多角形D上の点から任意の点への距離の二乗の平均値(グリーンの定理を用いて面積分を周回積分にして求める)

グリーンの定理(Green's theorem)

閉曲線$C$で囲まれた領域$D$を考える場合，$C^1$級凾数$P(x,y),Q(x,y)$について以下が成り立つ。 $$\begin{array}{rcl}{\iint }_D(\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y})\mathrm{d}x\mathrm{d}y& =& {\oint }_{C}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\end{array}$$

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面積分を周回積分へ変形し，区間毎の線積分に展開する

点$x,y$から点$(p,q)$までの距離の二乗は $$\begin{array}{rcl}⟨r^2⟩& =& {\left\{\sqrt{{(x-p)}^2+{(y-q)}^2}\right\}}^2\\ & =& {(x-p)}^2+{(y-q)}^2\end{array}$$ であり，多角形D上の点($x,y$)から任意の点($p,q$)への距離の二乗の平均値$⟨r^2⟩$は， $$\begin{array}{rcl}⟨r^2⟩& =& {\iint }_D\rho (x,y)\{{(x-p)}^2+{(y-q)}^2\};\mathrm{d}x\mathrm{d}y\\ & =& {\iint }_D\frac{1}{A}{(x-p)}^2+{(y-q)}^2\mathrm{d}x\mathrm{d}y\cdots \mathrm{密}\mathrm{度}\mathrm{は}\mathrm{均}\mathrm{一}\mathrm{と}\mathrm{仮}\mathrm{定}:\rho (x,y)=\frac{1}{A}(\mathrm{定}\mathrm{数},A:D\mathrm{の}\mathrm{面}\mathrm{積})\\ & =& \frac{1}{A}{\iint }_D{(x-p)}^2+{(y-q)}^2\mathrm{d}x\mathrm{d}y\cdots {\int }_{X}cA(x)\mathrm{d}x=c{\int }_{X}A(x)\mathrm{d}x\end{array}$$ である．

これをグリーンの定理の式で満たすために例えば $$\begin{array}{rcl}Q(x,y)& =& \int {(x-p)}^2\mathrm{d}x\\ P(x,y)& =& -\int {(y-q)}^2\mathrm{d}y\end{array}$$ とおく．これは $$\begin{array}{rcl}\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y}& =& \int {(x-p)}^2\mathrm{d}x-\{-\int {(y-q)}^2\mathrm{d}y\}\\ & =& {(x-p)}^2+{(y-q)}^2\end{array}$$ となり，グリーンの定理の面積分側の被積分凾数を表現できている．
この$P,Q$を用いて線積分側の 被積分凾数を求めると以下のようになる． $$\begin{array}{rcl}⟨r^2⟩& =& {\int }_D\rho (x,y)r{(x,y;p,q)}^2\mathrm{d}x\mathrm{d}y\\ & =& {\int }_D\rho (x,y)\{{(x-p)}^2+{(y-q)}^2\}\mathrm{d}x\mathrm{d}y\\ & =& {\int }_D\frac{1}{A}\{{(x-p)}^2+{(y-q)}^2\}\mathrm{d}x\mathrm{d}y\\ & =& \frac{1}{A}{\int }_D\{{(x-p)}^2+{(y-q)}^2\}\mathrm{d}x\mathrm{d}y\\ & =& \frac{1}{A}\left[{\int }_D[\frac{\partial }{\partial x}\{\int {(x-p)}^2\mathrm{d}x\}-\frac{\partial }{\partial y}\{-\int {(y-q)}^2\mathrm{d}y\}]\mathrm{d}x\mathrm{d}y\right]\\ & =& \frac{1}{A}[{\oint }_C\{\int {(x-p)}^2\mathrm{d}x\}\mathrm{d}y+\{-\int {(y-q)}^2\mathrm{d}y\}\mathrm{d}x]\\ & =& \frac{1}{A}\{{\oint }_C\frac{1}{3}{(x-p)}^3\mathrm{d}y-\frac{1}{3}{(y-q)}^3\mathrm{d}x\}\\ & =& \frac{1}{A}\{{\oint }_C\frac{1}{3}{(x-p)}^3\mathrm{d}y-{\oint }_C\frac{1}{3}{(y-q)}^3\mathrm{d}x\}\\ & =& \frac{1}{A}\{\frac{1}{3}{\oint }_C{(x-p)}^3\mathrm{d}y-\frac{1}{3}{\oint }_C{(y-q)}^3\mathrm{d}x\}\end{array}$$

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$\left\{\right\}$内の第一項 区間毎の線積分

$$\begin{array}{rcl}\frac{1}{3}{\oint }_C{(x-p)}^3\mathrm{d}y& =& \frac{1}{3}\sum _{k=1}^n{\int }_{y_{k-1}}^{y_k}{\{({ϵ}_{k}y+{\zeta }_k)-p\}}^3\mathrm{d}y\cdots x={ϵ}_{k}y+{\zeta }_k\mathrm{多}\mathrm{角}\mathrm{形}\mathrm{の}\mathrm{区}\mathrm{間}(\mathrm{頂}\mathrm{点}k-1\mathrm{か}\mathrm{ら}k)\mathrm{毎}\mathrm{の}\mathrm{直}\mathrm{線}\mathrm{の}\mathrm{式}\\ & =& \frac{1}{3}\sum _{k=1}^n{\int }_{y_{k-1}}^{y_k}\{{({ϵ}_{k}y+{\zeta }_k)}^3-3{({ϵ}_{k}y+{\zeta }_k)}^{2}p+3({ϵ}_{k}y+{\zeta }_k)p^2-p^3\}\mathrm{d}y\cdots (A-B{)}^3=A^3-3A^{2}B+3A{B}^2-B^3\\ & =& \frac{1}{3}\sum _{k=1}^n{[\frac{1}{4{ϵ}_k}{({ϵ}_{k}y+{\zeta }_k)}^4-3\frac{1}{3{ϵ}_k}{({ϵ}_{k}y+{\zeta }_k)}^{3}p+3\frac{1}{2{ϵ}_k}{({ϵ}_{k}y+{\zeta }_k)}^2{p}^2-y{p}^3]}_{y_{k-1}}^{y_k}\cdots \int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x,{\int }_a^b{x}^a\mathrm{d}x={\left[\frac{1}{a+1}{x}^{a+1}\right]}_a^b\\ & =& \frac{1}{3}\sum _{k=1}[\frac{1}{4{ϵ}_k}\{{({ϵ}_k{y}_k+{\zeta }_k)}^4-{({ϵ}_k{y}_{k-1}+{\zeta }_k)}^4\}-\frac{1}{{ϵ}_k}\{{({ϵ}_k{y}_k+{\zeta }_k)}^3-{({ϵ}_k{y}_{k-1}+{\zeta }_k)}^3\}p+\frac{3}{2{ϵ}_k}\{{({ϵ}_k{y}_k+{\zeta }_k)}^2-{({ϵ}_k{y}_{k-1}+{\zeta }_k)}^2\}{p}^2-(y_k-y_{k-1})p^3]\\ & =& \frac{1}{3}\sum _{k=1}^n[\frac{1}{4{ϵ}_k}\{{ϵ}_k^4(y_k^4-y_{k-1}^4)+4{ϵ}_k^3{\zeta }_k(y_k^3-y_{k-1}^3)+6{ϵ}_k^2{\zeta }_k^2(y_k^2-y_{k-1}^2)+4{ϵ}_k{\zeta }_k^3(y_k-y_{k-1})\}-\frac{1}{{ϵ}_k}\{{ϵ}_k^3(y_k^3-y_{k-1}^3)+3{ϵ}_k^2{\zeta }_k(y_k^2-y_{k-1}^2)+3{ϵ}_k{\zeta }_k^2(y_k-y_{k-1})\}p+\frac{3}{2{ϵ}_k}\{{ϵ}_k^2(y_k^2-y_{k-1}^2)+2{ϵ}_k{\zeta }_k(y_k-y_{k-1})\}{p}^2-(y_k-y_{k-1})p^3]\\ & & \cdots (ax+b{)}^4-(ay+b{)}^4=a^4{x}^4-a^4{y}^4+4a^{3}b{x}^3-4a^{3}b{y}^3+6a^2{b}^2{x}^2-6a^2{b}^2{y}^2+4a{b}^{3}x-4a{b}^{3}y=a^4(x^4-y^4)+4a^{3}b(x^3-y^3)+6a^2{b}^2(x^2-y^2)+4a{b}^3(x-y)\\ & & \cdots (ax+b{)}^3-(ay+b{)}^3=a^3{x}^3-a^3{y}^3+3a^{2}b{x}^2-3a^{2}b{y}^2+3a{b}^{2}x-3a{b}^{2}y=a^3(x^3-y^3)+3a^{2}b(x^2-y^2)+3a{b}^2(x-y)\\ & & \cdots (ax+b{)}^2-(ay+b{)}^2=a^2{x}^2-a^2{y}^2+2abx-2aby=a^2(x^2-y^2)+2ab(x-y)\\ & =& \frac{1}{3}\sum _{k=1}^n[\frac{1}{4{ϵ}_k}\{{ϵ}_k^4(y_k-y_{k-1})(y_k+y_{k-1})(y_k^2+y_{k-1}^2)+4{ϵ}_k^3{\zeta }_k(y_k-y_{k-1})(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+6{ϵ}_k^2{\zeta }_k^2(y_k-y_{k-1})(y_k+y_{k-1})+4{ϵ}_k{\zeta }_k^3(y_k-y_{k-1})\}-\frac{p}{{ϵ}_k}\{{ϵ}_k^3(y_k-y_{k-1})(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+3{ϵ}_k^2{\zeta }_k(y_k-y_{k-1})(y_k+y_{k-1})+3{ϵ}_k{\zeta }_k^2(y_k-y_{k-1})\}+\frac{3p^2}{2{ϵ}_k}\{{ϵ}_k^2(y_k-y_{k-1})(y_k+y_{k-1})+2{ϵ}_k{\zeta }_k(y_k-y_{k-1})\}-p^3(y_k-y_{k-1})]\\ & & \cdots (a^4-b^4)=(a-b)(a+b)(a^2+b^2)\\ & & \cdots (a^3-b^3)=(a-b)(a^2+ab+b^2)\\ & & \cdots (a^2-b^2)=(a-b)(a+b)\\ & =& \frac{1}{3}\sum _{k=1}^n(y_k-y_{k-1})[\frac{1}{4{ϵ}_k}\{{ϵ}_k^4(y_k+y_{k-1})(y_k^2+y_{k-1}^2)+4{ϵ}_k^3{\zeta }_k(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+6{ϵ}_k^2{\zeta }_k^2(y_k+y_{k-1})+4{ϵ}_k{\zeta }_k^3\}-\frac{p}{{ϵ}_k}\{{ϵ}_k^3(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+3{ϵ}_k^2{\zeta }_k(y_k+y_{k-1})+3{ϵ}_k{\zeta }_k^2\}+\frac{3p^2}{2{ϵ}_k}\{{ϵ}_k^2(y_k+y_{k-1})+2{ϵ}_k{\zeta }_k\}-p^3]\\ & =& \frac{1}{3}\sum _{k=1}^n(y_k-y_{k-1})[\frac{1}{4}\{{ϵ}_k^3(y_k+y_{k-1})(y_k^2+y_{k-1}^2)+4{ϵ}_k^2{\zeta }_k(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+6{ϵ}_k{\zeta }_k^2(y_k+y_{k-1})+4{\zeta }_k^3\}-p\{{ϵ}_k^2(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+3{ϵ}_k{\zeta }_k(y_k+y_{k-1})+3{\zeta }_k^2\}+\frac{3p^2}{2}\{{ϵ}_k(y_k+y_{k-1})+2{\zeta }_k\}-p^3]\end{array}$$

