【微分積分】8-1-3 積分順序の変更｜要点まとめ
このページでは、大学数学で重要な「積分順序の変更」について解説します。縦線集合・横線集合による積分領域の把握方法から、累次積分の順序を書き換える基本手順までをまとめています。直接積分するのが難しい2重積分を計算するための重要テクニックを、詳細な例題・解答付きで学べます。
積分順序の変更
【積分順序の変更】
関数\(f(x,y)\)が積分領域\(D\)上で連続し、
\(D=\{(x,y)|a\leqq x\leqq b,\varphi_1(x)\leqq y\leqq \varphi_2(x)\}\)
\(\ \ \ =\{(x,y)|c\leqq x\leqq d,\varphi_1(y)\leqq x\leqq \varphi_2(y)\}\)
のように\(x\)方向と\(y\)方向でともに縦線集合の場合には
\(\displaystyle \iint_Df(x,y)dxdy\)
\(\displaystyle \ \ \ =\int_a^b\left(\int_{\varphi_1(x)}^{\varphi_2(x)}f(x,y)dy\right)dx\)
\(\displaystyle \ \ \ =\int_c^d\left(\int_{\varphi_1(y)}^{\varphi_2(y)}f(x,y)dx\right)dy\)
が成り立つ。この2つの累次積分の一方から他方への書き換えを積分順序の変更という。
【例題】次の2重積分を求めなさい。
(1)\(\displaystyle \int_0^1\left(\int_x^1e^{y^2}dy\right)dx\)
積分領域を
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,x\leqq y\leqq 1\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq y\leqq 1,0\leqq x\leqq y\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\left(\int_x^1e^{y^2}dy\right)dx\)
\(\displaystyle =\int_0^1\left(\int_0^ye^{y^2}dx\right)dy\)
\(\displaystyle =\int_0^1ye^{y^2}dy\)
\(\displaystyle =\left[\frac{1}{2}e^{y^2}\right]_0^1\)
\(\displaystyle =\frac{e-1}{2}\)
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,x\leqq y\leqq 1\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq y\leqq 1,0\leqq x\leqq y\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\left(\int_x^1e^{y^2}dy\right)dx\)
\(\displaystyle =\int_0^1\left(\int_0^ye^{y^2}dx\right)dy\)
\(\displaystyle =\int_0^1ye^{y^2}dy\)
\(\displaystyle =\left[\frac{1}{2}e^{y^2}\right]_0^1\)
\(\displaystyle =\frac{e-1}{2}\)
(2)\(\displaystyle \int_0^1\left(\int_\sqrt{y}^1e^{x^3}dx\right)dy\)
積分領域を
\(\displaystyle D=\{(x,y)|0\leqq y\leqq 1,\sqrt{y}\leqq x\leqq 1\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq x\leqq 1,0\leqq y\leqq x^2\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\left(\int_\sqrt{y}^1e^{x^3}dx\right)dy\)
\(\displaystyle =\int_0^1\left(\int_0^{x^2}e^{x^3}dy\right)dx\)
\(\displaystyle =\int_0^1x^2e^{x^3}dx\)
\(\displaystyle =\left[\frac{1}{3}e^{x^3}\right]_0^1\)
\(\displaystyle =\frac{e-1}{3}\)
\(\displaystyle D=\{(x,y)|0\leqq y\leqq 1,\sqrt{y}\leqq x\leqq 1\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq x\leqq 1,0\leqq y\leqq x^2\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\left(\int_\sqrt{y}^1e^{x^3}dx\right)dy\)
\(\displaystyle =\int_0^1\left(\int_0^{x^2}e^{x^3}dy\right)dx\)
\(\displaystyle =\int_0^1x^2e^{x^3}dx\)
\(\displaystyle =\left[\frac{1}{3}e^{x^3}\right]_0^1\)
\(\displaystyle =\frac{e-1}{3}\)
