = -1 +2 \left\{ \dfrac{1}{2x} +\dfrac{x}{2} \left( 1 -2 \cdot \dfrac{1}{4x^2} \right) \right\} \quad ( \ \text{∵} \ [1] \ ) \\ & = x +\dfrac{1}{2x} -1 \geqq 2\sqrt{x \cdot \dfrac{1}{2x}} -1 \quad ( \ \text{∵} \ \text{ 相加相乗平均の関係} ) \\ & = \sqrt{2}-1 \lt \dfrac{1}{2} \end{align}\] 等号成立は, \(x = \dfrac{1}{2x}\) すなわち \(x = \dfrac{\sqrt{2}}{2}\) のとき
1* 2*より, 求める最小値は \[ \underline{\sqrt{2}-1} \]
(2)
(1) の過程より \[\begin{align} \displaystyle\int _ 0^1 f(x) \, dx & = \displaystyle\int _ 0^{\frac{1}{2}} (1-x) \, dx +\displaystyle\int _ {\frac{1}{2}}^1 \left( x +\dfrac{1}{2x} -1 \right) \, dx \\ & = \left[ x -\dfrac{x^2}{2} \right] _ 0^{\frac{1}{2}} +\left[ \dfrac{x^2}{2} +\dfrac{\log x}{2} -x \right] _ {\frac{1}{2}}^1 \\ & = \left( \dfrac{1}{2} -\dfrac{1}{8} \right) -\dfrac{1}{2} -\left( \dfrac{1}{8} -\dfrac{\log 2}{2} -\dfrac{1}{2} \right) \\ & = \underline{\dfrac{1}{4} +\dfrac{\log 2}{2}} \end{align}\]