${ϵ}_k,{\zeta }_k$の代入と整理

$$\begin{array}{rcl}{ϵ}_k& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}\\ {\zeta }_k& =& \frac{(x_{k-1}{y}_k-x_k{y}_{k-1})}{(y_k-y_{k-1})}\\ \\ & & {ϵ}_k^3(y_k+y_{k-1})(y_k^2+y_{k-1}^2)+4{ϵ}_k^2{\zeta }_k(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+6{ϵ}_k{\zeta }_k^2(y_k+y_{k-1})+4{\zeta }_k^3\\ & =& \frac{(x_k-x_{k-1}{)}^3}{(y_k-y_{k-1}{)}^3}(y_k+y_{k-1})(y_k^2+y_{k-1}^2)+4\frac{(x_k-x_{k-1}{)}^2}{(y_k-y_{k-1}{)}^2}\frac{(x_{k-1}{y}_k-x_k{y}_{k-1})}{(y_k-y_{k-1})}(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+6\frac{(x_k-x_{k-1})}{(y_k-y_{k-1})}\frac{(x_{k-1}{y}_k-x_k{y}_{k-1}{)}^2}{(y_k-y_{k-1}{)}^2}(y_k+y_{k-1})+4\frac{(x_{k-1}{y}_k-x_k{y}_{k-1}{)}^3}{(y_k-y_{k-1}{)}^3}\\ & =& \frac{1}{(y_k-y_{k-1}{)}^3}\{(x_k-x_{k-1}{)}^3(y_k+y_{k-1})(y_k^2+y_{k-1}^2)+4(x_k-x_{k-1}{)}^2(x_{k-1}{y}_k-x_k{y}_{k-1})(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+6(x_k-x_{k-1})(x_{k-1}{y}_k-x_k{y}_{k-1}{)}^2(y_k+y_{k-1})+4(x_{k-1}{y}_k-x_k{y}_{k-1}{)}^3\}\\ & =& (x_k+x_{k-1})(x_k^2+x_{k-1}^2)\\ \\ & & {ϵ}_k^2(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+3{ϵ}_k{\zeta }_k(y_k+y_{k-1})+3{\zeta }_k^2\\ & =& \frac{(x_k-x_{k-1}{)}^2}{(y_k-y_{k-1}{)}^2}(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+3\frac{(x_k-x_{k-1})}{(y_k-y_{k-1})}\frac{(x_{k-1}{y}_k-x_k{y}_{k-1})}{(y_k-y_{k-1})}(y_k+y_{k-1})+3\frac{(x_{k-1}{y}_k-x_k{y}_{k-1}{)}^2}{(y_k-y_{k-1}{)}^2}\\ & =& \frac{1}{(y_k-y_{k-1}{)}^2}\{(x_k-x_{k-1}{)}^2(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+3(x_k-x_{k-1})(x_{k-1}{y}_k-x_k{y}_{k-1})(y_k+y_{k-1})+3(x_{k-1}{y}_k-x_k{y}_{k-1}{)}^2\}\\ & =& x_k^2+x_k{x}_{k-1}+x_{k-1}^2\\ \\ & & {ϵ}_k(y_k+y_{k-1})+2{\zeta }_k\\ & =& \frac{(x_k-x_{k-1})}{(y_k-y_{k-1})}(y_k+y_{k-1})+2\frac{(x_{k-1}{y}_k-x_k{y}_{k-1})}{(y_k-y_{k-1})}\\ & =& \frac{1}{(y_k-y_{k-1})}\{(x_k-x_{k-1})(y_k+y_{k-1})+2(x_{k-1}{y}_k-x_k{y}_{k-1})\}\\ & =& x_k+x_{k-1}\end{array}$$

$\left\{\right\}$内の第一項　整理後

$$\begin{array}{rcl}\frac{1}{3}{\oint }_C{(x-p)}^3\mathrm{d}y& =& \frac{1}{3}\sum _{k=1}^n(y_k-y_{k-1})[\frac{1}{4}\{{ϵ}_k^3(y_k+y_{k-1})(y_k^2+y_{k-1}^2)+4{ϵ}_k^2{\zeta }_k(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+6{ϵ}_k{\zeta }_k^2(y_k+y_{k-1})+4{\zeta }_k^3\}-p\{{ϵ}_k^2(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+3{ϵ}_k{\zeta }_k(y_k+y_{k-1})+3{\zeta }_k^2\}+\frac{3p^2}{2}\{{ϵ}_k(y_k+y_{k-1})+2{\zeta }_k\}-p^3]\\ & =& \frac{1}{3}\sum _{k=1}^n(y_k-y_{k-1})[\frac{1}{4}\{(x_k+x_{k-1})(x_k^2+x_{k-1}^2)\}-p(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)+\frac{3p^2}{2}(x_k+x_{k-1})-p^3]\\ & =& \frac{1}{12}\sum _{k=1}^n(y_k-y_{k-1})\{(x_k+x_{k-1})(x_k^2+x_{k-1}^2)\}-\frac{p}{3}\sum _{k=1}^n(y_k-y_{k-1})(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)+\frac{p^2}{2}\sum _{k=1}^n(y_k-y_{k-1})(x_k+x_{k-1})-\frac{p^3}{3}\sum _{k=1}^n(y_k-y_{k-1})\\ & =& \frac{1}{12}\sum _{k=1}^n(y_k-y_{k-1})\{(x_k+x_{k-1})(x_k^2+x_{k-1}^2)\}-\frac{p}{3}\sum _{k=1}^n(y_k-y_{k-1})(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)+\frac{p^2}{2}\sum _{k=1}^n(y_k-y_{k-1})(x_k+x_{k-1})\cdots \sum _{k=1}^n(y_k-y_{k-1})=0(\mathrm{周}\mathrm{回}\mathrm{積}\mathrm{分}\mathrm{な}\mathrm{の}\mathrm{で}y\mathrm{軸}\mathrm{の}\mathrm{移}\mathrm{動}\mathrm{量}\mathrm{の}\mathrm{和}\mathrm{は}\mathrm{一}\mathrm{周}\mathrm{し}\mathrm{た}\mathrm{ら}\mathrm{元}\mathrm{に}\mathrm{戻}\mathrm{る})\end{array}$$

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$\left\{\right\}$内の第二項　整理後

$$\begin{array}{rcl}-\frac{1}{3}{\oint }_C{(y-q)}^3\mathrm{d}x& =& -[\frac{1}{12}\sum _{k=1}^n(x_k-x_{k-1})\{(y_k+y_{k-1})(y_k^2+y_{k-1}^2)\}-\frac{q}{3}\sum _{k=1}^n(x_k-x_{k-1})(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+\frac{q^2}{2}\sum _{k=1}^n(x_k-x_{k-1})(y_k+y_{k-1})]\end{array}$$

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項毎に計算したものをまとめる

$$\begin{array}{rcl}⟨r^2⟩& =& \frac{1}{A}\{\frac{1}{3}{\oint }_C{(x-p)}^3\mathrm{d}y-\frac{1}{3}{\oint }_C{(y-q)}^3\mathrm{d}x\}\\ & =& \frac{1}{A}[[\frac{1}{12}\sum _{k=1}^n(y_k-y_{k-1})\{(x_k+x_{k-1})(x_k^2+x_{k-1}^2)\}-\frac{p}{3}\sum _{k=1}^n(y_k-y_{k-1})(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)+\frac{p^2}{2}\sum _{k=1}^n(y_k-y_{k-1})(x_k+x_{k-1})]-[\frac{1}{12}\sum _{k=1}^n(x_k-x_{k-1})\{(y_k+y_{k-1})(y_k^2+y_{k-1}^2)\}-\frac{q}{3}\sum _{k=1}^n(x_k-x_{k-1})(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+\frac{q^2}{2}\sum _{k=1}^n(x_k-x_{k-1})(y_k+y_{k-1})]]\\ & =& \frac{1}{A}\left[[\frac{1}{12}\sum _{k=1}^n(y_k-y_{k-1})\{(x_k+x_{k-1})(x_k^2+x_{k-1}^2)\}-\frac{1}{12}\sum _{k=1}^n(x_k-x_{k-1})\{(y_k+y_{k-1})(y_k^2+y_{k-1}^2)\}-2p\frac{1}{6}\sum _{k=1}^n(y_k-y_{k-1})(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)-2q\{-\frac{1}{6}\sum _{k=1}^n(x_k-x_{k-1})(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)\}+p^2\frac{1}{2}\sum _{k=1}^n(y_k-y_{k-1})(x_k+x_{k-1})+q^2\{-\frac{1}{2}\sum _{k=1}^n(x_k-x_{k-1})(y_k+y_{k-1})\}]\right]\\ & =& \frac{1}{A}[\lambda +\mu -2pA{X}_G-2qA{Y}_G+p^{2}A+q^{2}A]\\ & & \cdots \lambda =\frac{1}{12}\sum _{k=1}^n(y_k-y_{k-1})\{(x_k+x_{k-1})(x_k^2+x_{k-1}^2)\}\\ & & \cdots \mu =-\frac{1}{12}\sum _{k=1}^n(x_k-x_{k-1})\{(y_k+y_{k-1})(y_k^2+y_{k-1}^2)\}\\ & & \cdots X_G=\frac{1}{6A}\sum _{k=1}^n(y_k-y_{k-1})(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)(\mathrm{重}\mathrm{心}X\mathrm{軸}\mathrm{位}\mathrm{置})\\ & & \cdots Y_G=-\frac{1}{6A}\sum _{k=1}^n(x_k-x_{k-1})(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)(\mathrm{重}\mathrm{心}Y\mathrm{軸}\mathrm{位}\mathrm{置})\\ & & \cdots A=\frac{1}{2}\sum _{k=1}^n(y_k-y_{k-1})(x_k+x_{k-1})(\mathrm{面}\mathrm{積})\\ & & \cdots A=-\frac{1}{2}\sum _{k=1}^n(x_k-x_{k-1})(y_k+y_{k-1})(\mathrm{面}\mathrm{積})\\ & =& \frac{\lambda +\mu }{A}-2p{X}_G-2q{Y}_G+p^2+q^2\\ & =& \frac{\lambda +\mu }{A}+X_G^2-2p{X}_G+p^2+Y_G^2-2q{Y}_G+q^2-X_G^2-Y_G^2\\ & =& \frac{\lambda +\mu }{A}+{(p-X_G)}^2+{(q-Y_G)}^2-X_G^2-Y_G^2\end{array}$$ よって$⟨r^2⟩$は多角形によって事前に決定される$A,X_G,Y_G,\lambda ,\mu$と，点$(p,q)$が与えられることで決まる．

$⟨r^2⟩$の最小値

$⟨r^2⟩$の最小値となる点$(u,v)$は，$(p-X_G),(p-X_G)$が0となる点$(X_G,Y_G)$である．

また，点$(X_G,Y_G)$から$⟨r^2⟩$の等しい点$(p,q)$は，以下の式を満たすことになる． $$\begin{array}{rcl}⟨r^2⟩& =& \frac{\lambda +\mu }{A}+{(p-X_G)}^2+{(q-Y_G)}^2-X_G^2-Y_G^2\\ {(p-X_G)}^2+{(q-Y_G)}^2& =& ⟨r^2⟩+X_G^2+Y_G^2-\frac{\lambda +\mu }{A}\\ \sqrt{{(p-X_G)}^2+{(q-Y_G)}^2}& =& \sqrt{⟨r^2⟩+X_G^2+Y_G^2-\frac{\lambda +\mu }{A}}\end{array}$$

多角形D上の点から任意の点への距離の平均値 (グリーンの定理を用いて面積分を周回積分にして求める)

グリーンの定理(Green's theorem)

閉曲線$C$で囲まれた領域$D$を考える場合，$C^1$級凾数$P(x,y),Q(x,y)$について以下が成り立つ。 $$\begin{array}{rcl}{\iint }_D(\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y})\mathrm{d}x\mathrm{d}y& =& {\oint }_{C}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\end{array}$$

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面積分を周回積分へ変形し，区間毎の積分に展開する

点$(x,y)$から点$(p,q)$までの距離は $$\begin{array}{rcl}r(x,y;p,q)& =& \sqrt{{(x-p)}^2+{(y-q)}^2}\end{array}$$ であり，多角形D上の点$x,y$から任意の点$p,q$への距離の平均値$\overline{r}$は， $$\begin{array}{rcl}\overline{r}& =& {\iint }_D\rho (x,y)\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x\mathrm{d}y\\ & =& {\iint }_D\frac{1}{A}\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x\mathrm{d}y\cdots \mathrm{密}\mathrm{度}\mathrm{は}\mathrm{均}\mathrm{一}\mathrm{と}\mathrm{仮}\mathrm{定}:\rho (x,y)=\frac{1}{A}(\mathrm{定}\mathrm{数},A:D\mathrm{の}\mathrm{面}\mathrm{積})\\ & =& \frac{1}{A}{\iint }_D\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x\mathrm{d}y\cdots {\int }_{X}cA(x)\mathrm{d}x=c{\int }_{X}A(x)\mathrm{d}x\end{array}$$ である．

