{"id":3490,"date":"2024-11-24T11:13:01","date_gmt":"2024-11-24T02:13:01","guid":{"rendered":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/?p=3490"},"modified":"2025-01-15T15:08:14","modified_gmt":"2025-01-15T06:08:14","slug":"%e6%95%b0%e5%ad%a6ii_4q%e6%a8%a1%e6%93%ac","status":"publish","type":"post","link":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/?p=3490","title":{"rendered":"\u6570\u5b66II_4Q\u6a21\u64ec"},"content":{"rendered":"\n\n\n\n\n\n
\u518d\u751f\u30ea\u30b9\u30c8<\/span><\/a><\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
4H1_\u524d\u534a1<\/a><\/td>\n4H1_\u5f8c\u534a1<\/a><\/td>\n2H1_1<\/a><\/td>\n4H2_\u524d\u534a1<\/a><\/td>\n4H2_\u5f8c\u534a1<\/a><\/td>\n2H2_1<\/a><\/td>\n<\/tr>\n
4H1_\u524d\u534a2<\/a><\/td>\n4H1_\u5f8c\u534a2<\/a><\/td>\n2H1_2<\/a><\/td>\n4H2_\u524d\u534a2<\/a><\/td>\n4H2_\u5f8c\u534a2<\/a><\/td>\n2H2_2<\/a><\/td>\n<\/tr>\n
4H11_pdf<\/a><\/td>\n4H12_pdf<\/a><\/td>\n2H1_pdf<\/a><\/td>\n4H21_pdf<\/a><\/td>\n4H22_pdf<\/a><\/td>\n2H2_pdf<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
4H1_\u524d\u534a<\/span>4H1_\u5f8c\u534a<\/span>2H1<\/span>4H2_\u524d\u534a<\/span>4H2_\u5f8c\u534a<\/span>2H2<\/span><\/div>
\n

\u4e0d\u5b9a\u7a4d\u5206\uff1a\u516c\u5f0f II<\/span><\/p>\n

1\uff0e\u6b21\u306e\u4e0d\u5b9a\u7a4d\u5206\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $\\displaystyle\\int\\dfrac{1}{x^2+25}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\displaystyle\\int\\dfrac{1}{x^2+a^2}\\,dx=\\dfrac{1}{a}\\mathrm{Tan}^{-1}\\dfrac{x}{a}+C\\;(a\\gt 0)$\u3000\u3092\u7528\u3044\u308b
\n$\\displaystyle\\int\\dfrac{1}{x^2+25}\\,dx=\\dfrac{1}{5}\\mathrm{Tan}^{-1}\\dfrac{x}{5}+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(2) $\\displaystyle\\int\\dfrac{1}{x^2-25}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\displaystyle\\int\\dfrac{1}{x^2-a^2}\\,dx=\\dfrac{1}{2a}\\mathrm{log}\\left|\\dfrac{x-a}{x+a}\\right|+C\\;(a\\gt 0)$\u3000\u3092\u7528\u3044\u308b
\n$\\displaystyle\\int\\dfrac{1}{x^2-25}\\,dx=\\dfrac{1}{10}\\mathrm{log}\\left|\\dfrac{x-5}{x+5}\\right|+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(3) $\\displaystyle\\int\\dfrac{1}{\\sqrt{x^2+25}}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\displaystyle\\int\\dfrac{1}{\\sqrt{x^2+A}}\\,dx=\\mathrm{log}\\left|x+\\sqrt{x^2+A}\\right|+C\\;(A\\neq 0)$\u3000\u3092\u7528\u3044\u308b
\n$\\displaystyle\\int\\dfrac{1}{\\sqrt{x^2+25}}\\,dx=\\mathrm{log}\\left|x+\\sqrt{x^2+25}\\right|+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(4) $\\displaystyle\\int\\dfrac{1}{\\sqrt{25-x^2}}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\displaystyle\\int\\dfrac{1}{\\sqrt{a^2-x^2}}\\,dx=\\mathrm{Sin}^{-1}\\dfrac{x}{a}+C\\;(a\\gt 0)$\u3000\u3092\u7528\u3044\u308b
\n$\\displaystyle\\int\\dfrac{1}{\\sqrt{25-x^2}}\\,dx=\\mathrm{Sin}^{-1}\\dfrac{x}{5}+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(5) $\\displaystyle\\int\\dfrac{1}{\\sqrt{x^2-25}}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\displaystyle\\int\\dfrac{1}{\\sqrt{x^2+A}}\\,dx=\\mathrm{log}\\left|x+\\sqrt{x^2+A}\\right|+C\\;(A\\neq 0)$\u3000\u3092\u7528\u3044\u308b
\n$\\displaystyle\\int\\dfrac{1}{\\sqrt{x^2-25}}\\,dx=\\mathrm{log}\\left|x+\\sqrt{x^2-25}\\right|+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(6) $\\displaystyle\\int\\sqrt{x^2+25}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\displaystyle\\int\\sqrt{x^2+A}\\,dx=\\dfrac{1}{2}\\left(x\\sqrt{x^2+A}+A\\mathrm{log}\\left|x+\\sqrt{x^2+A}\\right|\\right)+C\\;(A\\neq 0)$\u3000\u3092\u7528\u3044\u308b
\n$\\displaystyle\\int\\sqrt{x^2+25}\\,dx=\\dfrac{1}{2}\\left(x\\sqrt{x^2+25}+25\\mathrm{log}\\left|x+\\sqrt{x^2+25}\\right|\\right)+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(7) $\\displaystyle\\int\\sqrt{25-x^2}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\displaystyle\\int\\sqrt{a^2-x^2}\\,dx=\\dfrac{1}{2}\\left(x\\sqrt{a^2-x^2}+a^2\\mathrm{Sin}^{-1}\\dfrac{x}{a}\\right)+C\\;(a\\gt 0)$\u3000\u3092\u7528\u3044\u308b
\n$\\displaystyle\\int\\sqrt{25-x^2}\\,dx=\\dfrac{1}{2}\\left(x\\sqrt{25-x^2}+25 \\mathrm{Sin}^{-1}\\dfrac{x}{5}\\right)+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(8) $I=\\displaystyle\\int\\dfrac{1}{x^2+2x+5}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\displaystyle\\int\\dfrac{1}{(ax+b)^2+c^2}\\,dx=\\dfrac{1}{ac}\\mathrm{Tan}^{-1}\\dfrac{ax+b}{c}+C\\;(c\\gt 0, a\\neq 0)$\u3000\u3092\u7528\u3044\u308b
\n$I=\\displaystyle\\int\\dfrac{1}{(x^2+2x+1)-1+5}\\,dx=\\displaystyle\\int\\dfrac{1}{(x+1)^2+4}\\,dx=\\dfrac{1}{2}\\mathrm{Tan}^{-1}\\dfrac{x+1}{2}+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(9) $I=\\displaystyle\\int\\dfrac{1}{x^2+2x-3}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\displaystyle\\int\\dfrac{1}{(ax+b)^2-c^2}\\,dx=\\dfrac{1}{2ac}\\mathrm{log}\\left|\\dfrac{(ax+b)-c}{(ax+b)+c}\\right|+C\\;(c\\gt 0, a\\neq 0)$\u3000\u3092\u7528\u3044\u308b
\n$I=\\displaystyle\\int\\dfrac{1}{(x^2+2x+1)-1-3}\\,dx=\\displaystyle\\int\\dfrac{1}{(x+1)^2-4}\\,dx$
\n$\\quad=\\dfrac{1}{2\\cdot1\\cdot2}\\mathrm{log}\\left|\\dfrac{(x+1)-2}{(x+1)+2}\\right|+C=\\dfrac{1}{4}\\mathrm{log}\\left|\\dfrac{x-1}{x+3}\\right|+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(10) $I=\\displaystyle\\int\\dfrac{1}{\\sqrt{x^2+2x+5}}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\displaystyle\\int\\dfrac{1}{\\sqrt{(ax+b)^2+A}}\\,dx=\\dfrac{1}{a}\\mathrm{log}\\left|ax+b+\\sqrt{(ax+b)^2+A}\\right|+C\\;(A, a\\neq 0)$
\n$I=\\displaystyle\\int\\dfrac{1}{\\sqrt{(x^2+2x+1)-1+5}}\\,dx=\\displaystyle\\int\\dfrac{1}{\\sqrt{(x+1)^2+4}}\\,dx$
\n$\\quad=\\dfrac{1}{1}\\mathrm{log}\\left|(x+1)+\\sqrt{(x+1)^2+4}\\right|+C=\\mathrm{log}\\left|x+1+\\sqrt{x^2+2x+5}\\right|+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(11) $I=\\displaystyle\\int\\dfrac{1}{\\sqrt{3-x^2-2x}}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\displaystyle\\int\\dfrac{1}{\\sqrt{c^2-(ax+b)^2}}\\,dx=\\dfrac{1}{a}\\mathrm{Sin}^{-1}\\dfrac{ax+b}{c}+C\\;(c\\gt 0, a\\neq0)$\u3000\u3092\u7528\u3044\u308b
\n$I=\\displaystyle\\int\\dfrac{1}{\\sqrt{3+1-(x^2+2x+1)}}\\,dx=\\displaystyle\\int\\dfrac{1}{\\sqrt{4-(x+1)^2}}\\,dx=\\mathrm{Sin}^{-1}\\dfrac{x+1}{2}+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

