数Ⅲの部分積分を使った証明

　$\displaystyle \int_{\alpha}^{\beta}(x-\alpha)(x-\beta)\,dx$

$\displaystyle =\left[\dfrac{1}{2}(x-\alpha)^{2}(x-\beta)\right]_{\alpha}^{\beta}-\int_{\alpha}^{\beta}\dfrac{1}{2}(x-\alpha)^{2}\,dx$

$\displaystyle =-\left[\dfrac{1}{6}(x-\alpha)^{3}\right]_{\alpha}^{\beta}$

$\displaystyle =-\dfrac{1}{6}(\beta-\alpha)^{3}$

※ 数Ⅲ既習者はこちらの証明が楽でオススメです．

(Ⅰ)の証明

(ⅰ) $a > 0$ (2次関数が下)のとき

　$\displaystyle \int_{\alpha}^{\beta}\{1次関数 - (ax^{2}+bx+c)\}\,dx$

$\displaystyle =-a\int_{\alpha}^{\beta}(x-\alpha)(x-\beta)\,dx$

$\displaystyle =\dfrac{a}{6}(\beta-\alpha)^{3}$

(ⅱ) $a <0$ (2次関数が上)のとき

　$\displaystyle \int_{\alpha}^{\beta}\{ax^{2}+bx+c - (1次関数)\}\,dx$

$\displaystyle =a\int_{\alpha}^{\beta}(x-\alpha)(x-\beta)\,dx$

$\displaystyle =-\dfrac{a}{6}(\beta-\alpha)^{3}$

以上より求める面積は $\displaystyle \dfrac{|a|}{6}(\beta-\alpha)^3$

(Ⅱ)の証明

(ⅰ) $a_{1}-a_{2} > 0$ ( $y=a_{1}x^{2}+b_{1}x+c_{1}$ が下)のとき

　$\displaystyle \int_{\alpha}^{\beta}\{a_{2}x^{2}+b_{2}x+c_{2} - (a_{1}x^{2}+b_{1}x+c_{1})\}\,dx$

$\displaystyle =(a_{2}-a_{1})\int_{\alpha}^{\beta}(x-\alpha)(x-\beta)\,dx$

$\displaystyle =\dfrac{a_{1}-a_{2}}{6}(\beta-\alpha)^{3}$

(ⅱ) $a_{1}-a_{2} < 0$ ( $y=a_{2}x^{2}+b_{2}x+c_{2}$ が下)のとき

　$\displaystyle \int_{\alpha}^{\beta}\{a_{1}x^{2}+b_{1}x+c_{1} - (a_{2}x^{2}+b_{2}x+c_{2})\}\,dx$

$\displaystyle =(a_{1}-a_{2})\int_{\alpha}^{\beta}(x-\alpha)(x-\beta)\,dx$

$\displaystyle =-\dfrac{a_{1}-a_{2}}{6}(\beta-\alpha)^{3}$

以上より求める面積は $\displaystyle \dfrac{|a_{1}-a_{2}|}{6}(\beta-\alpha)^3$

解答　記述式を想定します．

(1)

交点は

$\dfrac{1}{2}x^{2}-3x+1=\dfrac{1}{2}x-4$

$\Longleftrightarrow x^{2}-7x+10=0$

$\therefore x=2，5$

$\displaystyle S=\int_{2}^{5}\left\{\dfrac{1}{2}x-4-\left(\dfrac{1}{2}x^{2}-3x+1\right)\right\}\,dx$

　$\displaystyle =-\dfrac{1}{2}\int_{2}^{5}(x-2)(x-5)\,dx$

　$\displaystyle =-\dfrac{1}{2}\color{red}{\boldsymbol{\left\{-\dfrac{1}{6}(5-2)^{3}\right\}}}$

　$\displaystyle =\boldsymbol{\dfrac{9}{4}}$

※ マークor答えのみの形式の解答：$S=\dfrac{\left|\frac{1}{2}\right|}{6}(5-2)^{3}=\dfrac{9}{4}$

(2)

交点の $x$ 座標を $\alpha$，$\beta$ $(\alpha < \beta)$ とする．

$x^{2}-ax+a^{2}=-x^{2}+a$

$\displaystyle \ \Longleftrightarrow 2x^{2}-ax+a^{2}-a=0$

必要条件として

$D=a^{2}-8(a^{2}-a)=-7a^{2}+8a > 0$

$\therefore \ 0< a < \dfrac{8}{7}$

この上で

$\alpha=\dfrac{a-\sqrt{D}}{4}$，$\beta=\dfrac{a+\sqrt{D}}{4}$

$\displaystyle T=\int_{\alpha}^{\beta}\{-x^{2}+a-(x^{2}-ax+a^{2})\}\,dx$

　$\displaystyle =-\int_{\alpha}^{\beta}(2x^{2}-ax+a^{2}-a)\,dx$

　$\displaystyle =-2\int_{\alpha}^{\beta}(x-\alpha)(x-\beta)\,dx$

　$\displaystyle =-2\color{red}{\boldsymbol{\left\{-\frac{1}{6}(\beta-\alpha)^{3}\right\}}}$

　$=\dfrac{1}{3}\left(\dfrac{\sqrt{D}}{2}\right)^{3}$

　$\displaystyle =\boldsymbol{\frac{1}{24}(-7a^{2}+8a)^{\frac{3}{2}}}$

　$=\dfrac{1}{24}\left\{-7\left(a-\dfrac{4}{7}\right)^{2}+\dfrac{16}{7}\right\}^{\frac{3}{2}}$

$0< a < \dfrac{8}{7}$ より $a=\dfrac{4}{7}$ のとき，最大値 $\dfrac{1}{24}\cdot\dfrac{64}{7\sqrt{7}}=\boldsymbol{\dfrac{8\sqrt{7}}{147}}$