これをグリーンの定理の式で満たすために例えば $$\begin{array}{rcl}Q(x,y)& =& \int \sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x\\ P(x,y)& =& 0\end{array}$$ とおく．これは $$\begin{array}{rcl}\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y}& =& \sqrt{{(x-p)}^2+{(y-q)}^2}-0\\ & =& \sqrt{{(x-p)}^2+{(y-q)}^2}\end{array}$$ となり，グリーンの定理の面積分側の被積分凾数を表現できている．
この$P,Q$を用いて周回積分側の 被積分凾数を求めると以下のようになる． $$\begin{array}{rcl}\overline{r}& =& \frac{1}{A}{\iint }_D\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x\mathrm{d}y\\ & =& \frac{1}{A}{\iint }_D(\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y})\mathrm{d}x\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_{C}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_{C}0\mathrm{d}x+\left({\int }_D\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x\right)\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_C(\int \sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x)\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_C(\int \sqrt{u^2+v^2}\mathrm{d}u)\mathrm{d}y\cdots u=x-p,v=y-q\\ & =& \frac{1}{A}{\oint }_C\frac{1}{2}\{u\sqrt{u^2+v^2}+v^2\ln |u+\sqrt{u^2+v^2}|\}\mathrm{d}y\cdots \int \sqrt{x^2+a^2}\mathrm{d}x=\frac{1}{2}\{x\sqrt{x^2+a^2}+a^2\ln |x+\sqrt{x^2+a^2}|\}+\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数}\\ & =& \frac{1}{2A}{\oint }_C\{(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}+{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\}\mathrm{d}y\end{array}$$ ここで周回積分を多角形として考える．多角形の各頂点の列を以下のように与えるとする． $$\begin{array}{r}(x_0,y_0),(x_1,y_1),(x_2,y_2),\cdots ,(x_n,y_n)=(x_0,y_0)\end{array}$$ 周回積分を多角形の区間ごとに分割し，区間毎の積分の和として以下のようになる． $$\begin{array}{rcl}\overline{r}& =& \frac{1}{2A}{\oint }_C\{(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}+{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\}\mathrm{d}y\\ & =& \frac{1}{2A}\sum _{k=1}^n{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}\{(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}+{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\}\mathrm{d}y\\ & =& \frac{1}{2A}\sum _{k=1}^n\{{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}y+{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\mathrm{d}y\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\{I_{1k}+I_{2k}\}\\ & & \cdots I_{1k}={\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}y\\ & & \cdots I_{2k}={\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\mathrm{d}y\end{array}$$

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多角形の頂点間線分(辺)に重なる直線の式を求める

多角形の各頂点において今の頂点$(x_k,y_k)$とその前の頂点$(x_{k-1},y_{k-1})$とを通る直線の式は以下のように求められる． $$\begin{array}{rcl}y-y_{k-1}& =& \frac{y_k-y_{k-1}}{x_k-x_{k-1}}(x-x_{k-1})\cdots \mathrm{傾}\mathrm{き}\theta \mathrm{の}\mathrm{直}\mathrm{線}\mathrm{に}\mathrm{お}\mathrm{け}\mathrm{る}\mathrm{原}\mathrm{点}\mathrm{を}，\mathrm{点}(p,q)\mathrm{へ}\mathrm{移}\mathrm{動}\mathrm{し}\mathrm{た}\mathrm{式}\mathrm{と}\mathrm{同}\mathrm{じ}．(y-q)=\theta (x-p)\\ x-x_{k-1}& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})\end{array}$$ $$\begin{array}{r}\\ x-x_{k-1}& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})\\ x& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})+x_{k-1}\\ x& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(-y_{k-1})+x_{k-1}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{(x_k-x_{k-1})(-y_{k-1})+x_{k-1}(y_k-y_{k-1})}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_k{y}_{k-1}+x_{k-1}{y}_{k-1}+x_{k-1}{y}_k-x_{k-1}{y}_{k-1}}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_k{y}_{k-1}+x_{k-1}{y}_k}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\\ & =& {ϵ}_{k}y+{\zeta }_k\end{array}$$

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区間毎の積分より距離の平均値$\overline{r}$を求める　第一項$I_{1k}$

$$\begin{array}{rcl}{I}_{1k}& =& {\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}y\\ & =& I_{1k}(y_k)-I_{1k}(y_{k-1})\\ I_{1k}(y)& =& \int (({ϵ}_{k}y+{\zeta }_k)-p)\sqrt{{(({ϵ}_{k}y+{\zeta }_k)-p)}^2+{(y-q)}^2}\mathrm{d}y\cdots x={ϵ}_{k}y+{\zeta }_k\\ & =& \int (({ϵ}_{k}y+{\zeta }_k)-p)\sqrt{({({ϵ}_{k}y+{\zeta }_k)}^2-2p({ϵ}_{k}y+{\zeta }_k)+p^2)+(y^2-2qy+q^2)}\mathrm{d}y\\ & =& \int ({ϵ}_{k}y+{\zeta }_k-p)\sqrt{({ϵ}_k^2{y}^2+2{ϵ}_k{\zeta }_{k}y+{\zeta }_k^2)-2p{ϵ}_{k}y-2p{\zeta }_k+p^2+y^2-2qy+q^2}\mathrm{d}y\\ & =& \int ({ϵ}_{k}y+{\zeta }_k-p)\sqrt{{ϵ}_k^2{y}^2+2{ϵ}_k{\zeta }_{k}y+{\zeta }_k^2-2p{ϵ}_{k}y-2p{\zeta }_k+p^2+y^2-2qy+q^2}\mathrm{d}y\\ & =& \int ({ϵ}_{k}y+{\zeta }_k-p)\sqrt{({ϵ}_k^2+1)y^2+2\{{ϵ}_k({\zeta }_k-p)-q\}y+{({\zeta }_k-p)}^2+q^2}\mathrm{d}y\\ & =& \int ({ϵ}_{k}y+{\zeta }_k-p)\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y\cdots {\alpha }_k={ϵ}_k^2+1,{\beta }_k=2\{{ϵ}_k({\zeta }_k-p)-q\},{\gamma }_k={({\zeta }_k-p)}^2+q^2\\ & =& \int {ϵ}_{k}y\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y+\int ({\zeta }_k-p)\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y\\ & =& {ϵ}_k\int y\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y+({\zeta }_k-p)\int \sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y\\ & =& {ϵ}_k\int y\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y+({\zeta }_k-p)\int \sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y\\ & =& {ϵ}_k\{\frac{1}{3{\alpha }_k}{({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}^{\frac{3}{2}}-\frac{{\beta }_k(2{\alpha }_{k}y+{\beta }_k)}{8{\alpha }_k^2}\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}+\frac{{\beta }_k({\beta }_k^2-4{\alpha }_k{\gamma }_k)}{16{\alpha }_k^{\frac{5}{2}}}\ln |(2{\alpha }_{k}y+{\beta }_k)+2\sqrt{{\alpha }_k({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}|+C_0\}\\ & & +({\zeta }_k-p)\{\frac{2{\alpha }_{k}y+{\beta }_k}{4{\alpha }_k}\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}+\frac{4{\alpha }_k{\gamma }_k-{\beta }_k^2}{8{\alpha }_k^{\frac{3}{2}}}\ln |(2{\alpha }_{k}y+{\beta }_k)+2\sqrt{{\alpha }_k({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}|+C_1\}\\ & & \cdots \int x\sqrt{a{x}^2+bx+c}\mathrm{d}x=\frac{1}{3a}{(a{x}^2+bx+c)}^{\frac{3}{2}}-\frac{b(2ax+b)}{8a^2}\sqrt{(a{x}^2+bx+c)}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|+C_0(C_0:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & & \cdots \int \sqrt{a{x}^2+bx+c}\mathrm{d}x=\frac{2ax+b}{4a}\sqrt{(a{x}^2+bx+c)}+\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|+C_1(C_1:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& {ϵ}_k\{\frac{1}{3{\alpha }_k}{({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}^{\frac{3}{2}}-\frac{{\beta }_k(2{\alpha }_{k}y+{\beta }_k)}{8{\alpha }_k^2}\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}+\frac{{\beta }_k({\beta }_k^2-4{\alpha }_k{\gamma }_k)}{16{\alpha }_k^{\frac{5}{2}}}\ln |(2{\alpha }_{k}y+{\beta }_k)+2\sqrt{{\alpha }_k({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}|\}\\ & & +({\zeta }_k-p)\{\frac{2{\alpha }_{k}y+{\beta }_k}{4{\alpha }_k}\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}+\frac{4{\alpha }_k{\gamma }_k-{\beta }_k^2}{8{\alpha }_k^{\frac{3}{2}}}\ln |(2{\alpha }_{k}y+{\beta }_k)+2\sqrt{{\alpha }_k({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}|\}+{ϵ}_k{C}_0+({\zeta }_k-p)C_1\\ & =& {ϵ}_k\{\frac{1}{3{\alpha }_k}{({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}^{\frac{3}{2}}-\frac{{\beta }_k(2{\alpha }_{k}y+{\beta }_k)}{8{\alpha }_k^2}\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}+\frac{{\beta }_k({\beta }_k^2-4{\alpha }_k{\gamma }_k)}{16{\alpha }_k^{\frac{5}{2}}}\ln |(2{\alpha }_{k}y+{\beta }_k)+2\sqrt{{\alpha }_k({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}|\}\\ & & +({\zeta }_k-p)\{\frac{2{\alpha }_{k}y+{\beta }_k}{4{\alpha }_k}\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}+\frac{4{\alpha }_k{\gamma }_k-{\beta }_k^2}{8{\alpha }_k^{\frac{3}{2}}}\ln |(2{\alpha }_{k}y+{\beta }_k)+2\sqrt{{\alpha }_k({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}|\}+C\cdots C={ϵ}_k{C}_0+({\zeta }_k-p)C_1(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\end{array}$$

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区間毎の積分より距離の平均値$\overline{r}$を求める　第二項$I_{2k}$

$$\begin{array}{rcl}{I}_{2k}& =& {\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\mathrm{d}y\\ & =& {\int }_{y_{k-1}}^{y_k}{(y-q)}^2\ln |(({ϵ}_{k}y+{\zeta }_k)-p)+\sqrt{{(({ϵ}_{k}y+{\zeta }_k)-p)}^2+{(y-q)}^2}|\mathrm{d}y\\ & =& {\int }_{y_{k-1}}^{y_k}{(y-q)}^2\ln |{ϵ}_{k}y+{\zeta }_k-p+\sqrt{({({ϵ}_{k}y+{\zeta }_k)}^2-2p({ϵ}_{k}y+{\zeta }_k)+p^2)+(y^2-2qy-q^2)}|\mathrm{d}y\\ & =& {\int }_{y_{k-1}}^{y_k}{(y-q)}^2\ln |{ϵ}_{k}y+{\zeta }_k-p+\sqrt{({ϵ}_k^2{y}^2+2{ϵ}_k{\zeta }_{k}y+{\zeta }_k^2)-2p{ϵ}_{k}y-2p{\zeta }_k+p^2+y^2-2qy-q^2}|\mathrm{d}y\\ & =& {\int }_{y_{k-1}}^{y_k}{(y-q)}^2\ln |{ϵ}_{k}y+{\zeta }_k-p+\sqrt{{ϵ}_k^2{y}^2+y^2+2{ϵ}_k{\zeta }_{k}y-2p{ϵ}_{k}y-2qy+{\zeta }_k^2-2p{\zeta }_k+p^2-q^2}|\mathrm{d}y\\ & =& {\int }_{y_{k-1}}^{y_k}{(y-q)}^2\ln |{ϵ}_{k}y+{\zeta }_k-p+\sqrt{({ϵ}_k^2+1)y^2+2\{{ϵ}_k({\zeta }_k-p)-q\}y+{({\zeta }_k-p)}^2-q^2}|\mathrm{d}y\\ & =& {\int }_{y_{k-1}}^{y_k}{(y-q)}^2\ln |{ϵ}_{k}y+{\zeta }_k-p+\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}|\mathrm{d}y\cdots {\alpha }_k={ϵ}_k^2+1,{\beta }_k=2\{{ϵ}_k({\zeta }_k-p)-q\},{\gamma }_k={({\zeta }_k-p)}^2-q^2\end{array}$$ →要:数値計算