\u4e0d\u5b9a\u7a4d\u5206\uff1a\u90e8\u5206\u5206\u6570\u5206\u89e3<\/span><\/p>\n

2\uff0e\u6b21\u306e\u4e0d\u5b9a\u7a4d\u5206\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $I=\\displaystyle\\int\\dfrac{5}{(x-1)(x+3)}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\dfrac{5}{(x-1)(x+3)}=\\dfrac{a}{x-1}+\\dfrac{b}{x+3}=\\dfrac{(a+b)x+3a-b}{(x-1)(x+3)}$\u3000\u3088\u308a
\n$a+b=0, 3a-b=5 \\Rightarrow a=\\dfrac{5}{4}, b=-\\dfrac{5}{4}$ \u3060\u304b\u3089
\n$I=\\dfrac{5}{4}\\displaystyle\\int\\left(\\dfrac{1}{x-1}-\\dfrac{1}{x+3}\\right)\\,dx=\\dfrac{5}{4}\\underbrace{\\mathrm{log}\\left|\\dfrac{x-1}{x+3}\\right|}_{\\mathrm{log}|x-1|-\\mathrm{log}|x+3|}+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(2) $I=\\displaystyle\\int\\dfrac{9x-4}{(3x+1)(x-2)}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\dfrac{9x-4}{(3x+1)(x-2)}=\\dfrac{a}{3x+1}+\\dfrac{b}{x-2}=\\dfrac{(a+3b)x-2a+b}{(3x+1)(x-2)}$\u3000\u3088\u308a
\n$a+3b=9, -2a+b=-4 \\Rightarrow a=3, b=2$ \u3060\u304b\u3089
\n$I=\\displaystyle\\int\\left(\\dfrac{3}{3x+1}+2\\cdot\\dfrac{1}{x-2}\\right)\\,dx=\\underbrace{\\mathrm{log}|3x+1|(x-2)^2}_{\\mathrm{log}|3x+1|+2\\mathrm{log}|x-2|}+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(3) $I=\\displaystyle\\int\\dfrac{x^2+3x}{(x^2+1)(x+1)}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\dfrac{x^2+3x}{(x^2+1)(x+1)}=\\dfrac{ax+b}{x^2+1}+\\dfrac{c}{x+1}=\\dfrac{(a+c)x^2+(a+b)x+(b+c)}{(x^2+1)(x+1)}$\u3000\u3088\u308a
\n$a+c=1, a+b=3, b+c=0 \\Rightarrow a=2, b=1, c=-1$ \u3060\u304b\u3089
\n$I=\\displaystyle\\int\\left(\\dfrac{2x+1}{x^2+1}-\\dfrac{1}{x+1}\\right)\\,dx=\\displaystyle\\int\\left(\\dfrac{1}{x^2+1}+\\dfrac{2x}{x^2+1}-\\dfrac{1}{x+1}\\right)\\,dx$
\n$\\quad=\\mathrm{Tan}^{-1}x+\\underbrace{\\mathrm{log}\\dfrac{x^2+1}{|x+1|}}_{\\mathrm{log}(x^2+1)-\\mathrm{log}|x-1|}+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(4) $I=\\displaystyle\\int\\dfrac{x-3}{(x-1)^2(x+1)}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\dfrac{x-3}{(x-1)^2(x+1)}=\\dfrac{a}{(x-1)^2}+\\dfrac{b}{x-1}+\\dfrac{c}{x+1}=\\dfrac{(b+c)x^2+(a-2c)x+(a-b+c)}{(x-1)^2(x+1)}$
\n\u3088\u308a\u3000$b+c=0, a-2c=1, a-b+c=3 \\Rightarrow a=-1, b=1, c=-1$ \u3060\u304b\u3089
\n$I=\\displaystyle\\int\\left(\\dfrac{-1}{(x-1)^2}+\\dfrac{1}{x-1}-\\dfrac{1}{x+1}\\right)\\,dx=\\displaystyle\\int\\left(-(x-1)^{-2}+\\dfrac{1}{x-1}-\\dfrac{1}{x+1}\\right)\\,dx$
\n$=\\underbrace{\\dfrac{1}{x-1}}_{\\dfrac{-1}{-1}(x-1)^{-1}}+\\underbrace{\\mathrm{log}\\left|\\dfrac{x-1}{x+1}\\right|}_{\\mathrm{log}|x-1|-\\mathrm{log}|x+1|}+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n<\/div>\n
\n

\u4e0d\u5b9a\u7a4d\u5206\uff1a\u4e09\u89d2\u95a2\u6570\u306e\u6709\u7406\u5f0f<\/span><\/p>\n

1\uff0e\u6b21\u306e\u4e0d\u5b9a\u7a4d\u5206\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $I=\\displaystyle\\int\\dfrac{1}{2\\mathrm{cos}x+3\\mathrm{sin}x+2}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$u=\\mathrm{tan}\\dfrac{x}{2} \\Rightarrow \\mathrm{cos}x=\\dfrac{1-u^2}{1+u^2}, \\mathrm{sin}x=\\dfrac{2u}{1+u^2}, dx=\\dfrac{2}{1+u^2}du$ \u3088\u308a
\n$I=\\displaystyle\\int\\dfrac{1}{\\left(2\\cdot\\dfrac{1-u^2}{1+u^2}+3\\cdot\\dfrac{2u}{1+u^2}+2 \\right)}\\dfrac{2}{1+u^2}\\,du$
\n$\\quad=\\displaystyle\\int\\underbrace{\\dfrac{2}{2\\cdot(1-u^2)+3\\cdot 2u+2(1+u^2)}}_{4+6u=2(3u+2)}\\,du$
\n$\\quad=\\displaystyle\\int\\dfrac{1}{3u+2}\\,du=\\dfrac{1}{3}\\displaystyle\\int\\dfrac{3}{3u+2}\\,du=\\dfrac{1}{3}\\mathrm{log}|3u+2|+C=\\dfrac{1}{3}\\mathrm{log}|3 \\mathrm{tan}\\dfrac{x}{2}+2|+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(2) $I=\\displaystyle\\int\\dfrac{1}{\\mathrm{cos}x+2\\mathrm{sin}x+3}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$u=\\mathrm{tan}\\dfrac{x}{2} \\Rightarrow \\mathrm{cos}x=\\dfrac{1-u^2}{1+u^2}, \\mathrm{sin}x=\\dfrac{2u}{1+u^2}, dx=\\dfrac{2}{1+u^2}du$ \u3088\u308a
\n$I=\\displaystyle\\int\\dfrac{1}{\\left(\\dfrac{1-u^2}{1+u^2}+2\\cdot\\dfrac{2u}{1+u^2}+3 \\right)}\\dfrac{2}{1+u^2}\\,du$
\n$\\quad=\\displaystyle\\int\\underbrace{\\dfrac{2}{(1-u^2)+2\\cdot 2u+3(1+u^2)}}_{2u^2+4u+4=2(u^2+2u+2)=2\\{(u+1)^2+1\\}}\\,du$
\n$\\quad=\\displaystyle\\int\\dfrac{1}{(u+1)^2+1}\\,du=\\mathrm{Tan}^{-1}(u+1)+C=\\mathrm{Tan}^{-1}\\left(\\mathrm{tan}\\dfrac{x}{2}+1\\right)+C$ (\u7b54)<\/div><\/div><\/span><\/p>\n

\u5b9a\u7a4d\u5206\uff1a\u5076\u95a2\u6570\u3068\u5947\u95a2\u6570<\/span><\/p>\n

2\uff0e\u6b21\u306e\u5b9a\u7a4d\u5206\u306e\u5024\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $I=\\displaystyle\\int_{-2}^{2}(x^5-2x^3+3x^2+5x+1)\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$x^5-2x^3+5x$\u306f\u5947\u95a2\u6570\u3060\u304b\u3089\u3001$\\displaystyle\\int_{-2}^{2}(x^5-2x^3+5x)\\,dx=0$ \u307e\u305f\u3001$3x^2+1$ \u306f\u5076\u95a2\u6570\u3060\u304b\u3089
\n$I=\\displaystyle\\int_{-2}^{2}(3x^2+1)\\,dx=2\\displaystyle\\int_{0}^{2}(3x^2+1)\\,dx=2\\lbrack x^3+x \\rbrack_{0}^{2}=2(8+2)=20$ (\u7b54)<\/span><\/div><\/div>\n

