【微分積分】4-3-1 積分公式の利用｜要点まとめ
このページでは、大学数学の微分積分で扱う「積分公式の利用」について整理します。既知の積分公式をどのように変形し、問題に応用するかという考え方を例題を通して解説します。また、誤差評価に用いられるテイラーの定理の積分剰余形についても説明し、積分計算の理解をより深めます。
積分公式を問題に応用する考え方
【例題】次の不定積分・定積分を求めなさい。
(1)\(\displaystyle \int_0^12^xdx\)
\(\displaystyle =\left[\frac{2^x}{\log2}\right]_0^1\)
\(\displaystyle =\frac{2-1}{\log2}\)
\(\displaystyle =\frac{1}{\log2}\)
\(\displaystyle =\frac{2-1}{\log2}\)
\(\displaystyle =\frac{1}{\log2}\)
(2)\(\displaystyle \int_0^2\frac{1}{x^2+4}dx\)
\(\displaystyle =\left[\frac{1}{2}\tan^{-1}\frac{x}{2}\right]_0^2\)
\(\displaystyle =\frac{1}{2}\tan^{-1}1-\frac{1}{2}\tan^{-1}0\)
\(\displaystyle =\frac{1}{2}・\frac{\pi}{4}-0\)
\(\displaystyle =\frac{\pi}{8}\)
\(\displaystyle =\frac{1}{2}\tan^{-1}1-\frac{1}{2}\tan^{-1}0\)
\(\displaystyle =\frac{1}{2}・\frac{\pi}{4}-0\)
\(\displaystyle =\frac{\pi}{8}\)
(3)\(\displaystyle \int_0^1\frac{1}{\sqrt{x^2+3}}dx\)
\(\displaystyle =[\log(x+\sqrt{x^2+3})]_0^1\)
\(\displaystyle =\log3-\log\sqrt{3}\)
\(\displaystyle =\frac{1}{2}\log3\)
\(\displaystyle =\log3-\log\sqrt{3}\)
\(\displaystyle =\frac{1}{2}\log3\)
(4)\(\displaystyle \int\frac{x^2}{(x^3+4)^{\frac{3}{2}}}dx\)
\(\displaystyle =\frac{1}{3}\int(x^3+4)'(x^3+4)^{-\frac{3}{2}}dx\)
\(\displaystyle =\frac{1}{3}\{-2(x^3+4)^{-\frac{1}{2}}\}+C\)
\(\displaystyle =-\frac{2}{3\sqrt{x^3+4}}+C\)
\(\displaystyle =\frac{1}{3}\{-2(x^3+4)^{-\frac{1}{2}}\}+C\)
\(\displaystyle =-\frac{2}{3\sqrt{x^3+4}}+C\)
(5)\(\displaystyle \int\tanh xdx\)
\(\displaystyle =\int\frac{\sinh x}{\cosh x}dx\)
\(\displaystyle =\int\frac{(\cosh x)'}{\cosh x}dx\)
\(=\log(\cosh x)+C\)
\(\displaystyle =\int\frac{(\cosh x)'}{\cosh x}dx\)
\(=\log(\cosh x)+C\)
(6)\(\displaystyle \int_0^\sqrt{3}\frac{1}{\sqrt{4-x^2}}dx\)
\(\displaystyle =\left[\sin^{-1}\frac{x}{2}\right]_0^\sqrt{3}\)
\(\displaystyle =\sin^{-1}\frac{\sqrt{3}}{2}-\sin^{-1}0\)
\(\displaystyle =\frac{\pi}{3}\)
\(\displaystyle =\sin^{-1}\frac{\sqrt{3}}{2}-\sin^{-1}0\)
\(\displaystyle =\frac{\pi}{3}\)
(7)\(\displaystyle \int\frac{1}{\sqrt{(2-x)(x-1)}}dx\)
\(\displaystyle =\int\frac{1}{\sqrt{\frac{1}{4}-(x-\frac{3}{2})^2}}dx\)
\(\displaystyle =\sin^{-1}\frac{x-\frac{3}{2}}{\frac{1}{2}}+C\)
\(\displaystyle =\sin^{-1}(2x-3)+C\)