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区間毎の積分より距離の平均値$\overline{r}$を求める　第一項$I_{1k}$+第二項$I_{2k}$

$$\begin{array}{rcl}\overline{r}& =& \frac{1}{2A}\sum _{k=1}^n\{{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}y+{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\mathrm{d}y\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\{I_{1k}+I_{2k}\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\{I_{1k}(y_k)-I_{1k}(y_{k-1})+I_{2k}\}\end{array}$$

x√(a x^2+b x +c)の積分

$$\begin{array}{rcl}\int x\sqrt{a{x}^2+bx+c}\mathrm{d}x& =& \int x\sqrt{{(\sqrt{a}x+\frac{b}{2\sqrt{a}})}^2-\frac{b^2}{4a}+c}\mathrm{d}x\cdots \mathrm{平}\mathrm{方}\mathrm{完}\mathrm{成}\\ & & \cdots a{x}^2+bx+c=(\sqrt{a}x+\alpha {)}^2-{\alpha }^2+c\\ & & \cdots (\sqrt{a}x+\alpha {)}^2=a{x}^2+2\sqrt{a}\alpha x+{\alpha }^2,2\sqrt{a}\alpha =b\mathrm{よ}\mathrm{り}\alpha =\frac{b}{2\sqrt{a}}\\ & =& \int \frac{2\sqrt{a}u-b}{2a}\sqrt{u^2-\frac{b^2}{4a}+c}\frac{1}{\sqrt{a}}\mathrm{d}u\\ & & \cdots u=\sqrt{a}x+\frac{b}{2\sqrt{a}}\\ & & \cdots x=\frac{u-\frac{b}{2\sqrt{a}}}{\sqrt{a}}=\frac{\frac{2\sqrt{a}u-b}{2\sqrt{a}}}{\sqrt{a}}=\frac{2\sqrt{a}u-b}{2a}\\ & & \cdots \frac{\mathrm{d}x}{\mathrm{d}u}=\frac{1}{\sqrt{a}},\mathrm{d}x=\frac{1}{\sqrt{a}}\mathrm{d}u\\ & =& \frac{1}{2a^{\frac{3}{2}}}\int (2\sqrt{a}u-b)\sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u\\ & =& \frac{1}{2a^{\frac{3}{2}}}\int 2\sqrt{a}u\sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u-\frac{1}{2a^{\frac{3}{2}}}\int b\sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u\\ & =& \frac{2\sqrt{a}}{2a^{\frac{3}{2}}}\int u\sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u-\frac{b}{2a^{\frac{3}{2}}}\int \sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u\\ & =& \frac{1}{a}\int u\sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u-\frac{b}{2a^{\frac{3}{2}}}\int \sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u\end{array}$$

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$$\begin{array}{rcl}\frac{1}{a}\int u\sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u& =& \frac{1}{a}\int u\sqrt{s}\frac{1}{2u}\mathrm{d}s\cdots s=u^2-\frac{b^2}{4a}+c,\frac{\mathrm{d}s}{\mathrm{d}u}=2u,\mathrm{d}u=\frac{1}{2u}\mathrm{d}s\\ & =& \frac{1}{2a}\int \sqrt{s}\mathrm{d}s\\ & =& \frac{1}{2a}[\frac{2}{3}{s}^{\frac{3}{2}}+C_0]\cdots \int x^{\frac{1}{2}}\mathrm{d}x=\frac{2}{3}{x}^{\frac{3}{2}}+C_0(C_0:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& \frac{1}{3a}{s}^{\frac{3}{2}}+\frac{1}{2a}{C}_0\\ & =& \frac{1}{3a}{(u^2-\frac{b^2}{4a}+c)}^{\frac{3}{2}}+\frac{1}{2a}{C}_0\\ & =& \frac{1}{3a}{\{{(\sqrt{a}x+\frac{b}{2\sqrt{a}})}^2-\frac{b^2}{4a}+c\}}^{\frac{3}{2}}+\frac{1}{2a}{C}_0\\ & =& \frac{1}{3a}{\{(a{x}^2+bx+\frac{b^2}{4a})-\frac{b^2}{4a}+c\}}^{\frac{3}{2}}+\frac{1}{2a}{C}_0\\ & =& \frac{1}{3a}{(a{x}^2+bx+c)}^{\frac{3}{2}}+\frac{1}{2a}{C}_0\end{array}$$

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$$\begin{array}{rcl}-\frac{b}{2a^{\frac{3}{2}}}\int \sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u& =& -\frac{b}{2a^{\frac{3}{2}}}\int \sqrt{(c-\frac{b^2}{4a})(t^2+1)}\sqrt{c-\frac{b^2}{4a}}\mathrm{d}t\\ & & \cdots t=\frac{u}{\sqrt{c-\frac{b^2}{4a}}},\frac{\mathrm{d}t}{\mathrm{d}u}=\frac{1}{\sqrt{c-\frac{b^2}{4a}}},\mathrm{d}u=\sqrt{c-\frac{b^2}{4a}}\mathrm{d}t\\ & & \cdots u^2+c-\frac{b^2}{4a}=(u^2+c-\frac{b^2}{4a})\frac{c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}}=(c-\frac{b^2}{4a})\frac{u^2+c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}}=(c-\frac{b^2}{4a})(\frac{u^2}{c-\frac{b^2}{4a}}+1)=(c-\frac{b^2}{4a})\{{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2+1\}=(c-\frac{b^2}{4a})(t^2+1)\\ & =& -\frac{b}{2a^{\frac{3}{2}}}(c-\frac{b^2}{4a})\int \sqrt{t^2+1}\mathrm{d}t\\ & =& -\frac{b}{2a^{\frac{3}{2}}}(c-\frac{b^2}{4a})\left\{\frac{1}{2}(t\sqrt{t^2+1}+\ln |t+\sqrt{t^2+1}|+C_1)\right\}\cdots \int \sqrt{x^2+a^2}\mathrm{d}x=\frac{1}{2}(x\sqrt{x^2+a^2}+a^2\ln |x+\sqrt{x^2+a^2}|+C_1)(C_1:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& (\frac{b^3}{16a^{\frac{5}{2}}}-\frac{bc}{4a^{\frac{3}{2}}})(t\sqrt{t^2+1}+\ln |t+\sqrt{t^2+1}|+C_1)\\ & =& (\frac{b^3}{16a^{\frac{5}{2}}}-\frac{4abc}{16a^{\frac{5}{2}}})(t\sqrt{t^2+1}+\ln |t+\sqrt{t^2+1}|+C_1)\\ & =& \frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}(t\sqrt{t^2+1}+\ln |t+\sqrt{t^2+1}|+C_1)\\ & =& \frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2+1}+\ln |\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2+1}|+C_1\}\\ & =& \frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\{\left(\frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{{\left(\frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2+1}+\ln |\left(\frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{{\left(\frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2+1}|+C_1\}\\ & =& \frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\{\frac{2ax+b}{\sqrt{4ac-b^2}}\sqrt{\frac{4a(a{x}^2+bx+c)}{4ac-b^2}}+\ln |\frac{2ax+b}{\sqrt{4ac-b^2}}+\sqrt{\frac{4a(a{x}^2+bx+c)}{4ac-b^2}}|+C_1\}\\ & & \cdots \frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}=\frac{2ax+b}{\sqrt{4ac-b^2}}\\ & & \cdots {\left(\frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2={\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)}^2=\frac{4a^2{x}^2+4abx+b^2}{4ac-b^2}\\ & & \cdots {\left(\frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2+1=\frac{4a^2{x}^2+4abx+b^2}{4ac-b^2}+1=\frac{4a^2{x}^2+4abx+b^2+4ac-b^2}{4ac-b^2}=\frac{4a^2{x}^2+4abx+4ac}{4ac-b^2}=\frac{4a(a{x}^2+bx+c)}{4ac-b^2}\\ & =& \frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\{\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{(a{x}^2+bx+c)}+\ln \left|\frac{1}{\sqrt{4ac-b^2}}\{(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}\}\right|+C_1\}\\ & =& \frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\{\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{a(a{x}^2+bx+c)}+\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|-\ln \sqrt{4ac-b^2}+C_1\}\\ & =& \frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{(a{x}^2+bx+c)}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|-\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln \sqrt{4ac-b^2}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}{C}_1\\ & =& -\frac{b(2ax+b)}{8a^{\frac{4}{2}}}\sqrt{(a{x}^2+bx+c)}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|-\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln \sqrt{4ac-b^2}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}{C}_1\\ & =& -\frac{b(2ax+b)}{8a^2}\sqrt{(a{x}^2+bx+c)}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|-\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln \sqrt{4ac-b^2}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}{C}_1\end{array}$$

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$$\begin{array}{rcl}\int x\sqrt{a{x}^2+bx+c}\mathrm{d}x& =& \frac{1}{a}\int u\sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u-\frac{b}{2a^{\frac{3}{2}}}\int \sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u\\ & =& \frac{1}{3a}{(a{x}^2+bx+c)}^{\frac{3}{2}}+\frac{1}{2a}{C}_0-\frac{b(2ax+b)}{8a^2}\sqrt{(a{x}^2+bx+c)}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|-\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln \sqrt{4ac-b^2}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}{C}_1\\ & & \cdots \frac{1}{a}\int u\sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u=\frac{1}{3a}{(a{x}^2+bx+c)}^{\frac{3}{2}}+\frac{1}{2a}{C}_0\\ & & \cdots -\frac{b}{2a^{\frac{3}{2}}}\int \sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u=-\frac{b(2ax+b)}{8a^2}\sqrt{(a{x}^2+bx+c)}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|-\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln \sqrt{4ac-b^2}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}{C}_1\\ & =& \frac{1}{3a}{(a{x}^2+bx+c)}^{\frac{3}{2}}-\frac{b(2ax+b)}{8a^2}\sqrt{(a{x}^2+bx+c)}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|+\frac{1}{2a}{C}_0-\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln \sqrt{4ac-b^2}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}{C}_1\\ & =& \frac{1}{3a}{(a{x}^2+bx+c)}^{\frac{3}{2}}-\frac{b(2ax+b)}{8a^2}\sqrt{(a{x}^2+bx+c)}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|+C\cdots C=\frac{1}{2a}{C}_0-\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln \sqrt{4ac-b^2}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}{C}_1(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\end{array}$$

√(a x^2+b x +c)の積分

$$\begin{array}{rcl}\int \sqrt{a{x}^2+bx+c}\mathrm{d}x& =& \int \sqrt{{(\sqrt{a}x+\frac{b}{2\sqrt{a}})}^2-\frac{b^2}{4a}+c}\mathrm{d}x\cdots \mathrm{平}\mathrm{方}\mathrm{完}\mathrm{成}\\ & & \cdots a{x}^2+bx+c=(\sqrt{a}x+\alpha {)}^2-{\alpha }^2+c\\ & & \cdots (\sqrt{a}x+\alpha {)}^2=a{x}^2+2\sqrt{a}\alpha x+{\alpha }^2,2\sqrt{a}\alpha =b\mathrm{よ}\mathrm{り}\alpha =\frac{b}{2\sqrt{a}}\\ & =& \int \sqrt{u^2-\frac{b^2}{4a}+c}\frac{1}{\sqrt{a}}\mathrm{d}u\\ & & \cdots u=\sqrt{a}x+\frac{b}{2\sqrt{a}}\\ & & \cdots \frac{\mathrm{d}u}{\mathrm{d}x}=\sqrt{a},\mathrm{d}x=\frac{1}{\sqrt{a}}\mathrm{d}u\\ & =& \frac{1}{\sqrt{a}}\int \sqrt{u^2-\frac{b^2}{4a}+c}\mathrm{d}u\\ & =& \frac{1}{\sqrt{a}}\int \sqrt{(c-\frac{b^2}{4a})(t^2+1)}\sqrt{c-\frac{b^2}{4a}}\mathrm{d}t\\ & & \cdots t=\frac{u}{\sqrt{c-\frac{b^2}{4a}}},\frac{\mathrm{d}t}{\mathrm{d}u}=\frac{1}{\sqrt{c-\frac{b^2}{4a}}},\mathrm{d}u=\sqrt{c-\frac{b^2}{4a}}\mathrm{d}t\\ & & \cdots u^2+c-\frac{b^2}{4a}=(u^2+c-\frac{b^2}{4a})\frac{c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}}=(c-\frac{b^2}{4a})\frac{u^2+c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}}=(c-\frac{b^2}{4a})(\frac{u^2}{c-\frac{b^2}{4a}}+1)=(c-\frac{b^2}{4a})\{{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2+1\}=(c-\frac{b^2}{4a})(t^2+1)\\ & =& \frac{4ac-b^2}{4a^{\frac{3}{2}}}\int \sqrt{t^2+1}\mathrm{d}t\\ & =& \frac{4ac-b^2}{4a^{\frac{3}{2}}}\left\{\frac{1}{2}(t\sqrt{t^2+1}+\ln |t+\sqrt{t^2+1}|+C_0)\right\}\cdots \int \sqrt{x^2+a^2}\mathrm{d}x=\frac{1}{2}(x\sqrt{x^2+a^2}+a^2\ln |x+\sqrt{x^2+a^2}|+C_0)(C_0:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& \frac{4ac-b^2}{8a^{\frac{3}{2}}}(t\sqrt{t^2+1}+\ln |t+\sqrt{t^2+1}|+C_0)\\ & =& \frac{4ac-b^2}{8a^{\frac{3}{2}}}(t\sqrt{t^2+1}+\ln |t+\sqrt{t^2+1}|+C_0)\\ & =& \frac{4ac-b^2}{8a^{\frac{3}{2}}}(t\sqrt{t^2+1}+\ln |t+\sqrt{t^2+1}|+C_0)\\ & =& \frac{4ac-b^2}{8a^{\frac{3}{2}}}\{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2+1}+\ln |\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2+1}|+C_0\}\\ & =& \frac{4ac-b^2}{8a^{\frac{3}{2}}}\{\left(\frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{{\left(\frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2+1}+\ln |\left(\frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{{\left(\frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2+1}|+C_0\}\\ & =& \frac{4ac-b^2}{8a^{\frac{3}{2}}}\{\frac{2ax+b}{\sqrt{4ac-b^2}}\sqrt{\frac{4a(a{x}^2+bx+c)}{4ac-b^2}}+\ln |\frac{2ax+b}{\sqrt{4ac-b^2}}+\sqrt{\frac{4a(a{x}^2+bx+c)}{4ac-b^2}}|+C_0\}\\ & & \cdots \frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}=\frac{2ax+b}{\sqrt{4ac-b^2}}\\ & & \cdots {\left(\frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2={\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)}^2=\frac{4a^2{x}^2+4abx+b^2}{4ac-b^2}\\ & & \cdots {\left(\frac{\sqrt{a}x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)}^2+1=\frac{4a^2{x}^2+4abx+b^2}{4ac-b^2}+1=\frac{4a^2{x}^2+4abx+b^2+4ac-b^2}{4ac-b^2}=\frac{4a^2{x}^2+4abx+4ac}{4ac-b^2}=\frac{4a(a{x}^2+bx+c)}{4ac-b^2}\\ & =& \frac{4ac-b^2}{8a^{\frac{3}{2}}}\{\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{(a{x}^2+bx+c)}+\ln \left|\frac{1}{\sqrt{4ac-b^2}}\{(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}\}\right|+C_0\}\\ & =& \frac{4ac-b^2}{8a^{\frac{3}{2}}}\{\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{a(a{x}^2+bx+c)}+\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|-\ln \sqrt{4ac-b^2}+C_0\}\\ & =& \frac{4ac-b^2}{8a^{\frac{3}{2}}}\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{(a{x}^2+bx+c)}+\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|-\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln \sqrt{4ac-b^2}+\frac{4ac-b^2}{8a^{\frac{3}{2}}}{C}_0\\ & =& \frac{2ax+b}{4a}\sqrt{(a{x}^2+bx+c)}+\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|-\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln \sqrt{4ac-b^2}+\frac{4ac-b^2}{8a^{\frac{3}{2}}}{C}_0\\ & =& \frac{2ax+b}{4a}\sqrt{(a{x}^2+bx+c)}+\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|+C\cdots C=-\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln \sqrt{4ac-b^2}+\frac{4ac-b^2}{8a^{\frac{3}{2}}}{C}_0(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\end{array}$$