(2) $I=\\displaystyle\\int_{-1}^{1}\\dfrac{x\\mathrm{cos}x+1}{x^2+1}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\displaystyle\\int_{-1}^{1}\\dfrac{x\\mathrm{cos}x}{x^2+1}\\,dx+\\displaystyle\\int_{-1}^{1}\\dfrac{1}{x^2+1}\\,dx$ \u3053\u3053\u3067\u3001$f(x)=\\dfrac{x\\mathrm{cos}x}{x^2+1}$ \u3068\u304a\u304f\u3068\u3001
\n$f(x)=\\dfrac{x\\mathrm{cos}x}{x^2+1} \\Rightarrow f(-x)=\\dfrac{-x\\mathrm{cos}(-x)}{(-x)^2+1}=\\dfrac{-x\\mathrm{cos}x}{x^2+1}=-f(x) $ \u3088\u308a\u5947\u95a2\u6570\u3060\u304b\u3089\u3001
\n$\\displaystyle\\int_{-1}^{1}\\dfrac{x\\mathrm{cos}x}{x^2+1}\\,dx=0$ \u307e\u305f\u3001$\\dfrac{1}{x^2+1}$ \u306f\u5076\u95a2\u6570\u3060\u304b\u3089\u3001
\n$I=\\displaystyle\\int_{-1}^{1}\\dfrac{1}{x^2+1}\\,dx=2\\displaystyle\\int_{0}^{1}\\dfrac{1}{x^2+1}\\,dx=2\\left\\lbrack \\mathrm{Tan}^{-1}x \\right\\rbrack _0^1 =2(\\underbrace{\\mathrm{Tan}^{-1}1}_{\\scriptsize \\dfrac{\\pi}{4}}-0)=\\dfrac{\\pi}{2}$<\/span><\/div><\/div>\n

\u5b9a\u7a4d\u5206\uff1a\u57fa\u672c<\/span><\/p>\n

3\uff0e\u6b21\u306e\u5b9a\u7a4d\u5206\u306e\u5024\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $I=\\displaystyle\\int_{-2}^{1}x^2\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\dfrac{1}{3}\\left\\lbrack x^3 \\right\\rbrack_{-2}^{1}=\\dfrac{1}{3}\\left\\lbrace 1^3 -(-2)^3 \\right \\rbrace =\\dfrac{9}{3}=3$ (\u7b54)<\/span><\/div><\/div>\n

(2) $I=\\displaystyle\\int_{0}^{7}\\dfrac{1}{\\sqrt[3]{x+1}}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\displaystyle\\int_{0}^{7}(x+1)^{\\scriptsize\\dfrac{-1}{3}}\\,dx=\\dfrac{3}{2}\\left\\lbrack (x+1)^{\\scriptsize\\dfrac{2}{3}} \\right\\rbrack_{0}^{7}=\\dfrac{3}{2}\\left\\lbrace 8^{\\scriptsize\\dfrac{2}{3}} -1^{\\scriptsize\\dfrac{2}{3}} \\right \\rbrace=\\dfrac{3}{2}\\left\\lbrace (2^3)^{\\scriptsize\\dfrac{2}{3}} -1^{\\scriptsize\\dfrac{2}{3}} \\right \\rbrace$
\n$\\quad =\\dfrac{3}{2}(2^2-1)=\\dfrac{9}{2}$ (\u7b54)<\/span><\/div><\/div>\n

(3) $I=\\displaystyle\\int_{0}^{1}\\dfrac{1}{x+1}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\left\\lbrack \\mathrm{log}|x+1| \\right\\rbrack_{0}^{1}=\\mathrm{log}2 -\\underbrace{\\mathrm{log}1}_{0} =\\mathrm{log}2$ (\u7b54)<\/span><\/div><\/div>\n

(4) $I=\\displaystyle\\int_{2}^{3}e^{2x-4}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\dfrac{1}{2}\\left\\lbrack e^{2x-4} \\right\\rbrack_{2}^{3}=\\dfrac{1}{2}\\left(e^2 -\\underbrace{e^0}_{1}\\right) =\\dfrac{e^2-1}{2}$ (\u7b54)<\/span><\/div><\/div>\n

(5) $I=\\displaystyle\\int_{0}^{\\scriptsize\\dfrac{\\pi}{4}}\\mathrm{sin}3x\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\dfrac{-1}{3}\\left\\lbrack \\mathrm{cos}3x \\right\\rbrack_{0}^{\\scriptsize\\dfrac{\\pi}{4}}=\\dfrac{-1}{3}\\left(\\mathrm{cos}\\dfrac{3\\pi}{4} -\\underbrace{\\mathrm{cos}0}_{1}\\right) =\\dfrac{-1}{3}\\left(\\dfrac{-\\sqrt{2}}{2} -1\\right) $
\n$\\quad=\\dfrac{1}{3}\\cdot\\dfrac{\\sqrt{2}+2}{2}=\\dfrac{\\sqrt{2}+2}{6}$ (\u7b54)<\/span><\/div><\/div>\n

(6) $I=\\displaystyle\\int_{0}^{\\scriptsize\\dfrac{\\pi}{4}}\\dfrac{1}{\\mathrm{cos}^2x}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\left\\lbrack \\mathrm{tan}x \\right\\rbrack_{0}^{\\scriptsize\\dfrac{\\pi}{4}}=\\underbrace{\\mathrm{tan}\\dfrac{\\pi}{4}}_{1}-\\underbrace{\\mathrm{tan}0}_{0}=1$ (\u7b54)<\/span><\/div><\/div>\n

(7) $I=\\displaystyle\\int_{0}^{3}\\dfrac{1}{x^2+3}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\dfrac{1}{\\sqrt{3}}\\left\\lbrack \\mathrm{Tan}^{-1}\\dfrac{x}{\\sqrt{3}} \\right\\rbrack_{0}^{3}=\\dfrac{1}{\\sqrt{3}}\\left(\\mathrm{Tan}^{-1}\\dfrac{3}{\\sqrt{3}} -\\underbrace{\\mathrm{Tan}^{-1}0}_{0}\\right) $
\n$\\quad =\\dfrac{1}{\\sqrt{3}}\\mathrm{Tan}^{-1}\\sqrt{3}=\\dfrac{1}{\\sqrt{3}}\\cdot\\dfrac{\\pi}{3}=\\dfrac{\\sqrt{3}\\pi}{9}$ (\u7b54)<\/span><\/div><\/div>\n

(8) $I=\\displaystyle\\int_{0}^{\\sqrt{2}}\\dfrac{1}{\\sqrt{4-x^2}}\\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\left\\lbrack \\mathrm{Sin}^{-1}\\dfrac{x}{2} \\right\\rbrack_{0}^{\\sqrt{2}}=\\mathrm{Sin}^{-1}\\dfrac{\\sqrt{2}}{2}-\\underbrace{\\mathrm{Sin}^{-1}0}_{0}=\\dfrac{\\pi}{4}$ (\u7b54)<\/span><\/div><\/div>\n<\/div>\n
\n

\u884c\u5217\u306e\u548c\u30fb\u5b9a\u6570\u500d\u304a\u3088\u3073\u7a4d\u3068\u9006\u884c\u5217<\/span><\/p>\n

1\uff0e$A=\\begin{pmatrix}3&-2\\\\1&-1\\end{pmatrix},\\;B=\\begin{pmatrix}5&1\\\\-2&1\\end{pmatrix}$ \u306e\u3068\u304d\u6b21\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $2A-B$

<\/span>\u89e3\u7b54<\/div>
$2A-B=2\\begin{pmatrix}3&-2\\\\1&-1\\end{pmatrix}-\\begin{pmatrix}5&1\\\\-2&1\\end{pmatrix}=\\begin{pmatrix}6&-4\\\\2&-2\\end{pmatrix}-\\begin{pmatrix}5&1\\\\-2&1\\end{pmatrix}=\\begin{pmatrix}1&-5\\\\4&-3\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

(2) $AB$

<\/span>\u89e3\u7b54<\/div>
$AB=\\begin{pmatrix}3&-2\\\\1&-1\\end{pmatrix} \\begin{pmatrix}5&1\\\\-2&1\\end{pmatrix}=\\begin{pmatrix}3\\cdot5+(-2)\\cdot(-2)&3\\cdot1+(-2)\\cdot1\\\\1\\cdot5+(-1)\\cdot(-2)&1\\cdot1+(-1)\\cdot1\\end{pmatrix}=\\begin{pmatrix}19&1\\\\7&0\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

(3) $A^2$

<\/span>\u89e3\u7b54<\/div>
$A^2=\\begin{pmatrix}3&-2\\\\1&-1\\end{pmatrix}
\n\\begin{pmatrix}3&-2\\\\1&-1\\end{pmatrix}=\\begin{pmatrix}3\\cdot3+(-2)\\cdot1 &3\\cdot(-2)+(-2)\\cdot(-1)\\\\1\\cdot3+(-1)\\cdot1 &1\\cdot(-2)+(-1)\\cdot(-1)\\end{pmatrix}=\\begin{pmatrix}7&-4\\\\2&-1\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

(4) $A^{-1}$

<\/span>\u89e3\u7b54<\/div>
$A^{-1}=\\dfrac{1}{3\\cdot(-1)-(-2)\\cdot1}\\begin{pmatrix}-1&-(-2)\\\\-1&3\\end{pmatrix}=\\dfrac{1}{-1}\\begin{pmatrix}-1&2\\\\-1&3\\end{pmatrix}=\\begin{pmatrix}1&-2\\\\1&-3\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

2\uff0e\u6b21\u306e\u884c\u5217\u306e\u7a4d\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $A=\\begin{pmatrix}1&0&-1\\\\2&1&0\\end{pmatrix} \\begin{pmatrix}1&2\\\\0&-1\\\\1&1\\end{pmatrix}$

