【微分積分】8-1-3 積分順序の変更｜問題集
1.次の積分順序の変更をしなさい。
(1)\(\displaystyle \int_0^1\int_{x^2}^xf(x,y)dydx\)
積分領域を
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,x^2\leqq y\leqq x\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq y\leqq 1,y\leqq x\leqq \sqrt{y}\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\int_{x^2}^xf(x,y)dydx\)
\(\displaystyle =\int_0^1\int_y^\sqrt{y}f(x,y)dxdy\)
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,x^2\leqq y\leqq x\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq y\leqq 1,y\leqq x\leqq \sqrt{y}\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\int_{x^2}^xf(x,y)dydx\)
\(\displaystyle =\int_0^1\int_y^\sqrt{y}f(x,y)dxdy\)
(2)\(\displaystyle \int_0^1\int_0^{y^2}f(x,y)dxdy\)
積分領域を
\(\displaystyle D=\{(x,y)|0\leqq y\leqq 1,0\leqq x\leqq y^2\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq x\leqq 1,\sqrt{x}\leqq y\leqq 1\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\int_0^{y^2}f(x,y)dxdy\)
\(\displaystyle =\int_0^1\int_\sqrt{x}^1f(x,y)dydx\)
\(\displaystyle D=\{(x,y)|0\leqq y\leqq 1,0\leqq x\leqq y^2\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq x\leqq 1,\sqrt{x}\leqq y\leqq 1\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\int_0^{y^2}f(x,y)dxdy\)
\(\displaystyle =\int_0^1\int_\sqrt{x}^1f(x,y)dydx\)
(3)\(\displaystyle \int_0^2\int_\frac{x}{2}^{3-x}f(x,y)dydx\)
積分領域を
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 2,\frac{x}{2}\leqq y\leqq 3-x\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq y\leqq 1,0\leqq x\leqq 2y\}\cup\{(x,y)|1\leqq y\leqq 3,0\leqq x\leqq 3-y\}\)
積分順序を変更すると
\(\displaystyle \int_0^2\int_\frac{x}{2}^{3-x}f(x,y)dydx\)
\(\displaystyle =\int_0^1\int_0^{2y}f(x,y)dxdy+\int_1^3\int_0^{3-y}f(x,y)dxdy\)
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 2,\frac{x}{2}\leqq y\leqq 3-x\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq y\leqq 1,0\leqq x\leqq 2y\}\cup\{(x,y)|1\leqq y\leqq 3,0\leqq x\leqq 3-y\}\)
積分順序を変更すると
\(\displaystyle \int_0^2\int_\frac{x}{2}^{3-x}f(x,y)dydx\)
\(\displaystyle =\int_0^1\int_0^{2y}f(x,y)dxdy+\int_1^3\int_0^{3-y}f(x,y)dxdy\)
(4)\(\displaystyle \int_0^1\int_{x^4}^{x^2}f(x,y)dydx\)
積分領域を
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,x^4\leqq y\leqq x^2\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq y\leqq 1,\sqrt{y}\leqq x\leqq y^\frac{1}{4}\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\int_{x^4}^{x^2}f(x,y)dydx\)
\(\displaystyle =\int_0^1\int_\sqrt{y}^{y^{\frac{1}{4}}}f(x,y)dxdy\)
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,x^4\leqq y\leqq x^2\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq y\leqq 1,\sqrt{y}\leqq x\leqq y^\frac{1}{4}\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\int_{x^4}^{x^2}f(x,y)dydx\)
\(\displaystyle =\int_0^1\int_\sqrt{y}^{y^{\frac{1}{4}}}f(x,y)dxdy\)
(5)\(\displaystyle \int_0^1\int_{-y}^yf(x,y)dxdy\)
積分領域を
\(\displaystyle D=\{(x,y)|0\leqq y\leqq 1,-y\leqq x\leqq y\}\)
\(\displaystyle \ \ \ =\{(x,y)|-1\leqq x\leqq 0,-x\leqq y\leqq 1\}\cup\{(x,y)|0\leqq x\leqq 1,x\leqq y\leqq 1\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\int_{-y}^yf(x,y)dxdy\)
\(\displaystyle =\int_{-1}^0\int_{-x}^1f(x,y)dydx+\int_0^1\int_x^1f(x,y)dydx\)
\(\displaystyle D=\{(x,y)|0\leqq y\leqq 1,-y\leqq x\leqq y\}\)
\(\displaystyle \ \ \ =\{(x,y)|-1\leqq x\leqq 0,-x\leqq y\leqq 1\}\cup\{(x,y)|0\leqq x\leqq 1,x\leqq y\leqq 1\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\int_{-y}^yf(x,y)dxdy\)
\(\displaystyle =\int_{-1}^0\int_{-x}^1f(x,y)dydx+\int_0^1\int_x^1f(x,y)dydx\)
(6)\(\displaystyle \int_1^4\int_x^{2x}f(x,y)dydx\)
積分領域を
\(\displaystyle D=\{(x,y)|1\leqq x\leqq 4,x\leqq y\leqq 2x\}\)