√(x^2+a^2)の積分

$$\begin{array}{rcl}\int \sqrt{x^2+a^2}\mathrm{d}x& =& \int \sqrt{a^2{\tan }^2\left(\theta \right)+a^2}a\frac{1}{{\cos }^2\left(\theta \right)}\mathrm{d}\theta \cdots x=a\tan \left(\theta \right),\frac{\mathrm{d}x}{\mathrm{d}\theta }=a\frac{\mathrm{d}}{\mathrm{d}\theta }\tan \left(\theta \right)=a\frac{1}{{\cos }^2\left(\theta \right)},\mathrm{d}x=a\frac{1}{{\cos }^2\left(\theta \right)}\mathrm{d}\theta \\ & =& a\int \sqrt{a^2\{{\tan }^2\left(\theta \right)+1\}}\frac{1}{{\cos }^2\left(\theta \right)}\mathrm{d}\theta \\ & =& a\int a\sqrt{{\tan }^2\left(\theta \right)+1}\frac{1}{{\cos }^2\left(\theta \right)}\mathrm{d}\theta \\ & =& a^2\int \sqrt{\frac{1}{{\cos }^2\left(\theta \right)}}\frac{1}{{\cos }^2\left(\theta \right)}\mathrm{d}\theta \cdots {\tan }^2\left(\theta \right)+1=\frac{{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}+1=\frac{{\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}=\frac{1}{{\cos }^2\left(\theta \right)}\\ & =& a^2\int \frac{1}{\cos \left(\theta \right)}\frac{1}{{\cos }^2\left(\theta \right)}\mathrm{d}\theta \\ & =& a^2\int \frac{1}{{\cos }^3\left(\theta \right)}\mathrm{d}\theta \\ & =& a^2[\frac{1}{2}\{\tan \left(\theta \right)\frac{1}{\cos \left(\theta \right)}+\ln |\tan \left(\theta \right)+\frac{1}{\cos \left(\theta \right)}|\}+C_0]\cdots \int \frac{1}{{\cos }^3\left(\theta \right)}\mathrm{d}\theta =\frac{1}{2}\{\tan \left(\theta \right)\frac{1}{\cos \left(\theta \right)}+\ln |\tan \left(\theta \right)+\frac{1}{\cos \left(\theta \right)}|\}+C_0(C_0:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& \frac{a^2}{2}\{\tan \left(\theta \right)\frac{1}{\cos \left(\theta \right)}+\ln |\tan \left(\theta \right)+\frac{1}{\cos \left(\theta \right)}|+C_0\}\\ & =& \frac{1}{2}\{a^2\tan \left(\theta \right)\frac{1}{\cos \left(\theta \right)}+a^2\ln |\tan \left(\theta \right)+\frac{1}{\cos \left(\theta \right)}|+a^2{C}_0\}\\ & =& \frac{1}{2}\{a^2\frac{1}{a^2}x\sqrt{x^2+a^2}+a^2\ln \left|\frac{1}{a}(x+\sqrt{x^2+a^2})\right|+a^2{C}_0\}\cdots \tan \left(\theta \right)\frac{1}{\cos \left(\theta \right)}=\frac{1}{a^2}x\sqrt{x^2+a^2},\tan \left(\theta \right)+\frac{1}{\cos \left(\theta \right)}=\frac{1}{a}(x+\sqrt{x^2+a^2})\\ & =& \frac{1}{2}[x\sqrt{x^2+a^2}+a^2\{\ln |x+\sqrt{x^2+a^2}|-\ln \left|a\right|\}+a^2{C}_0]\cdots \ln \frac{A}{B}=\ln A-\ln B\\ & =& \frac{1}{2}\{x\sqrt{x^2+a^2}+a^2\ln |x+\sqrt{x^2+a^2}|-a^2\ln \left|a\right|+a^2{C}_0\}\\ & =& \frac{1}{2}\{x\sqrt{x^2+a^2}+a^2\ln |x+\sqrt{x^2+a^2}|\}-\frac{a^2}{2}\ln \left|a\right|+\frac{a^2}{2}{C}_0\\ & =& \frac{1}{2}\{x\sqrt{x^2+a^2}+a^2\ln |x+\sqrt{x^2+a^2}|\}+C\cdots C=-\frac{a^2}{2}\ln \left|a\right|+\frac{a^2}{2}{C}_0(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\end{array}$$

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$$\begin{array}{rcl}\tan \left(\theta \right)\frac{1}{\cos \left(\theta \right)}& =& \tan \left(\theta \right)\sqrt{\frac{1}{{\cos }^2\left(\theta \right)}}\\ & =& \tan \left(\theta \right)\sqrt{{\tan }^2\left(\theta \right)+1}\cdots \frac{1}{{\cos }^2\left(\theta \right)}=\frac{{\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}=\frac{{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}+1={\tan }^2\left(\theta \right)+1\\ & =& \frac{x}{a}\sqrt{{\left(\frac{x}{a}\right)}^2+1}\cdots x=a\tan \left(\theta \right)\mathrm{よ}\mathrm{り}\tan \left(\theta \right)=\frac{x}{a}\\ & =& \frac{x}{a}\sqrt{\frac{x^2+a^2}{a^2}}\\ & =& \frac{x}{a}\sqrt{\frac{1}{a^2}(x^2+a^2)}\\ & =& \frac{x}{a}\frac{1}{a}\sqrt{x^2+a^2}\\ & =& \frac{1}{a^2}x\sqrt{x^2+a^2}\end{array}$$

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$$\begin{array}{rcl}\tan \left(\theta \right)+\frac{1}{\cos \left(\theta \right)}& =& \tan \left(\theta \right)+\sqrt{\frac{1}{{\cos }^2\left(\theta \right)}}\\ & =& \tan \left(\theta \right)+\sqrt{{\tan }^2\left(\theta \right)+1}\cdots \frac{1}{{\cos }^2\left(\theta \right)}=\frac{{\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}=\frac{{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}+1={\tan }^2\left(\theta \right)+1\\ & =& \frac{x}{a}+\sqrt{{\left(\frac{x}{a}\right)}^2+1}\cdots x=a\tan \left(\theta \right)\mathrm{よ}\mathrm{り}\tan \left(\theta \right)=\frac{x}{a}\\ & =& \frac{x}{a}+\sqrt{\frac{x^2+a^2}{a^2}}\\ & =& \frac{x}{a}+\sqrt{\frac{1}{a^2}(x^2+a^2)}\\ & =& \frac{x}{a}+\frac{1}{a}\sqrt{x^2+a^2}\\ & =& \frac{1}{a}(x+\sqrt{x^2+a^2})\end{array}$$

cosの逆数(sec)の三乗の積分

$$\begin{array}{rcl}\int \frac{1}{{\cos }^3\left(\theta \right)}\mathrm{d}\theta & =& \int {\sec }^3\left(\theta \right)\mathrm{d}\theta \cdots \sec \left(\theta \right)=\frac{1}{\cos \left(\theta \right)}\\ & =& \int \frac{1}{\cos \left(\theta \right)}\frac{1}{{\cos }^2\left(\theta \right)}\mathrm{d}\theta \\ & =& \int \frac{1}{\cos \left(\theta \right)}{\{\tan \left(\theta \right)\}}^{\prime }\mathrm{d}\theta \cdots {\{\tan \left(\theta \right)\}}^{\prime }={\left\{\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}\right\}}^{\prime }={\{\sin \left(\theta \right)\}}^{\prime }\frac{1}{\cos \left(\theta \right)}+\sin \left(\theta \right){\left\{\frac{1}{\cos \left(\theta \right)}\right\}}^{\prime }=\cos \left(\theta \right)\frac{1}{\cos \left(\theta \right)}+\sin \left(\theta \right)\frac{\sin \left(\theta \right)}{{\cos }^2\left(\theta \right)}=\frac{\cos \left(\theta \right)}{\cos \left(\theta \right)}+\frac{{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}=\frac{{\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}=\frac{1}{{\cos }^2\left(\theta \right)}\\ & =& \frac{\tan \left(\theta \right)}{\cos \left(\theta \right)}-\int \frac{\sin \left(\theta \right)}{{\cos }^2\left(\theta \right)}\tan \left(\theta \right)\mathrm{d}\theta \cdots \int f^{\prime }g\mathrm{d}x=\left[fg\right]-\int f{g}^{\prime }\mathrm{d}x\\ & =& \frac{\tan \left(\theta \right)}{\cos \left(\theta \right)}-\int \frac{1}{\cos \left(\theta \right)}{\tan }^2\left(\theta \right)\mathrm{d}\theta \\ & =& \frac{\tan \left(\theta \right)}{\cos \left(\theta \right)}-\int \frac{1}{\cos \left(\theta \right)}\frac{{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}\mathrm{d}\theta \\ & =& \frac{\tan \left(\theta \right)}{\cos \left(\theta \right)}-\int \frac{{\sin }^2\left(\theta \right)}{{\cos }^3\left(\theta \right)}\mathrm{d}\theta \\ & =& \frac{\tan \left(\theta \right)}{\cos \left(\theta \right)}-\int \frac{1-{\cos }^2\left(\theta \right)}{{\cos }^3\left(\theta \right)}\mathrm{d}\theta \\ & =& \frac{\tan \left(\theta \right)}{\cos \left(\theta \right)}-\int \{\frac{1}{{\cos }^3\left(\theta \right)}-\frac{{\cos }^2\left(\theta \right)}{{\cos }^3\left(\theta \right)}\}\mathrm{d}\theta \\ & =& \frac{\tan \left(\theta \right)}{\cos \left(\theta \right)}-\int \{\frac{1}{{\cos }^3\left(\theta \right)}-\frac{1}{\cos \left(\theta \right)}\}\mathrm{d}\theta \\ & =& \frac{\tan \left(\theta \right)}{\cos \left(\theta \right)}-\int \frac{1}{{\cos }^3\left(\theta \right)}\mathrm{d}\theta +\int \frac{1}{\cos \left(\theta \right)}\mathrm{d}\theta \\ 2\int \frac{1}{{\cos }^3\left(\theta \right)}\mathrm{d}\theta & =& \frac{\tan \left(\theta \right)}{\cos \left(\theta \right)}+\int \frac{1}{\cos \left(\theta \right)}\mathrm{d}\theta \\ & =& \frac{\tan \left(\theta \right)}{\cos \left(\theta \right)}+\frac{1}{2}\ln \left|\frac{1+\sin \left(\theta \right)}{1-\sin \left(\theta \right)}\right|\cdots \int \frac{1}{\cos \left(\theta \right)}\mathrm{d}\theta =\frac{1}{2}\ln \left|\frac{1+\sin \left(\theta \right)}{1-\sin \left(\theta \right)}\right|+C(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ \int \frac{1}{{\cos }^3\left(\theta \right)}\mathrm{d}\theta & =& \frac{1}{2}\{\frac{\tan \left(\theta \right)}{\cos \left(\theta \right)}+\frac{1}{2}\ln \left|\frac{1+\sin \left(\theta \right)}{1-\sin \left(\theta \right)}\right|\}+C\cdots C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数}\\ & =& \frac{1}{2}\{\frac{\tan \left(\theta \right)}{\cos \left(\theta \right)}+\ln |\frac{1}{\cos \left(\theta \right)}+\tan \left(\theta \right)|\}+C\cdots \frac{1+\sin \left(\theta \right)}{1-\sin \left(\theta \right)}={\{\frac{1}{\cos \left(\theta \right)}+\tan \left(\theta \right)\}}^2\\ & =& \frac{1}{2}\{\sec \left(\theta \right)\tan \left(\theta \right)+\ln |\sec \left(\theta \right)+\tan \left(\theta \right)|\}+C\cdots \sec \left(\theta \right)=\frac{1}{\cos \left(\theta \right)}\end{array}$$