<\/span>\u89e3\u7b54<\/div>
$A=\\begin{pmatrix}1\\cdot1+0\\cdot0+(-1)\\cdot1 & 1\\cdot2+0\\cdot(-1)+(-1)\\cdot1
\n\\\\2\\cdot1+1\\cdot0+0\\cdot1
\n& 2\\cdot2+1\\cdot(-1)+0\\cdot1 \\end{pmatrix}=\\begin{pmatrix}0&1\\\\2&3\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

(2) $B=\\begin{pmatrix}1&2\\\\0&-1\\\\1&1\\end{pmatrix}\\begin{pmatrix}1&0&-1\\\\2&1&0\\end{pmatrix} $

<\/span>\u89e3\u7b54<\/div>
$B=\\begin{pmatrix}1\\cdot1+2\\cdot2&1\\cdot0+2\\cdot1&1\\cdot(-1)+2\\cdot0\\\\0\\cdot1+(-1)\\cdot2&0\\cdot0+(-1)\\cdot1&0\\cdot(-1)+(-1)\\cdot0\\\\1\\cdot1+1\\cdot2&1\\cdot0+1\\cdot1&1\\cdot(-1)+1\\cdot0\\end{pmatrix}=\\begin{pmatrix}5&2&-1\\\\-2&-1&0\\\\3&1&-1\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

\u884c\u5217\u306e\u51aa\u4e57\u3068\u30b1\u30fc\u30ea\u30fc\u30fb\u30cf\u30df\u30eb\u30c8\u30f3\u306e\u5b9a\u7406<\/span><\/p>\n

3\uff0e$A=\\begin{pmatrix}2&-6\\\\1&-2\\end{pmatrix},\\;B=\\begin{pmatrix}3&1\\\\-4&-2\\end{pmatrix}$ \u306e\u3068\u304d\u6b21\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $A^5$

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $A^2-(\\mathrm{tr})A+|A|E=0$ (\u30b1\u30fc\u30ec\u30fc\u30fb\u30cf\u30df\u30eb\u30c8\u30f3\u306e\u5b9a\u7406) \u3092\u7528\u3044\u308b\u3002
\n\u305f\u3060\u3057\u3001$E=\\begin{pmatrix}1&0\\\\0&1\\end{pmatrix}$ (\u5358\u4f4d\u884c\u5217)\u3001$A=\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}$ \u306e\u3068\u304d\u3001$\\mathrm{tr}A=a+d,\\;|A|=ad-bc$
\n$\\mathrm{tr}A=0,\\;|A|=2$ \u3088\u308a\u30b1\u30fc\u30ea\u30fc\u30fb\u30cf\u30df\u30eb\u30c8\u30f3\u306e\u5b9a\u7406\u304b\u3089 $A^2-0A+2E=0 \\Rightarrow A^2=-2E$
\n\u3088\u3063\u3066 $A^5=(A^2)^2A=(-2E)^2A=(-2)^2E^2A=4A=\\begin{pmatrix}8&-24\\\\4&-8\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

(2) $B^6$

<\/span>\u89e3\u7b54<\/div>
\u884c\u5217 $A$ \u306e\u56fa\u6709\u65b9\u7a0b\u5f0f $\\varphi_A(\\lambda)=\\lambda^2-(\\mathrm{tr}A)\\lambda+|A|=0$ \u306e\u89e3(\u56fa\u6709\u5024)\u304c $\\alpha\\neq \\beta$ \u306e\u3068\u304d
\n\u516c\u5f0f\u3000$A^n=\\dfrac{\\alpha^n-\\beta^n}{\\alpha-\\beta}A+\\dfrac{\\alpha\\beta^n-\\beta\\alpha^n}{\\alpha-\\beta}E$\u3000($E$ \u306f\u5358\u4f4d\u884c\u5217)\u3000\u3092\u7528\u3044\u308b
\n$\\mathrm{tr}A=1,\\;|A|=-2$ \u3088\u308a\u56fa\u6709\u65b9\u7a0b\u5f0f\u306f $\\lambda^2-\\lambda-2=(\\lambda-2)(\\lambda+1)=0$
\n\u3088\u3063\u3066\u3001\u56fa\u6709\u5024\u306f $\\lambda=2, -1$ \u3060\u304b\u3089\u3001
\n$B^6=\\dfrac{2^6-(-1)^6}{2-(-1)}B+\\dfrac{2\\cdot(-1)^6-(-1)\\cdot2^6}{2-(-1)}E=21B+22E$
\n$\\quad=21\\begin{pmatrix}5&1\\\\-2&1\\end{pmatrix}+22\\begin{pmatrix}1&0\\\\0&1\\end{pmatrix}=\\begin{pmatrix}85&21\\\\-84&-20\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

\u884c\u5217\u3068\u9023\u7acb\u65b9\u7a0b\u5f0f<\/span><\/p>\n

4. $A=\\begin{pmatrix}3&-2\\\\1&-1\\end{pmatrix},\\;B=\\begin{pmatrix}5&1\\\\-2&1\\end{pmatrix}$ \u306e\u3068\u304d\u6b21\u3092\u307f\u305f\u3059\u884c\u5217 $C,\\;D$ \u3092\u6c42\u3081\u3088\u3002
\n(1) $AC=B$

<\/span>\u89e3\u7b54<\/div>
\u4e21\u8fba\u306b\u5de6\u304b\u3089 $A^{-1}=\\dfrac{1}{-3-(-2)}\\begin{pmatrix}-1&2\\\\-1&3\\end{pmatrix}=\\begin{pmatrix}1&-2\\\\1&-3\\end{pmatrix}$ \u3092\u639b\u3051\u3066\u3001
\n$C=A^{-1}B=\\begin{pmatrix}1&-2\\\\1&-3\\end{pmatrix}\\begin{pmatrix}5&1\\\\-2&1\\end{pmatrix}=\\begin{pmatrix}9&-1\\\\11&-2\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

(2) $DA=B$

<\/span>\u89e3\u7b54<\/div>
\u4e21\u8fba\u306b\u53f3\u304b\u3089 $A^{-1}=\\dfrac{1}{-3-(-2)}\\begin{pmatrix}-1&2\\\\-1&3\\end{pmatrix}=\\begin{pmatrix}1&-2\\\\1&-3\\end{pmatrix}$ \u3092\u639b\u3051\u3066\u3001
\n$D=BA^{-1}=\\begin{pmatrix}5&1\\\\-2&1\\end{pmatrix}\\begin{pmatrix}1&-2\\\\1&-3\\end{pmatrix}=\\begin{pmatrix}6&-13\\\\-1&1\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

5. \u9023\u7acb\u65b9\u7a0b\u5f0f $\\begin{cases}2x-3y = 3\\\\2x-5y = 1\\end{cases}$ \u3092\u884c\u5217\u8868\u793a\u3057\u3066\u3001(\u9006)\u884c\u5217\u3092\u7528\u3044\u3066\u89e3\u3051\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\begin{pmatrix}2&-3\\\\2&-5\\end{pmatrix}\\begin{pmatrix}x\\\\y\\end{pmatrix}=\\begin{pmatrix}3\\\\1\\end{pmatrix}$ \u3088\u308a
\n$\\begin{pmatrix}x\\\\y\\end{pmatrix}=\\begin{pmatrix}2&-3\\\\2&-5\\end{pmatrix}^{-1}\\begin{pmatrix}3\\\\1\\end{pmatrix}=\\dfrac{1}{2\\cdot(-5)-(-3)\\cdot2}\\begin{pmatrix}-5&-(-3)\\\\-2&2\\end{pmatrix}\\begin{pmatrix}3\\\\1\\end{pmatrix}$
\n$\\hspace{24pt}=\\dfrac{1}{-4}\\begin{pmatrix}-12\\\\-4\\end{pmatrix}=\\begin{pmatrix}3\\\\1\\end{pmatrix}$ \u3088\u3063\u3066 $x=3, y=1$ (\u7b54)<\/span><\/div><\/div>\n<\/div>\n
\n

\u5b9a\u7a4d\u5206\uff1a\u4e09\u89d2\u95a2\u6570\u306e\u51aa\u4e57<\/span><\/p>\n

1\uff0e \u6b21\u306e\u5b9a\u7a4d\u5206\u306e\u5024\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $I=\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} \\left(\\mathrm{sin}^4x+\\mathrm{cos}^5x \\right) \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} \\mathrm{sin}^nx \\,dx=\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} \\mathrm{cos}^nx \\,dx=\\begin{cases}\\dfrac{n-1}{n}\\cdot\\dfrac{n-3}{n-2}\\cdots\\dfrac{3}{4}\\cdot\\dfrac{1}{2}\\cdot\\dfrac{\\pi}{2}\\;(n\uff1a\u5076\u6570)\\\\\\dfrac{n-1}{n}\\cdot\\dfrac{n-3}{n-2}\\cdots\\dfrac{4}{5}\\cdot\\dfrac{2}{3}\\cdot1\\quad(n\uff1a\u5947\u6570)\\end{cases}$ \u3092\u7528\u3044\u308b
\n$I=\\dfrac{3}{4}\\cdot\\dfrac{1}{2}\\cdot\\dfrac{\\pi}{2}+\\dfrac{4}{5}\\cdot\\dfrac{2}{3}\\cdot1=\\dfrac{3}{16}\\pi+\\dfrac{8}{15}$ (\u7b54)<\/span><\/div><\/div>\n