\(\displaystyle \ \ \ =\{(x,y)|1\leqq y\leqq 2,1\leqq x\leqq y\}\)
\(\displaystyle \ \ \ \ \ \ \cup\{(x,y)|2\leqq y\leqq 4,\frac{y}{2}\leqq x\leqq y\}\)
\(\displaystyle \ \ \ \ \ \ \cup\{(x,y)|4\leqq y\leqq 8,\frac{y}{2}\leqq x\leqq 4\}\)
積分順序を変更すると
\(\displaystyle \int_1^4\int_x^{2x}f(x,y)dydx=\int_1^2\int_1^yf(x,y)dxdy\)
\(\displaystyle \ \ \ +\int_2^4\int_\frac{y}{2}^yf(x,y)dxdy+\int_4^8\int_\frac{y}{2}^4f(x,y)dxdy\)
\(\displaystyle D=\{(x,y)|1\leqq x\leqq 4,x\leqq y\leqq 2x\}\)
\(\displaystyle \ \ \ =\{(x,y)|1\leqq y\leqq 2,1\leqq x\leqq y\}\)
\(\displaystyle \ \ \ \ \ \ \cup\{(x,y)|2\leqq y\leqq 4,\frac{y}{2}\leqq x\leqq y\}\)
\(\displaystyle \ \ \ \ \ \ \cup\{(x,y)|4\leqq y\leqq 8,\frac{y}{2}\leqq x\leqq 4\}\)
積分順序を変更すると
\(\displaystyle \int_1^4\int_x^{2x}f(x,y)dydx=\int_1^2\int_1^yf(x,y)dxdy\)
\(\displaystyle \ \ \ +\int_2^4\int_\frac{y}{2}^yf(x,y)dxdy+\int_4^8\int_\frac{y}{2}^4f(x,y)dxdy\)
2.次の2重積分を求めなさい。
(1)\(\displaystyle \int_0^1\int_y^1e^\frac{y}{x}dxdy\)
積分領域を
\(\displaystyle D=\{(x,y)|0\leqq y\leqq 1,y\leqq x\leqq 1\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq x\leqq 1,0\leqq y\leqq x\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\int_y^1e^\frac{y}{x}dxdy\)
\(\displaystyle =\int_0^1\int_0^xe^\frac{y}{x}dydx\)
\(\displaystyle =\int_0^1[xe^\frac{y}{x}]dx\)
\(\displaystyle =\int_0^1(xe-x)dx\)
\(\displaystyle =\left[\frac{(e-1)x^2}{2}\right]_0^1\)
\(\displaystyle =\frac{e-1}{2}\)
\(\displaystyle D=\{(x,y)|0\leqq y\leqq 1,y\leqq x\leqq 1\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq x\leqq 1,0\leqq y\leqq x\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\int_y^1e^\frac{y}{x}dxdy\)
\(\displaystyle =\int_0^1\int_0^xe^\frac{y}{x}dydx\)
\(\displaystyle =\int_0^1[xe^\frac{y}{x}]dx\)
\(\displaystyle =\int_0^1(xe-x)dx\)
\(\displaystyle =\left[\frac{(e-1)x^2}{2}\right]_0^1\)
\(\displaystyle =\frac{e-1}{2}\)
(2)\(\displaystyle \int_0^1\int_x^1e^{y^2}dydx\)
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
\(\displaystyle \int_0^1\int_x^1e^{y^2}dydx\)
\(\displaystyle =\int_0^1\int_0^ye^{y^2}dxdy\)
\(\displaystyle =\int_0^1[xe^{y^2}]_0^ydy\)
\(\displaystyle =\int_0^1ye^{y^2}dy\)
\(\displaystyle =\left[\frac{1}{2}e^{y^2}\right]_0^1\)
\(\displaystyle =\frac{e-1}{2}\)
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,x\leqq y\leqq 1\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq y\leqq 1,0\leqq x\leqq y\}\)
積分順序を変更すると
\(\displaystyle \int_0^1\int_x^1e^{y^2}dydx\)
\(\displaystyle =\int_0^1\int_0^ye^{y^2}dxdy\)
\(\displaystyle =\int_0^1[xe^{y^2}]_0^ydy\)
\(\displaystyle =\int_0^1ye^{y^2}dy\)
\(\displaystyle =\left[\frac{1}{2}e^{y^2}\right]_0^1\)
\(\displaystyle =\frac{e-1}{2}\)
(3)\(\displaystyle \int_0^1dy\int_y^\sqrt{y}\frac{\sin x}{x}dx\)
積分領域を
\(\displaystyle D=\{(x,y)|0\leqq y\leqq 1,y\leqq x\leqq \sqrt{y}\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq x\leqq 1,x^2\leqq y\leqq x\}\)
積分順序を変更すると
\(\displaystyle \int_0^1dy\int_y^\sqrt{y}\frac{\sin x}{x}dx\)
\(\displaystyle =\int_0^1\int_{x^2}^x\frac{\sin x}{x}dydx\)
\(\displaystyle =\int_0^1[\frac{y\sin x}{x}]_{x^2}^xdx\)
\(\displaystyle =\int_0^1(\sin x-x\sin x)dx\)
\(\displaystyle =[-\cos x+x\cos x-\sin x]_0^1\)
\(\displaystyle =1-\sin1\)
\(\displaystyle D=\{(x,y)|0\leqq y\leqq 1,y\leqq x\leqq \sqrt{y}\}\)
\(\displaystyle \ \ \ =\{(x,y)|0\leqq x\leqq 1,x^2\leqq y\leqq x\}\)
積分順序を変更すると
\(\displaystyle \int_0^1dy\int_y^\sqrt{y}\frac{\sin x}{x}dx\)
\(\displaystyle =\int_0^1\int_{x^2}^x\frac{\sin x}{x}dydx\)
\(\displaystyle =\int_0^1[\frac{y\sin x}{x}]_{x^2}^xdx\)
\(\displaystyle =\int_0^1(\sin x-x\sin x)dx\)
\(\displaystyle =[-\cos x+x\cos x-\sin x]_0^1\)
\(\displaystyle =1-\sin1\)
次の学習に進もう！