cosの逆数(sec)の積分

$$\begin{array}{rcl}\int \frac{1}{\cos \left(\theta \right)}\mathrm{d}\theta & =& \int \sec \left(\theta \right)\mathrm{d}\theta \cdots \sec \left(\theta \right)=\frac{1}{\cos \left(\theta \right)}\\ & =& \int \frac{1}{\cos \left(\theta \right)}\frac{\cos \left(\theta \right)}{\cos \left(\theta \right)}\mathrm{d}\theta \\ & =& \int \frac{\cos \left(\theta \right)}{{\cos }^2\left(\theta \right)}\mathrm{d}\theta \\ & =& \int \frac{\cos \left(\theta \right)}{1-{\sin }^2\left(\theta \right)}\mathrm{d}\theta \cdots {\cos }^2\left(\theta \right)=1-{\sin }^2\left(\theta \right)\\ & =& \int \frac{1}{2}\{\frac{\cos \left(\theta \right)}{1+\sin \left(\theta \right)}+\frac{\cos \left(\theta \right)}{1-\sin \left(\theta \right)}\}\mathrm{d}\theta \\ & & \cdots \frac{\cos \left(\theta \right)}{1-{\sin }^2\left(\theta \right)}=\frac{\alpha }{1+\sin \left(\theta \right)}+\frac{\beta }{1-\sin \left(\theta \right)}\\ & & \cdots \cos \left(\theta \right)=\alpha (1-\sin \left(\theta \right))+\beta (1+\sin \left(\theta \right))=(\alpha +\beta )+(\alpha -\beta )\sin \left(\theta \right)\\ & & \cdots \{\begin{array}{c}\cos \left(\theta \right)=\alpha +\beta \\ 0=(\alpha -\beta )\sin \left(\theta \right)\end{array}\\ & & \cdots \sin \left(\theta \right)\mathrm{は}\theta \mathrm{に}\mathrm{よ}\mathrm{っ}\mathrm{て}\mathrm{は}0\mathrm{と}\mathrm{は}\mathrm{な}\mathrm{ら}\mathrm{な}\mathrm{い}\mathrm{の}\mathrm{で},\alpha -\beta \mathrm{が}\mathrm{常}\mathrm{に}0,\mathrm{よ}\mathrm{っ}\mathrm{て}\alpha =\beta .\\ & & \cdots \mathrm{従}\mathrm{っ}\mathrm{て}\alpha +\beta =\alpha +\alpha =2\alpha =\cos \left(\theta \right).\\ & & \cdots \alpha =\beta =\frac{1}{2}\cos \left(\theta \right)\\ & & \cdots \frac{\cos \left(\theta \right)}{1-{\sin }^2\left(\theta \right)}=\frac{\frac{1}{2}\cos \left(\theta \right)}{1+\sin \left(\theta \right)}+\frac{\frac{1}{2}\cos \left(\theta \right)}{1-\sin \left(\theta \right)}\\ & =& \frac{1}{2}\int \{\frac{\cos \left(\theta \right)}{1+\sin \left(\theta \right)}+\frac{\cos \left(\theta \right)}{1-\sin \left(\theta \right)}\}\mathrm{d}\theta \\ & =& \frac{1}{2}\{\int \frac{\cos \left(\theta \right)}{1+\sin \left(\theta \right)}\mathrm{d}\theta +\int \frac{\cos \left(\theta \right)}{1-\sin \left(\theta \right)}\mathrm{d}\theta \}\\ & =& \frac{1}{2}\{\int \frac{\cos \left(\theta \right)}{u}\frac{1}{\cos \left(\theta \right)}\mathrm{d}u+\int \frac{\cos \left(\theta \right)}{v}\frac{-1}{\cos \left(\theta \right)}\mathrm{d}v\}\\ & & \cdots u=1+\sin \left(\theta \right),\frac{\mathrm{d}u}{\mathrm{d}\theta }=\cos \left(\theta \right),\mathrm{d}\theta =\frac{1}{\cos \left(\theta \right)}\mathrm{d}u\\ & & \cdots v=1-\sin \left(\theta \right),\frac{\mathrm{d}v}{\mathrm{d}\theta }=-\cos \left(\theta \right),\mathrm{d}\theta =\frac{-1}{\cos \left(\theta \right)}\mathrm{d}v\\ & =& \frac{1}{2}(\int \frac{1}{u}\mathrm{d}u-\int \frac{1}{v}\mathrm{d}v)\\ & =& \frac{1}{2}(\ln \left|u\right|-\ln \left|v\right|)\cdots \int \frac{1}{x}\mathrm{d}x=\ln \left|x\right|+C(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& \frac{1}{2}\ln \frac{\left|u\right|}{\left|v\right|}\cdots \log A-\log B=\log \frac{A}{B}\\ & =& \frac{1}{2}\ln \left|\frac{u}{v}\right|\cdots \frac{\left|A\right|}{\left|B\right|}=\left|\frac{A}{B}\right|\\ & =& \frac{1}{2}\ln \left|\frac{1+\sin \left(\theta \right)}{1-\sin \left(\theta \right)}\right|+C\cdots C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数}\\ \mathrm{も}\mathrm{し}\mathrm{く}\mathrm{は}，\\ & =& \frac{1}{2}\ln \left|\frac{1+\sin \left(\theta \right)}{1-\sin \left(\theta \right)}\right|+C\cdots A\log B=\log B^A\\ & =& \frac{1}{2}\ln \left|{\{\frac{1}{\cos \left(\theta \right)}+\tan \left(\theta \right)\}}^2\right|+C\cdots \frac{1+\sin \left(\theta \right)}{1-\sin \left(\theta \right)}={\{\frac{1}{\cos \left(\theta \right)}+\tan \left(\theta \right)\}}^2\mathrm{式}\mathrm{変}\mathrm{形}\mathrm{は}\mathrm{最}\mathrm{後}\mathrm{に}\mathrm{記}\mathrm{載}\\ & =& \frac{1}{2}2\ln |\frac{1}{\cos \left(\theta \right)}+\tan \left(\theta \right)|+C\cdots {\left(A^B\right)}^{\frac{1}{B}}=A^{\left(B\frac{1}{B}\right)}=A^1=A\\ & =& \ln |\frac{1}{\cos \left(\theta \right)}+\tan \left(\theta \right)|+C\\ & =& \ln |\sec \left(\theta \right)+\tan \left(\theta \right)|+C\cdots \frac{1}{\cos \left(\theta \right)}=\sec \left(\theta \right)\end{array}$$

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$$\begin{array}{rcl}\frac{1+\sin \left(\theta \right)}{1-\sin \left(\theta \right)}& =& \frac{1+\sin \left(\theta \right)}{1-\sin \left(\theta \right)}\frac{1+\sin \left(\theta \right)}{1+\sin \left(\theta \right)}\\ & =& \frac{{\{1+\sin \left(\theta \right)\}}^2}{1-{\sin }^2\left(\theta \right)}\cdots (A+B)(A-B)=A^2-B^2\\ & =& \frac{1+2\sin \left(\theta \right)+{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}\cdots 1-{\sin }^2\left(\theta \right)={\cos }^2\left(\theta \right),(A+B{)}^2=A^2+2AB+B^2\\ & =& \frac{1}{{\cos }^2\left(\theta \right)}+2\frac{\sin \left(\theta \right)}{{\cos }^2\left(\theta \right)}+\frac{{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}\\ & =& \frac{1}{{\cos }^2\left(\theta \right)}+2\frac{1}{\cos \left(\theta \right)}\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}+\frac{{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}\\ & =& \frac{1}{{\cos }^2\left(\theta \right)}+2\frac{1}{\cos \left(\theta \right)}\tan \left(\theta \right)+{\tan }^2\left(\theta \right)\cdots \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}=\tan \left(\theta \right)\\ & =& {\{\frac{1}{\cos \left(\theta \right)}+\tan \left(\theta \right)\}}^2\cdots A^2+2AB+B^2=(A+B{)}^2\end{array}$$

多角形の頂点座標から重心を求める (グリーンの定理を用いて面積分を周回積分にして求める)

グリーンの定理(Green's theorem)

閉曲線$C$で囲まれた領域$D$を考える場合，$C^1$級凾数$P(x,y),Q(x,y)$について以下が成り立つ。 $$\begin{array}{rcl}{\iint }_D(\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y})\mathrm{d}x\mathrm{d}y& =& {\oint }_{C}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\end{array}$$

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面積分を周回積分へ変形し，区間毎の積分に展開する

Dの重心のX座標$X_G$は $$\begin{array}{rcl}{X}_G& =& {\iint }_D\rho (x,y)x\mathrm{d}x\mathrm{d}y\\ & =& {\iint }_D\frac{1}{A}x\mathrm{d}x\mathrm{d}y\cdots \mathrm{密}\mathrm{度}\mathrm{は}\mathrm{均}\mathrm{一}\mathrm{と}\mathrm{仮}\mathrm{定}:\rho (x,y)=\frac{1}{A}(\mathrm{定}\mathrm{数},A:D\mathrm{の}\mathrm{面}\mathrm{積})\\ & =& \frac{1}{A}{\iint }_{D}x\mathrm{d}x\mathrm{d}y\cdots {\int }_{X}cA(x)\mathrm{d}x=c{\int }_{X}A(x)\mathrm{d}x\end{array}$$ であり，これをグリーンの定理の式で満たすために例えば $$\begin{array}{rcl}Q(x,y)& =& \frac{1}{2}{x}^2\\ P(x,y)& =& 0\end{array}$$ とおく(それぞれ一階微分可能で連続($C^1$級凾数))．これは $$\begin{array}{rcl}\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y}& =& x-0\\ & =& x\end{array}$$ となり，グリーンの定理の面積分側の被積分凾数を表現できている．
この$P,Q$を用いて周回積分側の 被積分凾数を求めると以下のようになる． $$\begin{array}{rcl}{X}_G& =& \frac{1}{A}{\iint }_{D}x\mathrm{d}x\mathrm{d}y\\ & =& \frac{1}{A}{\iint }_D(\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y})\mathrm{d}x\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_{C}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_{C}0\mathrm{d}x+\frac{1}{2}{x}^2\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_C\frac{1}{2}{x}^2\mathrm{d}y\\ & =& \frac{1}{2A}{\oint }_C{x}^2\mathrm{d}y\end{array}$$ ここで周回積分を多角形として考える．多角形の各頂点の列を以下のように与えるとする． $$\begin{array}{r}(x_0,y_0),(x_1,y_1),(x_2,y_2),\cdots ,(x_n,y_n)=(x_0,y_0)\end{array}$$  

 

 