(2) $I=\\displaystyle \\int_{0}^{\\pi} \\mathrm{sin}^6x \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} \\mathrm{sin}^nx \\,dx=\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} \\mathrm{cos}^nx \\,dx=\\begin{cases}\\dfrac{n-1}{n}\\cdot\\dfrac{n-3}{n-2}\\cdots\\dfrac{3}{4}\\cdot\\dfrac{1}{2}\\cdot\\dfrac{\\pi}{2}\\;(n\uff1a\u5076\u6570)\\\\\\dfrac{n-1}{n}\\cdot\\dfrac{n-3}{n-2}\\cdots\\dfrac{4}{5}\\cdot\\dfrac{2}{3}\\cdot1\\quad(n\uff1a\u5947\u6570)\\end{cases}$ \u3092\u7528\u3044\u308b
\n$I=2\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} \\mathrm{sin}^6x \\,dx=2\\cdot\\dfrac{5}{6}\\cdot\\dfrac{3}{4}\\cdot\\dfrac{1}{2}\\cdot\\dfrac{\\pi}{2}=\\dfrac{5}{16}\\pi$ (\u7b54)<\/span><\/div><\/div>\n

(3) $I=\\displaystyle \\int_{0}^{\\pi} \\mathrm{cos}^6x \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} \\mathrm{sin}^nx \\,dx=\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} \\mathrm{cos}^nx \\,dx=\\begin{cases}\\dfrac{n-1}{n}\\cdot\\dfrac{n-3}{n-2}\\cdots\\dfrac{3}{4}\\cdot\\dfrac{1}{2}\\cdot\\dfrac{\\pi}{2}\\;(n\uff1a\u5076\u6570)\\\\\\dfrac{n-1}{n}\\cdot\\dfrac{n-3}{n-2}\\cdots\\dfrac{4}{5}\\cdot\\dfrac{2}{3}\\cdot1\\quad(n\uff1a\u5947\u6570)\\end{cases}$ \u3092\u7528\u3044\u308b
\n$I=2\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} \\mathrm{cos}^6x \\,dx=2\\cdot\\dfrac{5}{6}\\cdot\\dfrac{3}{4}\\cdot\\dfrac{1}{2}\\cdot\\dfrac{\\pi}{2}=\\dfrac{5}{16}\\pi$ (\u7b54)<\/span><\/div><\/div>\n

(4) $I=\\displaystyle \\int_{0}^{\\pi} \\mathrm{cos}^7x \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$J_1=\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} \\mathrm{cos}^7x \\,dx, J_2=\\displaystyle \\int_{\\scriptsize\\dfrac{\\pi}{2}}^{\\pi}$ \u3068\u304a\u304f\u3068\u3001$y=\\mathrm{cos}x$ \u306e\u30b0\u30e9\u30d5\u306e\u5bfe\u79f0\u6027\u3088\u308a\u3001
\n$7$(=\u5947\u6570)\u4e57\u3067\u3042\u308b\u3053\u3068\u306b\u6ce8\u610f\u308c\u3070\u3001$J_2=-J_1$ \u3060\u304b\u3089 $I=J_1+J_2=J_1+(-J_1)=0$ (\u7b54)<\/span><\/div><\/div>\n

\u5b9a\u7a4d\u5206:\u90e8\u5206\u7a4d\u5206<\/span><\/p>\n

2\uff0e\u6b21\u306e\u5b9a\u7a4d\u5206\u306e\u5024\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $I=\\displaystyle \\int_{0}^{1} 4xe^{2x} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\left\\lbrack (2x-1)e^{2x}\\right\\rbrack _{0}^{1}=e^2-(-e^0)=e^2+1$(\u7b54)<\/span><\/p>\n

\"\"<\/div><\/div>\n

(2) $I=\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{10}} 25x\\mathrm{sin}5x \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\left \\lbrack -5x\\mathrm{cos}5x+\\mathrm{sin}5x \\right\\rbrack _{0}^{\\scriptsize\\dfrac{\\pi}{10}}=\\dfrac{-\\pi}{2}\\underbrace{\\mathrm{cos}\\dfrac{\\pi}{2}}_{0}+\\underbrace{\\mathrm{sin}\\dfrac{\\pi}{2}}_{1}-\\left( 0+\\underbrace{\\mathrm{sin}0}_{0} \\right)=1$ (\u7b54)<\/span><\/p>\n

\"\"<\/div><\/div>\n

(3) $I=\\displaystyle \\int_{1}^{e} x\\mathrm{log}x \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\left \\lbrack \\dfrac{1}{2}x^2\\mathrm{log}x-\\dfrac{1}{2}\\displaystyle \\int x \\,dx \\right\\rbrack _{0}^{e}=\\left \\lbrack \\dfrac{1}{2}x^2\\mathrm{log}x-\\dfrac{1}{4}x^2 \\right\\rbrack _{0}^{e}$
\n$\\quad=\\underbrace{\\dfrac{1}{2}e^2\\underbrace{\\mathrm{log}e}_{1}-\\dfrac{1}{4}e^2}_{{\\scriptsize\\dfrac{1}{4}}e^2}-\\left(\\dfrac{1}{2}\\underbrace{\\mathrm{log}1}_{0}-\\dfrac{1}{4}\\right)=\\dfrac{e^2+1}{4}$ (\u7b54)<\/span><\/p>\n

\"\"<\/div><\/div>\n

\u5b9a\u7a4d\u5206:\u7f6e\u63db\u7a4d\u5206<\/span><\/p>\n

3\uff0e\u6b21\u306e\u5b9a\u7a4d\u5206\u306e\u5024\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $I=\\displaystyle \\int_{0}^{3} \\dfrac{x}{\\sqrt{x+1}} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$t=\\sqrt{x+1} \\to t^2=x+1 \\to \\begin{array}{c} 2tdt=dx \\\\ x=t^2-1\\end{array}\\quad\\begin{array}{c|cc} x & 0 & 3\\\\ \\hline t & 1 &\u30002\u3000\\end{array}$ \u3088\u308a
\n$I=\\displaystyle \\int_{1}^{2} \\dfrac{t^2-1}{t} 2t\\,dt=2\\displaystyle \\int_{1}^{2} (t^2-1) \\,dt = 2\\left\\lbrack \\dfrac{1}{3}t^3-t \\right\\rbrack_1^2 $
\n$\\quad= 2 \\left\\lbrace \\left(\\dfrac{8}{3}-2\\right)-\\left(\\dfrac{1}{3}-1\\right) \\right \\rbrace =\\dfrac{8}{3}$ (\u7b54)<\/span><\/div><\/div>\n

(2) $I=\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} (\\mathrm{sin}x+1)\\mathrm{cos}x \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$t=\\mathrm{sin}x \\to dt=\\mathrm{cos}x\\,dx \\to \\begin{array}{c|cc} x & 0 & \\dfrac{\\pi}{2}\\\\ \\hline t=\\mathrm{sin}x & 0 &\u30001\u3000\\end{array}$ \u3088\u308a
\n$I=\\displaystyle \\int_{0}^{1} (t+1) \\,dt = \\left\\lbrack \\dfrac{1}{2}t^2+t \\right\\rbrack_0^1 = \\dfrac{1}{2}+1=\\dfrac{3}{2}$ (\u7b54)<\/span><\/div><\/div>\n

(3) $I=\\displaystyle \\int_{0}^{1} \\dfrac{2x+1}{x^2+x+1} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$t=x^2+x+1 \\to dt=(2x+1)\\,dx \\to \\begin{array}{c|cc} x & 0 & 1 \\\\ \\hline t & 1 &\u30003\u3000\\end{array}$ \u3088\u308a
\n$I=\\displaystyle \\int_{1}^{3} \\dfrac{1}{t} \\,dt = \\left\\lbrack \\mathrm{log}|t| \\right\\rbrack_1^3 =\\mathrm{log}|3|-\\underbrace{\\mathrm{log}|1|}_{0}=\\mathrm{log}3$ (\u7b54)<\/span><\/div><\/div>\n

(4) $I=\\displaystyle \\int_{1}^{e} \\dfrac{\\mathrm{log}x}{x} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$t=\\mathrm{log}x \\to dt=\\dfrac{1}{x}\\,dx \\to \\begin{array}{c|cc} x & 1 & e \\\\ \\hline t=\\mathrm{log}x & 0 &\u30001\u3000\\end{array}$ \u3088\u308a
\n$I=\\displaystyle \\int_{0}^{1} t \\,dt = \\dfrac{1}{2}\\left\\lbrack t^2 \\right\\rbrack_0^1 =\\dfrac{1}{2}\\left(1^2-0^2\\right)=\\dfrac{1}{2}$ (\u7b54)<\/span><\/div><\/div>\n<\/div>\n
\n