\(\displaystyle =\sin^{-1}\frac{x-\frac{3}{2}}{\frac{1}{2}}+C\)
\(\displaystyle =\sin^{-1}(2x-3)+C\)
(8)\(\displaystyle \int\frac{x}{x^2+3}dx\)
\(\displaystyle =\frac{1}{2}\int\frac{2x}{x^2+3}dx\)
\(\displaystyle =\frac{1}{2}\int\frac{(x^2+3)'}{x^2+3}dx\)
\(\displaystyle =\frac{1}{2}\log(x^2+3)+C\)
\(\displaystyle =\frac{1}{2}\int\frac{(x^2+3)'}{x^2+3}dx\)
\(\displaystyle =\frac{1}{2}\log(x^2+3)+C\)
(9)\(\displaystyle \int\frac{e^x}{e^x-1}dx\)
\(\displaystyle =\int\frac{(e^x-1)'}{e^x-1}dx\)
\(\displaystyle =\log|e^x-1|+C\)
\(\displaystyle =\log|e^x-1|+C\)
(10)\(\displaystyle \int\sin^2xdx\)
\(\displaystyle =\int\frac{1-\cos2x}{2}dx\)
\(\displaystyle =\frac{x}{2}-\frac{\sin2x}{4}+C\)
\(\displaystyle =\frac{x}{2}-\frac{\sin2x}{4}+C\)
(11)\(\displaystyle \int\sin x\cos xdx\)
\(\displaystyle =\int\frac{\sin2x}{2}dx\)
\(\displaystyle =-\frac{\cos2x}{4}+C\)
\(\displaystyle =-\frac{\cos2x}{4}+C\)
(12)\(\displaystyle \int\cos^3xdx\)
\(\displaystyle =\int(1-\sin^2x)\cos xdx\)
\(\displaystyle =\sin x-\frac{\sin^3x}{3}+C\)
\(\displaystyle =\sin x-\frac{\sin^3x}{3}+C\)
(13)\(\displaystyle \int\sin2x\cos3xdx\)
\(\displaystyle =\int\frac{\sin5x-\sin x}{2}dx\)
\(\displaystyle =-\frac{\cos5x}{10}+\frac{\cos x}{2}+C\)
\(\displaystyle =-\frac{\cos5x}{10}+\frac{\cos x}{2}+C\)
(14)\(\displaystyle \int\sin x\cos^nxdx\)
\(\displaystyle =-\int(\cos x)'\cos^nxdx\)
\(\displaystyle =-\frac{\cos^{n+1}x}{n+1}+C\)
\(\displaystyle =-\frac{\cos^{n+1}x}{n+1}+C\)
(15)\(\displaystyle \int\tan^2xdx\)
\(\displaystyle =\int\left(\frac{1}{\cos^2x}-1\right)\)
\(\displaystyle =\tan x-x+C\)
\(\displaystyle =\tan x-x+C\)
(16)\(\displaystyle \int x\sin2xdx\)
\(\displaystyle =x・\frac{-\cos2x}{2}+\frac{1}{2}\int\cos2xdx\)
\(\displaystyle =-\frac{1}{2}x\cos2x+\frac{1}{4}\sin2x+C\)
\(\displaystyle =-\frac{1}{2}x\cos2x+\frac{1}{4}\sin2x+C\)
(17)\(\displaystyle \int(\log x)^2dx\)
\(\displaystyle =x(\log x)^2-\int x・2(\log x)\frac{1}{x}dx\)
\(\displaystyle =x(\log x)^2-2\int\log xdx\)
\(\displaystyle =x(\log x)^2-2x\log x+2x+C\)
\(\displaystyle =x(\log x)^2-2\int\log xdx\)
\(\displaystyle =x(\log x)^2-2x\log x+2x+C\)
(18)\(\displaystyle \int\tan^{-1}xdx\)
\(\displaystyle =x\tan^{-1}x-\int x・\frac{1}{x^2+1}dx\)