 周回積分を多角形の区間ごとに分割し，区間毎の積分の和として以下のようになる． $$\begin{array}{rcl}{X}_G& =& \frac{1}{2A}{\oint }_C{x}^2\mathrm{d}y\\ & =& \frac{1}{2A}\sum _{k=1}^n{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}{x}^2\mathrm{d}y\end{array}$$ 同様にDの重心のY座標$Y_G$は $$\begin{array}{rcl}{Y}_G& =& {\iint }_D\rho (x,y)y\mathrm{d}x\mathrm{d}y\\ & =& {\iint }_D\frac{1}{A}y\mathrm{d}x\mathrm{d}y\cdots \mathrm{密}\mathrm{度}\mathrm{は}\mathrm{均}\mathrm{一}\mathrm{と}\mathrm{仮}\mathrm{定}:\rho (x,y)=\frac{1}{A}(\mathrm{定}\mathrm{数},A:D\mathrm{の}\mathrm{面}\mathrm{積})\\ & =& \frac{1}{A}{\iint }_{D}y\mathrm{d}x\mathrm{d}y\cdots {\int }_{X}cA(x)\mathrm{d}x=c{\int }_{X}A(x)\mathrm{d}x\end{array}$$ であり，これをグリーンの定理の式で満たすために例えば $$\begin{array}{rcl}Q(x,y)& =& 0\\ P(x,y)& =& -\frac{1}{2}{y}^2\end{array}$$ とおく(それぞれ一階微分可能で連続($C^1$級凾数))．これは $$\begin{array}{rcl}\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y}& =& 0-(-y)\\ & =& y\end{array}$$ となり，やはりグリーンの定理の面積分側の被積分凾数を表現できている．
この$P,Q$を用いて周回積分側の 被積分凾数を求めると以下のようになる． $$\begin{array}{rcl}{Y}_G& =& \frac{1}{A}{\iint }_{D}y\mathrm{d}x\mathrm{d}y\\ & =& \frac{1}{A}{\iint }_D(\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y})\mathrm{d}x\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_{C}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_C-\frac{1}{2}{y}^2\mathrm{d}x+0\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_C-\frac{1}{2}{y}^2\mathrm{d}x\\ & =& -\frac{1}{2A}{\oint }_C{y}^2\mathrm{d}x\end{array}$$ X座標同様に周回積分を多角形の区間ごとに分割し，区間毎の積分の和として以下のようになる． $$\begin{array}{rcl}{Y}_G& =& -\frac{1}{2A}{\oint }_C{y}^2\mathrm{d}x\\ & =& -\frac{1}{2A}\sum _{k=1}^n{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}{y}^2\mathrm{d}x\end{array}$$

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多角形の頂点間線分(辺)に重なる直線の式を求める

多角形の各頂点において今の頂点$(x_k,y_k)$とその前の頂点$(x_{k-1},y_{k-1})$とを通る直線の式は以下のように求められる． $$\begin{array}{rcl}y-y_{k-1}& =& \frac{y_k-y_{k-1}}{x_k-x_{k-1}}(x-x_{k-1})\cdots \mathrm{傾}\mathrm{き}\theta \mathrm{の}\mathrm{直}\mathrm{線}\mathrm{に}\mathrm{お}\mathrm{け}\mathrm{る}\mathrm{原}\mathrm{点}\mathrm{を}，\mathrm{点}(p,q)\mathrm{へ}\mathrm{移}\mathrm{動}\mathrm{し}\mathrm{た}\mathrm{式}\mathrm{と}\mathrm{同}\mathrm{じ}．(y-q)=\theta (x-p)\\ x-x_{k-1}& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})\end{array}$$ $$\begin{array}{r}\\ x-x_{k-1}& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})\\ x& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})+x_{k-1}\\ x& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(-y_{k-1})+x_{k-1}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{(x_k-x_{k-1})(-y_{k-1})+x_{k-1}(y_k-y_{k-1})}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_k{y}_{k-1}+x_{k-1}{y}_{k-1}+x_{k-1}{y}_k-x_{k-1}{y}_{k-1}}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_k{y}_{k-1}+x_{k-1}{y}_k}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\end{array}$$ $$\begin{array}{rcl}y-y_{k-1}& =& \frac{y_k-y_{k-1}}{x_k-x_{k-1}}(x-x_{k-1})\\ y& =& \frac{y_k-y_{k-1}}{x_k-x_{k-1}}(x-x_{k-1})+y_{k-1}\\ & =& \frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{y_k-y_{k-1}}{x_k-x_{k-1}}(-x_{k-1})+y_{k-1}\\ & =& \frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{(-x_{k-1})(y_k-y_{k-1})+(x_k-x_{k-1})y_{k-1}}{x_k-x_{k-1}}\\ & =& \frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{-x_{k-1}{y}_k+x_{k-1}{y}_{k-1}+x_k{y}_{k-1}-x_{k-1}{y}_{k-1}}{x_k-x_{k-1}}\\ & =& \frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{-x_{k-1}{y}_k+x_k{y}_{k-1}}{x_k-x_{k-1}}\\ & =& \frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{x_k{y}_{k-1}-x_{k-1}{y}_k}{x_k-x_{k-1}}\end{array}$$

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区間毎の積分より重心を求める

区間毎の積分の和の式における被積分凾数に，先ほど求めた直線の式を代入し積分を行う．まずX座標$X_G$について進める． $$\begin{array}{rcl}{X}_G& =& \frac{1}{2A}\sum _{k=1}^n{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}{x}^2\mathrm{d}y\\ & =& \frac{1}{2A}\sum _{k=1}^n{\int }_{y_{k-1}}^{y_k}{(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}})}^2\mathrm{d}y\cdots x=\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\\ & =& \frac{1}{2A}\sum _{k=1}^n{\int }_{y_{k-1}}^{y_k}\{{\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y\right)}^2+2\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y\right)\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right)+{\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right)}^2\}\mathrm{d}y\cdots x=(A+B{)}^2=A^2+2AB+B^2\\ & =& \frac{1}{2A}\sum _{k=1}^n\{{\int }_{y_{k-1}}^{y_k}{\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y\right)}^2\mathrm{d}y+{\int }_{y_{k-1}}^{y_k}2\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y\right)\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right)\mathrm{d}y+{\int }_{y_{k-1}}^{y_k}{\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right)}^2\mathrm{d}y\}\cdots {\int }_X(A(x)+B(x)+C(x))\mathrm{d}x={\int }_{X}A(x)\mathrm{d}x+{\int }_{X}B(x)\mathrm{d}x+{\int }_{X}C(x)\mathrm{d}x\\ & =& \frac{1}{2A}\sum _{k=1}^n\{{\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\right)}^2{\int }_{y_{k-1}}^{y_k}{y}^2\mathrm{d}y+2\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\right)\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right){\int }_{y_{k-1}}^{y_k}y\mathrm{d}y+{\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right)}^2{\int }_{y_{k-1}}^{y_k}\mathrm{d}y\}\cdots {\int }_{X}cA(x)\mathrm{d}x=c{\int }_{X}A(x)\mathrm{d}x\\ & =& \frac{1}{2A}\sum _{k=1}^n\{{\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\right)}^2{\left[\frac{1}{3}{y}^3\right]}_{y_{k-1}}^{y_k}+2\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\right)\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right){\left[\frac{1}{2}{y}^2\right]}_{y_{k-1}}^{y_k}+{\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right)}^2{\left[y\right]}_{y_{k-1}}^{y_k}\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\{\frac{1}{3}{\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\right)}^2{\left[y^3\right]}_{y_{k-1}}^{y_k}+\frac{1}{2}2\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\right)\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right){\left[y^2\right]}_{y_{k-1}}^{y_k}+{\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right)}^2{\left[y\right]}_{y_{k-1}}^{y_k}\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\{\frac{1}{3}{\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\right)}^2(y_k^3-y_{k-1}^3)+\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\right)\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right)(y_k^2-y_{k-1}^2)+{\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right)}^2(y_k-y_{k-1})\}\end{array}$$ $\left\{\right\}$の中をまとめる． $$\begin{array}{rcl}{X}_G& =& \frac{1}{2A}\sum _{k=1}^n\{\frac{1}{3}{\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\right)}^2(y_k^3-y_{k-1}^3)+\left(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\right)\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right)(y_k^2-y_{k-1}^2)+{\left(\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\right)}^2(y_k-y_{k-1})\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{1}{(y_k-y_{k-1}{)}^2}\{\frac{1}{3}{(x_k-x_{k-1})}^2(y_k^3-y_{k-1}^3)+(x_k-x_{k-1})(x_{k-1}{y}_k-x_k{y}_{k-1})(y_k^2-y_{k-1}^2)+{(x_{k-1}{y}_k-x_k{y}_{k-1})}^2(y_k-y_{k-1})\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{1}{3(y_k-y_{k-1}{)}^2}\{{(x_k-x_{k-1})}^2(y_k-y_{k-1})(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+3(x_k-x_{k-1})(x_{k-1}{y}_k-x_k{y}_{k-1})(y_k-y_{k-1})(y_k+y_{k-1})+3{(x_{k-1}{y}_k-x_k{y}_{k-1})}^2(y_k-y_{k-1})\}\\ & & \cdots A^3-B^3=(A-B)(A^2+AB+B^2),A^2-B^2=(A-B)(A+B)\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{(y_k-y_{k-1})}{3(y_k-y_{k-1}{)}^2}\{{(x_k-x_{k-1})}^2(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+3(x_k-x_{k-1})(x_{k-1}{y}_k-x_k{y}_{k-1})(y_k+y_{k-1})+3{(x_{k-1}{y}_k-x_k{y}_{k-1})}^2\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{1}{3(y_k-y_{k-1})}\{{(x_k-x_{k-1})}^2(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+3(x_{k-1}{y}_k-x_k{y}_{k-1})\{(x_k-x_{k-1})(y_k+y_{k-1})+(x_{k-1}{y}_k-x_k{y}_{k-1})\}\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{1}{3(y_k-y_{k-1})}\{{(x_k-x_{k-1})}^2(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+3(x_{k-1}{y}_k-x_k{y}_{k-1})(x_k{y}_k-x_{k-1}{y}_{k-1})\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{1}{3(y_k-y_{k-1})}\{(x_k^2-2x_k{x}_{k-1}+x_{k-1}^2)(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)+3(x_{k-1}{y}_k-x_k{y}_{k-1})(x_k{y}_k-x_{k-1}{y}_{k-1})\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{1}{3(y_k-y_{k-1})}(x_k^2{y}_k^2+x_k^2{y}_k{y}_{k-1}+x_k^2{y}_{k-1}^2-2x_k{x}_{k-1}{y}_k^2-2x_k{x}_{k-1}{y}_k{y}_{k-1}-2x_k{x}_{k-1}{y}_{k-1}^2+x_{k-1}^2{y}_k^2+x_{k-1}^2{y}_k{y}_{k-1}+x_{k-1}^2{y}_{k-1}^2+3x_k{x}_{k-1}{y}_k^2-3x_{k-1}^2{y}_k{y}_{k-1}-3x_k^2{y}_k{y}_{k-1}+3x_k{x}_{k-1}{y}_{k-1}^2)\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{1}{3(y_k-y_{k-1})}(x_k^2{y}_k^2+x_k^2{y}_{k-1}^2+x_{k-1}^2{y}_k^2+x_{k-1}^2{y}_{k-1}^2+x_k{x}_{k-1}{y}_k^2+x_k{x}_{k-1}{y}_{k-1}^2-2x_k^2{y}_k{y}_{k-1}-2x_{k-1}^2{y}_k{y}_{k-1}-2x_k{x}_{k-1}{y}_k{y}_{k-1})\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{1}{3(y_k-y_{k-1})}(x_k^2(y_k^2+y_{k-1}^2)+x_{k-1}^2(y_k^2+y_{k-1}^2)+x_k{x}_{k-1}(y_k^2+y_{k-1}^2)-2y_k{y}_{k-1}(x_k^2+x_k{x}_{k-1}+x_{k-1}^2))\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{1}{3(y_k-y_{k-1})}\{(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)(y_k^2+y_{k-1}^2)-2y_k{y}_{k-1}(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{1}{3(y_k-y_{k-1})}(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)(y_k^2-2y_k{y}_{k-1}+y_{k-1}^2)\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{1}{3(y_k-y_{k-1})}(x_k^2+x_k{x}_{k-1}+x_{k-1}^2){(y_k-y_{k-1})}^2\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{y_k-y_{k-1}}{3(y_k-y_{k-1})}(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)(y_k-y_{k-1})\\ & =& \frac{1}{2A}\sum _{k=1}^n\frac{1}{3}(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)(y_k-y_{k-1})\\ & =& \frac{1}{2A}\frac{1}{3}\sum _{k=1}^n(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)(y_k-y_{k-1})\\ & =& \frac{1}{6A}\sum _{k=1}^n(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)(y_k-y_{k-1})\end{array}$$ 多角形の頂点座標から，その重心のX座標$X_G$が求められた．
同様に重心のY座標$Y_G$は以下のようになる． $$\begin{array}{rcl}{Y}_G& =& -\frac{1}{2A}\sum _{k=1}^n{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}{y}^2\mathrm{d}y\\ & =& -\frac{1}{2A}\sum _{k=1}^n{\int }_{x_{k-1}}^{x_k}{(\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{x_k{y}_{k-1}-x_{k-1}{y}_k}{x_k-x_{k-1}})}^2\mathrm{d}y\cdots y=\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{x_k{y}_{k-1}-x_{k-1}{y}_k}{x_k-x_{k-1}}\\ & =& -\frac{1}{6A}\sum _{k=1}^n(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)(x_k-x_{k-1})\end{array}$$