\u533a\u5206\u6c42\u7a4d\u6cd5\u3068\u6975\u9650<\/span><\/p>\n

1\uff0e\u6b21\u306e\u6975\u9650\u5024\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $I=\\displaystyle \\lim_{n\\to\\infty}\\dfrac{1}{n^4}\\left( 1^3+2^3+\\cdots+(n-1)^3+n^3\\right)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\displaystyle \\lim_{n\\to\\infty}\\left( \\left(\\dfrac{1}{n}\\right)^3+\\cdots+\\left(\\dfrac{n-1}{n}\\right)^3+\\left(\\dfrac{n}{n}\\right)^3+\\left(\\dfrac{2}{n}\\right)^3\\right)\\dfrac{1}{n}$
\n$\\quad=\\underbrace{\\displaystyle \\lim_{n\\to\\infty} \\sum_k^n}_{\\int_{0}^{1}} \\quad \\underbrace{\\left( \\dfrac{k}{n} \\right)^3}_{x^3} \\; \\cdot \\; \\underbrace{\\dfrac{1}{n}}_{dx}=\\displaystyle \\int_{0}^{1} x^3 \\,dx=\\dfrac{1}{4}\\left\\lbrack x^4 \\right\\rbrack_0^1=\\dfrac{1}{4}(1-0)=\\dfrac{1}{4}$ (\u7b54)<\/span><\/div><\/div>\n

(2) $I=\\displaystyle \\lim_{n\\to\\infty}\\left( \\dfrac{n}{n^2+1^2}+\\dfrac{n}{n^2+2^2}+\\cdots+\\dfrac{n}{n^2+(n-1)^2}+\\dfrac{n}{n^2+n^2}\\right)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I=\\displaystyle \\lim_{n\\to\\infty}\\left( \\dfrac{1}{1+\\left(\\dfrac{1}{n}\\right)^2}+\\dfrac{1}{1+\\left(\\dfrac{1}{n}\\right)^2}+\\cdots+\\dfrac{1}{1+\\left(\\dfrac{1}{n}\\right)^2}+\\dfrac{1}{1+\\left(\\dfrac{1}{n}\\right)^2} \\right)\\dfrac{1}{n}$
\n$\\quad=\\underbrace{\\displaystyle \\lim_{n\\to\\infty} \\sum_k^n}_{\\int_{0}^{1}} \\quad \\underbrace{ \\dfrac{1}{1+\\left( \\dfrac{k}{n} \\right)^2}}_{\\scriptsize\\dfrac{1}{1+x^2}} \\; \\cdot \\; \\underbrace{\\dfrac{1}{n}}_{dx}=\\displaystyle \\int_{0}^{1} \\dfrac{1}{1+x^2} \\,dx=\\dfrac{1}{4}\\left\\lbrack \\mathrm{Tan}^{-1}x \\right\\rbrack_0^1$
\n$\\quad=\\mathrm{Tan}^{-1}1-\\underbrace{\\mathrm{Tan}^{-1}0}_{0}=\\dfrac{\\pi}{4}$ (\u7b54)<\/span><\/div><\/div>\n

\u9762\u7a4d<\/span><\/p>\n

2\uff0e\u6b21\u306e\u66f2\u7dda\u3084\u76f4\u7dda\u3067\u56f2\u307e\u308c\u305f\u56f3\u5f62\u306e\u9762\u7a4d $S$ \u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $y=\\mathrm{cos}x\\;\\left(-\\dfrac{\\pi}{2}\\leqq x \\leqq \\dfrac{\\pi}{2}\\right)$ \u3068 $x$ \u8ef8<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$S=2\\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} \\mathrm{cos}x \\,dx=2\\left\\lbrack \\mathrm{sin}x \\right\\rbrack_0^{\\scriptsize\\dfrac{\\pi}{2}}=2\\left( \\underbrace{\\mathrm{sin}\\dfrac{\\pi}{2}}_{1}- \\underbrace{\\mathrm{sin}0}_{0} \\right)=2$ (\u7b54)<\/span><\/div><\/div>\n

(2) $y=\\sqrt{1-x}$ \u3068 $x$ \u8ef8\u304a\u3088\u3073 $y$ \u8ef8<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$x=1$ \u306e\u3068\u304d $x$ \u8ef8\u3068\u4ea4\u308f\u308b\u306e\u3067
\n$S=\\displaystyle \\int_{0}^{1} \\sqrt{1-x} \\,dx=\\displaystyle \\int_{0}^{1} (1-x)^{\\scriptsize\\dfrac{1}{2}} \\,dx=\\dfrac{2}{-1\\cdot3} \\left\\lbrack
\n(1-x)^{\\scriptsize\\dfrac{3}{2}} \\right\\rbrack_0^{1}=\\dfrac{-2}{3}(0- 1)=\\dfrac{2}{3}$ (\u7b54)<\/span><\/div><\/div>\n

(3) \u66f2\u7dda $y=x^2-2$ \u3068 \u76f4\u7dda $y=x$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $S=\\displaystyle \\int_{a}^{b} \\left(\\underbrace{y}_{\u4e0a}-\\underbrace{y}_{\u4e0b}\\right) \\,dx$ \u3092\u7528\u3044\u308b
\n\u9023\u7acb\u65b9\u7a0b\u5f0f $\\begin{cases}y=x^2-2\\\\y=x\\end{cases}\\;$ \u3092\u89e3\u3044\u3066\u3001\u4ea4\u70b9\u306e $x$ \u5ea7\u6a19\u306f $x=-1, 2$ \u3060\u304b\u3089
\n$S=\\displaystyle \\int_{-1}^{2} \\left\\lbrace \\underbrace{x}_{\u4e0a}-\\underbrace{\\left(x^2-2\\right)}_{\u4e0b} \\right\\rbrace \\,dx=\\left\\lbrack
\n\\dfrac{x^2}{2}-\\dfrac{x^3}{3}+2x \\right\\rbrack_{-1}^{2}$
\n$\\quad=\\left(2-\\dfrac{8}{3}+4\\right)-\\left(\\dfrac{1}{2}-\\dfrac{-1}{3}-2\\right)=\\dfrac{9}{2}$ (\u7b54)<\/span><\/div><\/div>\n

\u4f53\u7a4d<\/span><\/p>\n

3\uff0e\u6b21\u306e\u66f2\u7dda\u3084\u76f4\u7dda\u3067\u56f2\u307e\u308c\u305f\u56f3\u5f62\u3092 $x$ \u8ef8\u306e\u5468\u308a\u306b\u56de\u8ee2\u3057\u3066\u3067\u304d\u308b\u56de\u8ee2\u4f53\u306e\u4f53\u7a4d$V$ \u3092\u3081\u3088\u3002<\/p>\n

(1) $y=\\mathrm{cos}x\\;\\left(-\\dfrac{\\pi}{2}\\leqq x \\leqq \\dfrac{\\pi}{2}\\right)$ \u3068 $x$ \u8ef8<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $V=\\pi\\displaystyle \\int_{a}^{b} y^2 \\,dx$ \u3092\u7528\u3044\u308b
\n$V=2\\cdot\\pi \\displaystyle \\int_{0}^{\\scriptsize\\dfrac{\\pi}{2}} \\mathrm{cos}^2x \\,dx=2\\pi\\cdot\\dfrac{1}{2}\\dfrac{\\pi}{2}=\\dfrac{\\pi^2}{2}$ (\u7b54)<\/span><\/div><\/div>\n

(2) $y=\\sqrt{1-x}$ \u3068 $x$ \u8ef8\u304a\u3088\u3073 $y$ \u8ef8<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $V=\\pi\\displaystyle \\int_{a}^{b} y^2 \\,dx$ \u3092\u7528\u3044\u308b
\n$S=\\displaystyle \\int_{0}^{1} \\left(\\sqrt{1-x}\\right)^2 \\,dx=\\pi\\displaystyle \\int_{0}^{1} (1-x) \\,dx=\\pi \\left\\lbrack
\nx-\\dfrac{x^2}{2} \\right\\rbrack_0^{1}=\\dfrac{\\pi}{2}$ (\u7b54)<\/span><\/div><\/div>\n

\u66f2\u7dda\u306e\u9577\u3055<\/span><\/p>\n

4\uff0e\u66f2\u7dda $y=\\dfrac{1}{8}x^4+\\dfrac{1}{4x^4}\\;\\left(1\\leqq x \\leqq 2\\right)$ \u306e\u9577\u3055 $L$ \u3092\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $L=\\displaystyle \\int_{a}^{b} \\sqrt{1+\\left(y{\\prime}\\right)^2} \\,dx$ \u3092\u7528\u3044\u308b
\n$1+\\left(y{\\prime}\\right)^2=1+\\left( \\dfrac{x^3}{2}-\\dfrac{1}{2x^2}\\right)^2=1+\\left(\\dfrac{x^3}{2}\\right)^3-\\dfrac{1}{2}+\\left( \\dfrac{1}{2x^2}\\right)$
\n$\\quad=\\left(\\dfrac{x^3}{2}\\right)^3+\\dfrac{1}{2}+\\left( \\dfrac{1}{2x^2}\\right)=\\left( \\dfrac{x^3}{2}+\\dfrac{1}{2x^2}\\right)^2$ \u3088\u308a
\n$L=\\displaystyle \\int_{1}^{2} \\sqrt{1+\\left(y{\\prime}\\right)^2} \\,dx=\\pi\\displaystyle \\int_{1}^{2} \\left(\\dfrac{x^3}{2}+\\dfrac{1}{2x^3}\\right) \\,dx= \\left\\lbrack
\n\\dfrac{x^4}{8}-\\dfrac{1}{4x^2} \\right\\rbrack_1^{2}$
\n$\\quad=\\left(\\dfrac{16}{8}-\\dfrac{1}{16}\\right)-\\left(\\dfrac{1}{8}-\\dfrac{1}{4}\\right)=\\dfrac{33}{16}$ (\u7b54)<\/span><\/div><\/div>\n