(3)\(\displaystyle \int_1^\sqrt{2}\left(\int_{x^2}^2\frac{1}{y^2}e^{\frac{x}{\sqrt{y}}}dy\right)dx\)
積分領域を
\(\displaystyle D=\{(x,y)|1\leqq x\leqq \sqrt{2},x^2\leqq y\leqq 2\}\)
\(\displaystyle \ \ \ =\{(x,y)|1\leqq y\leqq 2,1\leqq x\leqq \sqrt{y}\}\)
積分順序を変更すると
\(\displaystyle \int_1^\sqrt{2}\left(\int_{x^2}^2\frac{1}{y^2}e^{\frac{x}{\sqrt{y}}}dy\right)dx\)
\(\displaystyle =\int_1^2\left(\int_1^\sqrt{y}\frac{1}{y^2}e^{\frac{x}{\sqrt{y}}}dx\right)dy\)
\(\displaystyle =\int_1^2\left[\frac{1}{y\sqrt{y}}e^{\frac{x}{\sqrt{y}}}\right]_1^\sqrt{y}dy\)
\(\displaystyle =\int_1^2\frac{1}{y\sqrt{y}}(e-e^{\frac{1}{\sqrt{y}}})dy\)
\(\displaystyle =\left[-\frac{2e}{\sqrt{y}}+2e^{\frac{1}{\sqrt{y}}}\right]_1^2\)
\(\displaystyle =2e^{\frac{1}{\sqrt{2}}}-\sqrt{2}e\)
\(\displaystyle D=\{(x,y)|1\leqq x\leqq \sqrt{2},x^2\leqq y\leqq 2\}\)
\(\displaystyle \ \ \ =\{(x,y)|1\leqq y\leqq 2,1\leqq x\leqq \sqrt{y}\}\)
積分順序を変更すると
\(\displaystyle \int_1^\sqrt{2}\left(\int_{x^2}^2\frac{1}{y^2}e^{\frac{x}{\sqrt{y}}}dy\right)dx\)
\(\displaystyle =\int_1^2\left(\int_1^\sqrt{y}\frac{1}{y^2}e^{\frac{x}{\sqrt{y}}}dx\right)dy\)
\(\displaystyle =\int_1^2\left[\frac{1}{y\sqrt{y}}e^{\frac{x}{\sqrt{y}}}\right]_1^\sqrt{y}dy\)
\(\displaystyle =\int_1^2\frac{1}{y\sqrt{y}}(e-e^{\frac{1}{\sqrt{y}}})dy\)
\(\displaystyle =\left[-\frac{2e}{\sqrt{y}}+2e^{\frac{1}{\sqrt{y}}}\right]_1^2\)
\(\displaystyle =2e^{\frac{1}{\sqrt{2}}}-\sqrt{2}e\)
(4)\(\displaystyle \int_0^1\left(\int_0^\sqrt{1-x^2}(1-y^2)^\frac{3}{2}dy\right)dx\)
積分領域を
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,0\leqq y\leqq \sqrt{1-x^2}\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq y\leqq 1,0\leqq x\leqq \sqrt{1-y^2}\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\left(\int_0^\sqrt{1-x^2}(1-y^2)^\frac{3}{2}dy\right)dx\)
\(\displaystyle =\int_0^1\left(\int_0^\sqrt{1-y^2}(1-y^2)^\frac{3}{2}dx\right)dy\)
\(\displaystyle =\int_0^1(1-y^2)^2dy\)
\(\displaystyle =\int_0^1(y^4-2y^2+1)dy\)
\(\displaystyle =\left[\frac{1}{5}y^5-\frac{2}{3}y^3+y\right]_0^1\)
\(\displaystyle =\frac{8}{15}\)
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,0\leqq y\leqq \sqrt{1-x^2}\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq y\leqq 1,0\leqq x\leqq \sqrt{1-y^2}\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\left(\int_0^\sqrt{1-x^2}(1-y^2)^\frac{3}{2}dy\right)dx\)
\(\displaystyle =\int_0^1\left(\int_0^\sqrt{1-y^2}(1-y^2)^\frac{3}{2}dx\right)dy\)
\(\displaystyle =\int_0^1(1-y^2)^2dy\)
\(\displaystyle =\int_0^1(y^4-2y^2+1)dy\)
\(\displaystyle =\left[\frac{1}{5}y^5-\frac{2}{3}y^3+y\right]_0^1\)
\(\displaystyle =\frac{8}{15}\)
次の学習に進もう！