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例:正四角形

$$\begin{array}{rcl}(x_0,y_0),(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4=x_0,y_4=y_0)& =& (0,0),(1,0),(1,1),(0,1),(0,0)\\ A& =& 1\end{array}$$ $$\begin{array}{rcl}{X}_G& =& \frac{1}{6A}\sum _{k=1}^n(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)(y_k-y_{k-1})\\ & =& \frac{1}{6\cdot 1}\{(1^2+1\cdot 0+0^2)(0-0)+(1^2+1\cdot 1+1^2)(1-0)+(0^2+0\cdot 1+1^2)(1-1)+(0^2+0\cdot 0+0^2)(0-1)\}\\ & =& \frac{1}{6}(0+3+0+0)\\ & =& \frac{1}{6}3\\ & =& \frac{1}{2}\end{array}$$ $$\begin{array}{rcl}{Y}_G& =& -\frac{1}{6A}\sum _{k=1}^n(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)(x_k-x_{k-1})\\ & =& -\frac{1}{6\cdot 1}\{(0^2+0\cdot 0+0^2)(1-0)+(1^2+1\cdot 0+0^2)(1-1)+(1^2+1\cdot 1+1^2)(0-1)+(0^2+0\cdot 1+1^2)(0-0)\}\\ & =& -\frac{1}{6}(0+0-3+0)\\ & =& -\frac{1}{6}(-3)\\ & =& \frac{1}{2}\end{array}$$ よって重心は以下の点になる． $$\begin{array}{rcl}(X_G,Y_G)& =& (\frac{1}{2},\frac{1}{2})\end{array}$$

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例:二等辺三角形

$$\begin{array}{rcl}(x_0,y_0),(x_1,y_1),(x_2,y_2),(x_3=x_0,y_3=y_0)& =& (0,0),(1,0),(\frac{1}{2},1),(0,0)\\ A& =& \frac{1}{2}\end{array}$$ $$\begin{array}{rcl}{X}_G& =& \frac{1}{6A}\sum _{k=1}^n(x_k^2+x_k{x}_{k-1}+x_{k-1}^2)(y_k-y_{k-1})\\ & =& \frac{1}{6\cdot \frac{1}{2}}[(1^2+1\cdot 0+0^2)(0-0)+\{{\left(\frac{1}{2}\right)}^2+\frac{1}{2}\cdot 1+1^2\}(1-0)+\{0^2+0\cdot \frac{1}{2}+{\left(\frac{1}{2}\right)}^2\}(0-1)]\\ & =& \frac{1}{3}(0+\frac{7}{4}-\frac{1}{4})\\ & =& \frac{1}{3}\frac{6}{4}\\ & =& \frac{1}{2}\end{array}$$ $$\begin{array}{rcl}{Y}_G& =& -\frac{1}{6A}\sum _{k=1}^n(y_k^2+y_k{y}_{k-1}+y_{k-1}^2)(x_k-x_{k-1})\\ & =& -\frac{1}{6\cdot \frac{1}{2}}\{(0^2+0\cdot 0+0^2)(1-0)+(1^2+1\cdot 0+0^2)(\frac{1}{2}-1)+(0^2+0\cdot 1+1^2)(0-\frac{1}{2})\}\\ & =& -\frac{1}{3}(0-\frac{1}{2}-\frac{1}{2})\\ & =& -\frac{1}{3}(-1)\\ & =& \frac{1}{3}\end{array}$$ よって重心は以下の点になる． $$\begin{array}{rcl}(X_G,Y_G)& =& (\frac{1}{2},\frac{1}{3})\end{array}$$

多角形の頂点座標から面積を求める (グリーンの定理を用いて面積分を周回積分にして求める)

グリーンの定理(Green's theorem)

閉曲線$C$で囲まれた領域$D$を考える場合，$C^1$級凾数$P(x,y),Q(x,y)$について以下が成り立つ。 $$\begin{array}{rcl}{\iint }_D(\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y})\mathrm{d}x\mathrm{d}y& =& {\oint }_{C}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\end{array}$$

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面積分を周回積分へ変形し，区間毎の積分に展開する

Dの面積Aは面積分で表すと $$\begin{array}{rcl}A& =& {\iint }_{D}1\mathrm{d}x\mathrm{d}y\end{array}$$ であり，これをグリーンの定理の式で満たすために例えば $$\begin{array}{rcl}Q(x,y)& =& x\\ P(x,y)& =& 0\end{array}$$ とおく(それぞれ一階微分可能で連続($C^1$級凾数))．これは $$\begin{array}{rcl}\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y}& =& 1-0\\ & =& 1\end{array}$$ となり，グリーンの定理の面積分側の被積分凾数を表現できている．
この$P,Q$を用いて周回積分側の 被積分凾数を求めると以下のようになる． $$\begin{array}{rcl}A& =& {\iint }_{D}1\mathrm{d}x\mathrm{d}y\\ & =& {\iint }_D(\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y})\mathrm{d}x\mathrm{d}y\\ & =& {\oint }_{C}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\\ & =& {\oint }_{C}0\mathrm{d}x+x\mathrm{d}y\\ & =& {\oint }_{C}x\mathrm{d}y\end{array}$$ ここで周回積分を多角形として考える．多角形の各頂点の列を以下のように与えるとする． $$\begin{array}{r}(x_0,y_0),(x_1,y_1),(x_2,y_2),\cdots ,(x_n,y_n)=(x_0,y_0)\end{array}$$ 

周回積分を多角形の区間ごとに分割し，区間毎の積分の和として以下のようになる． $$\begin{array}{rcl}A& =& {\oint }_{C}x\mathrm{d}y\\ & =& \sum _{k=1}^n{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}x\mathrm{d}y\end{array}$$

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多角形の頂点間線分(辺)に重なる直線の式を求める

多角形の各頂点において今の頂点$(x_k,y_k)$とその前の頂点$(x_{k-1},y_{k-1})$とを通る直線の式は以下のように求められる． $$\begin{array}{rcl}{y}_{-}{y}_{k-1}& =& \frac{y_k-y_{k-1}}{x_k-x_{k-1}}(x-x_{k-1})\cdots \mathrm{傾}\mathrm{き}\theta \mathrm{の}\mathrm{直}\mathrm{線}\mathrm{に}\mathrm{お}\mathrm{け}\mathrm{る}\mathrm{原}\mathrm{点}\mathrm{を}，\mathrm{点}(p,q)\mathrm{へ}\mathrm{移}\mathrm{動}\mathrm{し}\mathrm{た}\mathrm{式}\mathrm{と}\mathrm{同}\mathrm{じ}．(y-q)=\theta (x-p)\\ \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})& =& x-x_{k-1}\\ x-x_{k-1}& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})\\ x& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})+x_{k-1}\\ x& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(-y_{k-1})+x_{k-1}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{(x_k-x_{k-1})(-y_{k-1})+x_{k-1}(y_k-y_{k-1})}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_k{y}_{k-1}+x_{k-1}{y}_{k-1}+x_{k-1}{y}_k-x_{k-1}{y}_{k-1}}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_k{y}_{k-1}+x_{k-1}{y}_k}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\end{array}$$

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区間毎の積分より面積を求める

区間毎の積分の和の式における被積分凾数に，先ほど求めた直線の式を代入する． $$\begin{array}{rcl}A& =& \sum _{k=1}^n{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}x\mathrm{d}y\\ & =& \sum _{k=1}^n{\int }_{y_{k-1}}^{y_k}(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}})\mathrm{d}y\cdots x=\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\\ & =& \sum _{k=1}^n\{{\int }_{y_{k-1}}^{y_k}\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y\mathrm{d}y+{\int }_{y_{k-1}}^{y_k}\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\mathrm{d}y\}\cdots {\int }_X(A(x)+B(x))\mathrm{d}x={\int }_{X}A(x)\mathrm{d}x+{\int }_{X}B(x)\mathrm{d}x\\ & =& \sum _{k=1}^n\{\frac{x_k-x_{k-1}}{y_k-y_{k-1}}{\int }_{y_{k-1}}^{y_k}y\mathrm{d}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}{\int }_{y_{k-1}}^{y_k}\mathrm{d}y\}\cdots {\int }_{X}cA(x)\mathrm{d}x=c{\int }_{X}A(x)\mathrm{d}x\\ & =& \sum _{k=1}^n\{\frac{x_k-x_{k-1}}{y_k-y_{k-1}}{\left[\frac{1}{2}{y}^2\right]}_{y_{k-1}}^{y_k}+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}{\left[y\right]}_{y_{k-1}}^{y_k}\}\\ & =& \sum _{k=1}^n\{\frac{1}{2}\frac{x_k-x_{k-1}}{y_k-y_{k-1}}{\left[y^2\right]}_{y_{k-1}}^{y_k}+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}{\left[y\right]}_{y_{k-1}}^{y_k}\}\\ & =& \sum _{k=1}^n\{\frac{1}{2}\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y_k^2-y_{k-1}^2)+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}(y_k-y_{k-1})\}\\ & =& \sum _{k=1}^n\{\frac{1}{2}\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y_k+y_{k-1})(y_k-y_{k-1})+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}(y_k-y_{k-1})\}\cdots A^2-B^2=(A+B)(A-B)\\ & =& \sum _{k=1}^n\{\frac{1}{2}(x_k-x_{k-1})(y_k+y_{k-1})+x_{k-1}{y}_k-x_k{y}_{k-1}\}\\ & =& \sum _{k=1}^n\{\frac{1}{2}(x_k{y}_k+x_k{y}_{k-1}-x_{k-1}{y}_k-x_{k-1}{y}_{k-1})+x_{k-1}{y}_k-x_k{y}_{k-1}\}\\ & =& \sum _{k=1}^n\{\frac{1}{2}{x}_k{y}_k+(\frac{1}{2}-1)x_k{y}_{k-1}+(-\frac{1}{2}+1)x_{k-1}{y}_k-\frac{1}{2}{x}_{k-1}{y}_{k-1}\}\\ & =& \sum _{k=1}^n\{\frac{1}{2}{x}_k{y}_k-\frac{1}{2}{x}_k{y}_{k-1}+\frac{1}{2}{x}_{k-1}{y}_k-\frac{1}{2}{x}_{k-1}{y}_{k-1}\}\\ & =& \sum _{k=1}^n\left\{\frac{1}{2}(x_k{y}_k-x_k{y}_{k-1}+x_{k-1}{y}_k-x_{k-1}{y}_{k-1})\right\}\\ & =& \sum _{k=1}^n\left[\frac{1}{2}\{x_k(y_k-y_{k-1})+x_{k-1}(y_k-y_{k-1})\}\right]\\ & =& \sum _{k=1}^n\frac{1}{2}(x_k+x_{k-1})(y_k-y_{k-1})\\ & =& \frac{1}{2}\sum _{k=1}^n(x_k+x_{k-1})(y_k-y_{k-1})\end{array}$$ 多角形の頂点座標から，その面積が求められた．

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例:正四角形

$$\begin{array}{rcl}(x_0,y_0),(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4=x_0,y_4=y_0)& =& (0,0),(1,0),(1,1),(0,1),(0,0)\end{array}$$ $$\begin{array}{rcl}A& =& \frac{1}{2}\sum _{k=1}^n(x_k+x_{k-1})(y_k-y_{k-1})\\ & =& \frac{1}{2}\{(x_1+x_0)(y_1-y_0)+(x_2+x_1)(y_2-y_1)+(x_3+x_2)(y_3-y_2)+(x_4+x_3)(y_4-y_3)\}\\ & =& \frac{1}{2}\{(1+0)(0-0)+(1+1)(1-0)+(0+1)(1-1)+(0+0)(0-1)\}\\ & =& \frac{1}{2}(0+2+0+0)\\ & =& \frac{1}{2}2\\ & =& 1\end{array}$$

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例:二等辺三角形

$$\begin{array}{rcl}(x_0,y_0),(x_1,y_1),(x_2,y_2),(x_3=x_0,y_3=y_0)& =& (0,0),(1,0),(\frac{1}{2},1),(0,0)\end{array}$$ $$\begin{array}{rcl}A& =& \frac{1}{2}\sum _{k=1}^n(x_k+x_{k-1})(y_k-y_{k-1})\\ & =& \frac{1}{2}\{(x_1+x_0)(y_1-y_0)+(x_2+x_1)(y_2-y_1)+(x_3+x_2)(y_3-y_2)\}\\ & =& \frac{1}{2}\{(1+0)(0-0)+(\frac{1}{2}+1)(1-0)+(0+\frac{1}{2})(0-1)\}\\ & =& \frac{1}{2}(0+\frac{3}{2}-\frac{1}{2})\\ & =& \frac{1}{2}1\\ & =& \frac{1}{2}\end{array}$$