\u5e83\u7fa9\u7a4d\u5206<\/span><\/p>\n

5\uff0e\u6b21\u306e\u5e83\u7fa9\u7a4d\u5206\u306e\u5024\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $I=\\displaystyle \\int_{0}^{1} \\dfrac{1}{\\sqrt{1-x^2}} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I_{\\epsilon}=\\displaystyle \\int_{0}^{1-\\epsilon} \\dfrac{1}{\\sqrt{1-x^2}} \\,dx=\\left\\lbrack
\n\\mathrm{Sin}^{-1}x \\right\\rbrack_0^{1-\\epsilon}=\\mathrm{Sin}^{-1}(1-\\epsilon)$ \u3088\u308a
\n$I=\\displaystyle \\lim_{\\epsilon \\to +0} I_{\\epsilon}=\\displaystyle \\lim_{\\epsilon \\to +0} \\mathrm{Sin}^{-1}(1-\\epsilon)=\\mathrm{Sin}^{-1}1=\\dfrac{\\pi}{2}$ (\u7b54)<\/span><\/div><\/div>\n

(2) $I=\\displaystyle \\int_{1}^{\\infty} \\dfrac{1}{x^2} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I_M=\\displaystyle \\int_{1}^{M} \\dfrac{1}{x^2} \\,dx=\\left\\lbrack
\n\\dfrac{1}{-x} \\right\\rbrack_0^{M}=\\dfrac{1}{-M}-\\dfrac{1}{-1}$ \u3088\u308a
\n$I=\\displaystyle \\lim_{M \\to \\infty} I_{M}=\\displaystyle \\lim_{M \\to \\infty} \\left(\\dfrac{-1}{M} +1 \\right)=0+1=1$ (\u7b54)<\/span><\/div><\/div>\n

(3) $I=\\displaystyle \\int_{\\sqrt{3}}^{\\infty} \\dfrac{1}{x^2+1} \\,dx$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$I_M=\\displaystyle \\int_{\\sqrt{3}}^{M} \\dfrac{1}{x^2+1} \\,dx=\\left\\lbrack
\n\\mathrm{Tan}^{-1}x \\right\\rbrack_1^{M}=\\mathrm{Tan}^{-1}M-\\mathrm{Tan}^{-1}\\sqrt{3}$ \u3088\u308a
\n$I=\\displaystyle \\lim_{M \\to \\infty} I_{M}=\\displaystyle \\lim_{M \\to \\infty} \\left(\\mathrm{Tan}^{-1}M-\\mathrm{Tan}^{-1}\\sqrt{3} \\right)=\\underbrace{\\mathrm{Tan}^{-1}\\infty}_{\\scriptsize\\dfrac{\\pi}{2}}-\\underbrace{\\mathrm{Tan}^{-1}\\sqrt{3}}_{\\scriptsize\\dfrac{\\pi}{3}}=\\dfrac{\\pi}{6}$ (\u7b54)<\/span><\/div><\/div>\n<\/div>\n
\n

\u5bfe\u79f0\u5909\u63db<\/span><\/p>\n

1. \u6b21\u306e\u5bfe\u79f0\u5909\u63db\u3092\u8868\u3059\u884c\u5217\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $x$ \u8ef8\u5bfe\u79f0\u5909\u63db<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\begin{cases}x^{\\prime}=x+0\\cdot y\\\\y^{\\prime}=0\\cdot x-y\\end{cases}\\;\\Rightarrow\\;\\begin{pmatrix}1&0\\\\0&-1\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

(2) $y$ \u8ef8\u5bfe\u79f0\u5909\u63db<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\begin{cases}x^{\\prime}=-x+0\\cdot y\\\\y^{\\prime}=0\\cdot x+y\\end{cases}\\;\\Rightarrow\\;\\begin{pmatrix}-1&0\\\\0&1\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

(3) \u539f\u70b9\u5bfe\u79f0\u5909\u63db<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\begin{cases}x^{\\prime}=-x+0\\cdot y\\\\y^{\\prime}=0\\cdot x-y\\end{cases}\\;\\Rightarrow\\;\\begin{pmatrix}-1&0\\\\0&-1\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

(4) \u76f4\u7dda $y=x$ \u306b\u95a2\u3059\u308b\u5bfe\u79f0\u5909\u63db<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\begin{cases}x^{\\prime}=0\\cdot x+y\\\\y^{\\prime}=x+0\\cdot y\\end{cases}\\;\\Rightarrow\\;\\begin{pmatrix}0&1\\\\1&0\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

1\u6b21\u5909\u63db\u306e\u50cf\u3068\u539f\u50cf(\uff1d\u9006\u50cf)<\/span><\/p>\n

2. \u4e00\u6b21\u5909\u63db $\\begin{cases}x^{\\prime}=2x+3y\\\\y^{\\prime}=x+2y\\end{cases}$ \u306b\u3064\u3044\u3066\u3001\u6b21\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) \u70b9 $\\mathrm{P}(1, 2)$ \u306e\u50cf $\\mathrm{P}^{\\prime}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\begin{pmatrix}2&3\\\\1&2\\end{pmatrix}\\begin{pmatrix}1\\\\2\\end{pmatrix}=\\begin{pmatrix}8\\\\5\\end{pmatrix}\\;\\Rightarrow\\;\\mathrm{P}^{\\prime}(8, 5)$ (\u7b54)<\/span><\/div><\/div>\n

(2) \u70b9 $\\mathrm{Q}^{\\prime}(3, 1)$ \u306e\u539f\u50cf(=\u9006\u50cf) $\\mathrm{Q}(x, y)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}^{-1}=\\dfrac{1}{ad-bc}\\begin{pmatrix}d&-b\\\\-c&a\\end{pmatrix}$ \u3092\u7528\u3044\u308b
\n$\\begin{pmatrix}2&3\\\\1&2\\end{pmatrix}\\begin{pmatrix}x\\\\y\\end{pmatrix}=\\begin{pmatrix}3\\\\1\\end{pmatrix}\\;\\Rightarrow\\;\\begin{pmatrix}x\\\\y\\end{pmatrix}=\\begin{pmatrix}2&3\\\\1&2\\end{pmatrix}^{-1}\\begin{pmatrix}3\\\\1\\end{pmatrix}=\\dfrac{1}{4-3}\\begin{pmatrix}2&-3\\\\-1&2\\end{pmatrix}\\begin{pmatrix}3\\\\1\\end{pmatrix}=\\begin{pmatrix}3\\\\-1\\end{pmatrix}$
\n$\\mathrm{Q}(3, -1)$ (\u7b54)<\/span><\/div><\/div>\n

\u539f\u70b9\u5468\u308a\u306e\u56de\u8ee2<\/span><\/p>\n

3. \u70b9 ${\\mathrm{P}}(\\sqrt{3}, 1)$ \u3092\u539f\u70b9\u306e\u5468\u308a\u306b$120^{\\circ}$ \u56de\u8ee2\u3057\u305f\u70b9 $\\mathrm{P}^{\\prime}$ \u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>

\n(\u539f\u70b9\u5468\u308a\u306e $\\theta$ \u56de\u8ee2\u306e\u884c\u5217) $\\begin{pmatrix}\\mathrm{cos}\\theta&-\\mathrm{sin}\\theta\\\\\\mathrm{sin}\\theta&\\mathrm{cos}\\theta\\end{pmatrix}$ \u3092\u7528\u3044\u308b
\n$\\begin{pmatrix}\\mathrm{cos}120^{\\circ}&-\\mathrm{sin}120^{\\circ}\\\\\\mathrm{sin}120^{\\circ}&\\mathrm{cos}120^{\\circ}\\end{pmatrix}\\begin{pmatrix}\\sqrt{3}\\\\1\\end{pmatrix}=\\dfrac{1}{2}\\begin{pmatrix}-1 & -\\sqrt{3} \\\\ \\sqrt{3} & -1\\end{pmatrix}\\begin{pmatrix}\\sqrt{3} \\\\ 1\\end{pmatrix}=\\dfrac{1}{2}\\begin{pmatrix}-2\\sqrt{3}\\\\ 2\\end{pmatrix}=\\begin{pmatrix}-\\sqrt{3}\\\\1\\end{pmatrix}$
\n$\\mathrm{P}^{\\prime}(-\\sqrt{3}, 1)$ (\u7b54)<\/span><\/div><\/div>\n