\(\displaystyle =x\tan^{-1}x-\frac{1}{2}\int\frac{2x}{x^2+1}dx\)
\(\displaystyle =x\tan^{-1}x-\frac{1}{2}\log(x^2+1)+C\)
\(\displaystyle =x\tan^{-1}x-\frac{1}{2}\int\frac{2x}{x^2+1}dx\)
\(\displaystyle =x\tan^{-1}x-\frac{1}{2}\log(x^2+1)+C\)
(19)\(\displaystyle \int\sin^{-1}xdx\)
\(\displaystyle =x\sin^{-1}x-\int x・\frac{1}{\sqrt{1-x^2}}dx\)
\(\displaystyle =x\sin^{-1}x+\int\frac{-2x}{2\sqrt{1-x^2}}dx\)
\(\displaystyle =x\sin^{-1}x+\sqrt{1-x^2}+C\)
\(\displaystyle =x\sin^{-1}x+\int\frac{-2x}{2\sqrt{1-x^2}}dx\)
\(\displaystyle =x\sin^{-1}x+\sqrt{1-x^2}+C\)
(20)\(\displaystyle \int x\sqrt{x^2+2}dx\)
\(t=x^2+2\)とおくと、\(\displaystyle dx=\frac{1}{2x}dt\)
\(\displaystyle =\int\frac{\sqrt{t}}{2}dt\)
\(\displaystyle =\frac{1}{3}t^{\frac{3}{2}}+C\)
\(\displaystyle =\frac{1}{3}(x^2+2)^{\frac{3}{2}}+C\)
\(\displaystyle =\int\frac{\sqrt{t}}{2}dt\)
\(\displaystyle =\frac{1}{3}t^{\frac{3}{2}}+C\)
\(\displaystyle =\frac{1}{3}(x^2+2)^{\frac{3}{2}}+C\)
(21)\(\displaystyle \int_0^2x\sqrt{3x^2+4}dx\)
\(t=3x^2+4\)とおくと、\(\displaystyle dx=\frac{1}{6x}dt\)
\(\displaystyle =\int_4^{16}\frac{\sqrt{t}}{6}dt\)
\(\displaystyle =\left[\frac{1}{9}t^{\frac{3}{2}}\right]_4^{16}\)
\(\displaystyle =\frac{1}{9}(64-8)\)
\(\displaystyle =\frac{56}{9}\)
\(\displaystyle =\int_4^{16}\frac{\sqrt{t}}{6}dt\)
\(\displaystyle =\left[\frac{1}{9}t^{\frac{3}{2}}\right]_4^{16}\)
\(\displaystyle =\frac{1}{9}(64-8)\)
\(\displaystyle =\frac{56}{9}\)
(22)\(\displaystyle \int_0^\sqrt{2}x^3e^{x^2}dx\)
\(t=x^2\)とおくと、\(\displaystyle dx=\frac{1}{2x}dt\)
\(\displaystyle =\frac{1}{2}\int_0^2te^tdt\)
\(\displaystyle =\frac{1}{2}[te^t]_0^2-\frac{1}{2}\int_0^2e^tdt\)
\(\displaystyle =e^2-\frac{1}{2}[e^t]_0^2\)
\(\displaystyle =\frac{e^2+1}{2}\)
\(\displaystyle =\frac{1}{2}\int_0^2te^tdt\)
\(\displaystyle =\frac{1}{2}[te^t]_0^2-\frac{1}{2}\int_0^2e^tdt\)
\(\displaystyle =e^2-\frac{1}{2}[e^t]_0^2\)
\(\displaystyle =\frac{e^2+1}{2}\)
テイラーの定理における積分剰余の意味
【テイラーの定理の積分剰余形】
関数\(f(x)\)は点\(a\)を含む開区間\(I\)上で\(C^n\)級とする。このとき、各\(x\in I\)に対して
\(\displaystyle R_n(x)=\frac{1}{(n-1)!}\int_a^xf^{(n)}(t)(x-t)^{n-1}dt\)
のとき、
\(\displaystyle f(x)=\sum_{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a)^k+R_n(x)\)
が成り立つ。
次の学習に進もう！