1\u6b21\u5909\u63db\u306e\u5408\u6210\u3068\u9006\u5909\u63db<\/span><\/p>\n

4\uff0e\u4e00\u6b21\u5909\u63db $f:\\begin{cases}x^{\\prime}=2x+3y\\\\y^{\\prime}=x+2y\\end{cases},g:\\begin{cases}x^{\\prime}=x-2y\\\\y^{\\prime}=x+3y\\end{cases}\\;$ \u306b\u3064\u3044\u3066\u3001\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088\u3002<\/p>\n

(1) $f$ \u306e\u9006\u5909\u63db $f^{-1}$ \u3092\u8868\u3059\u884c\u5217\u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}^{-1}=\\dfrac{1}{ad-bc}\\begin{pmatrix}d&-b\\\\-c&a\\end{pmatrix}$ \u3092\u7528\u3044\u308b
\n$\\underbrace{\\begin{pmatrix}2&3\\\\1&2\\end{pmatrix}^{-1}}_{f^{-1}}=\\dfrac{1}{4-3}\\begin{pmatrix}2&-3\\\\-1&2\\end{pmatrix}=\\begin{pmatrix}2&-3\\\\-1&2\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

(2) \u5408\u6210 $f^{-1}\\circ g$ \u3092\u8868\u3059\u884c\u5217\u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\underbrace{\\begin{pmatrix}2&-3\\\\-1&2\\end{pmatrix}}_{f^{-1}}\\;\\underbrace{\\begin{pmatrix}1&-2\\\\1&3\\end{pmatrix}}_{g}=\\begin{pmatrix}-1&-13\\\\1&8\\end{pmatrix}$ (\u7b54)<\/span><\/div><\/div>\n

1\u6b21\u5909\u63db\u306b\u3088\u308b\u76f4\u7dda\u306e\u50cf<\/span><\/p>\n

5\uff0e\u6b21\u306e1\u6b21\u5909\u63db\u306b\u3088\u308b\u76f4\u7dda $y=2x+1$ \u306e\u50cf\u3092\u6c42\u3081\u3088\u3002<\/p>\n

(1) $f:\\begin{cases}x^{\\prime}=-2x+2y\\\\y^{\\prime}=x+y\\end{cases}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\begin{cases}x^{\\prime}=-2x+2(2x+1)=2x+2\\\\y^{\\prime}=x+(2x+1)=3x+1\\end{cases}\\;\\Rightarrow\\;\\underbrace{3x^{\\prime}}_{6x+6}-\\underbrace{2y^{\\prime}}_{-6x+2}=4\\;\\Rightarrow\\;2y^{\\prime}=3x^{\\prime}-4$ \u3088\u308a
\n\u76f4\u7dda $y=\\dfrac{3}{2}x-2$ (\u7b54)<\/span><\/div><\/div>\n

(2) $g:\\begin{cases}x^{\\prime}=3x+6y\\\\y^{\\prime}=x+2y\\end{cases}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\begin{cases}x^{\\prime}=3x+6(2x+1)=15x+6\\\\y^{\\prime}=x+2(2x+1)=5x+2\\end{cases}\\;\\Rightarrow\\;\\underbrace{y^{\\prime}}_{5x+2}=\\dfrac{1}{3}\\underbrace{x^{\\prime}}_{15x+6}$ \u3088\u308a\u3001\u76f4\u7dda $y=\\dfrac{1}{3}x$ (\u7b54)<\/span><\/div><\/div>\n

(3) $h:\\begin{cases}x^{\\prime}=2x-y\\\\y^{\\prime}=-4x+2y\\end{cases}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\begin{cases}x^{\\prime}=2x-(2x+1)=-1\\\\y^{\\prime}=-4x+2(2x+1)=2\\end{cases}\\;$ \u3088\u308a\u30011\u70b9 $(-1, 2)$ (\u7b54)<\/span><\/div><\/div>\n

1\u6b21\u5909\u63db\u306e\u6c7a\u5b9a\u554f\u984c<\/span><\/p>\n

6. 1\u6b21\u5909\u63db $f$ \u306b\u3088\u308b\u70b9 $\\mathrm{P}(2, 1),\\;\\mathrm{Q}(1, 1)$ \u306e\u50cf\u304c $\\mathrm{P}^{\\prime}(-1, 3),\\;\\mathrm{Q}^{\\prime}(0, 2)$ \u306e\u3068\u304d\u3001$f$ \u3092\u8868\u3059\u884c\u5217 $A$ \u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}^{-1}=\\dfrac{1}{ad-bc}\\begin{pmatrix}d&-b\\\\-c&a\\end{pmatrix}$ \u3088\u308a $\\begin{pmatrix}2&1\\\\1&1\\end{pmatrix}^{-1}=\\dfrac{1}{2-1}\\begin{pmatrix}1&-1\\\\-1&2\\end{pmatrix}$ \u306b\u6ce8\u610f\u3059\u308b
\n$A\\begin{pmatrix}2&1\\\\1&1\\end{pmatrix}=\\begin{pmatrix}-1&0\\\\3&2\\end{pmatrix}$
\n$\\Rightarrow\\;A=\\begin{pmatrix}-1&0\\\\3&2\\end{pmatrix}\\begin{pmatrix}2&1\\\\1&1\\end{pmatrix}^{-1}=\\begin{pmatrix}-1&0\\\\3&2\\end{pmatrix}\\begin{pmatrix}1&-1\\\\-1&2\\end{pmatrix}=\\begin{pmatrix}-1&1\\\\1&1\\end{pmatrix}$(\u7b54)<\/span><\/div><\/div>\n

\u884c\u5217\u306e\u6b63\u5247\u6761\u4ef6<\/span><\/p>\n

7\uff0e\u884c\u5217 $A=\\begin{pmatrix}2&x\\\\3&6\\end{pmatrix}$ \u304c\u6b63\u5247\u3067\u306a\u3044\u3068\u304d\u3001$x$ \u3092\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$|A|=\\mathrm{det}A=\\begin{array}{|cc|}2&x\\\\3&6\\end{array}=\\mathrm{det}\\begin{pmatrix}2&x\\\\3&6\\end{pmatrix}=12-3x=0\\;\\Rightarrow\\;x=4$ (\u7b54)<\/span><\/div><\/div>\n

\u884c\u5217\u5f0f\u3068\u9762\u7a4d<\/span><\/p>\n

8. $x^2+y^2=1$ \u306e1\u6b21\u5909\u63db $f:\\begin{cases}x^{\\prime}=2x \\\\ y^{\\prime}=3y\\end{cases} $ \u306b\u3088\u308b\u50cf\u3092\u6c42\u3081\u3001\u50cf\u304c\u56f2\u3080\u9762\u7a4d\u3082\u6c42\u3081\u3088\u3002<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f $\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}^{-1}=\\dfrac{1}{ad-bc}\\begin{pmatrix}d&-b\\\\-c&a\\end{pmatrix}$ \u3088\u308a
\n$\\begin{pmatrix}2&0\\\\0&3\\end{pmatrix}^{-1}=\\dfrac{1}{6-0}\\begin{pmatrix}3&0\\\\0&2\\end{pmatrix}=\\begin{pmatrix}\\dfrac{1}{2}&0\\\\0&\\dfrac{1}{3}\\end{pmatrix}$ \u306b\u6ce8\u610f\u3059\u308b
\n$\\begin{pmatrix}2&0\\\\0&3\\end{pmatrix}\\begin{pmatrix}x\\\\y\\end{pmatrix}=\\begin{pmatrix}x^{\\prime}\\\\y^{\\prime}\\end{pmatrix}\\;\\Rightarrow\\;\\begin{pmatrix}x\\\\y\\end{pmatrix}=\\begin{pmatrix}2&0\\\\0&3\\end{pmatrix}^{-1}\\begin{pmatrix}x^{\\prime}\\\\y^{\\prime}\\end{pmatrix}=\\begin{pmatrix}\\dfrac{1}{2}&0\\\\0&\\dfrac{1}{3}\\end{pmatrix}\\begin{pmatrix}x^{\\prime}\\\\y^{\\prime}\\end{pmatrix}=\\begin{pmatrix}\\dfrac{x^{\\prime}}{2}\\\\ \\dfrac{y^{\\prime}}{3}\\end{pmatrix}$ \u3088\u308a
\n\u50cf\u306f $\\;\\underbrace{\\left(
\n\\dfrac{x^{\\prime}}{2} \\right)^{2}}_{x^2}+\\underbrace{\\left( \\dfrac{y^{\\prime}}{3} \\right)^{2}}_{y^2}=1\\;\\Rightarrow\\;$ \u6955\u5186 $\\;\\dfrac{x^2}{4}+\\dfrac{y^2}{9}=1$ (\u7b54)
\n\u9762\u7a4d\u306f $\\underbrace{(\\text{\u534a\u5f841\u306e\u5186\u306e\u9762\u7a4d})}_{\\pi}\\times\\underbrace{\\left|\\mathrm{det}\\begin{pmatrix}2&0\\\\0&3\\end{pmatrix}\\right|}_{6}=6\\pi$ (\u7b54)<\/span><\/div><\/div>\n<\/div><\/div><\/div>\n","protected":false},"excerpt":{"rendered":"

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