高専の数学解説

第8章 8.2

(1)

∫12∫1x2(x2+2y) dy dx=∫12[x2y+y2]1x2dx=∫12x4+x4−x2−1 dx=∫122x4−x2−1 dx\int_1^2\int_1^{x^2} (x^2+2y)\,dy\,dx = \int_1^2 \left[x^2y+y^2\right]_1^{x^2} dx = \int_1^2 x^4+x^4-x^2-1\,dx = \int_1^2 2x^4-x^2-1\,dx

=[25x5−13x3−x]12=13615= \left[\frac{2}{5}x^5-\frac{1}{3}x^3-x\right]_1^2 = \frac{136}{15}

順序入れ替え

∫14∫y2(x2+2y) dx dy=∫14[13x3+2xy]y2dy=∫1483+4y−13y32−2y32 dy\int_1^4\int_{\sqrt{y}}^2 (x^2+2y)\,dx\,dy = \int_1^4 \left[\frac{1}{3}x^3+2xy\right]_{\sqrt{y}}^2 dy = \int_1^4 \frac{8}{3}+4y-\frac{1}{3}y^{\frac{3}{2}}-2y^{\frac{3}{2}}\,dy

=∫1483+4y−73y32 dy=[83y+2y2−1415y52]14=13615= \int_1^4 \frac{8}{3}+4y-\frac{7}{3}y^{\frac{3}{2}}\,dy = \left[\frac{8}{3}y+2y^2-\frac{14}{15}y^{\frac{5}{2}}\right]_1^4 = \frac{136}{15}

(2)

∫24∫4x2xy2 dy dx=∫24[−xy]4x2dx=∫24−1x+14x dx=[−log⁡x+18x2]24=32−log⁡2\int_2^4\int_4^{x^2} \frac{x}{y^2}\,dy\,dx = \int_2^4 \left[-\frac{x}{y}\right]_4^{x^2} dx = \int_2^4 -\frac{1}{x}+\frac{1}{4}x\,dx = \left[-\log x+\frac{1}{8}x^2\right]_2^4 = \frac{3}{2}-\log 2

順序入れ替え

∫416∫y4xy2 dx dy=∫416[12x2y2]y4dy=∫4168y2−12y dy=[−8y−12log⁡y]416=32−log⁡2\int_4^{16}\int_{\sqrt{y}}^4 \frac{x}{y^2}\,dx\,dy = \int_4^{16} \left[\frac{1}{2}\frac{x^2}{y^2}\right]_{\sqrt{y}}^4 dy = \int_4^{16} \frac{8}{y^2}-\frac{1}{2y}\,dy = \left[-\frac{8}{y}-\frac{1}{2}\log y\right]_4^{16} = \frac{3}{2}-\log 2

(3)

∫03∫02−23x(1−x+y3)dy dx=∫03[y−x3y−16y2]02−23xdx=∫0343−89x+427x2 dx\int_0^3\int_0^{2-\frac{2}{3}x} \left(1-\frac{x+y}{3}\right) dy\,dx = \int_0^3 \left[y-\frac{x}{3}y-\frac{1}{6}y^2\right]_0^{2-\frac{2}{3}x} dx = \int_0^3 \frac{4}{3}-\frac{8}{9}x+\frac{4}{27}x^2\,dx

=[43x−49x2+481x3]03=43= \left[\frac{4}{3}x-\frac{4}{9}x^2+\frac{4}{81}x^3\right]_0^3 = \frac{4}{3}

順序入れ替え

∫02∫03−32y(1−x+y3)dx dy=∫02[x−16x2−y3x]03−32ydy=∫0232−y+18y2 dy\int_0^2\int_0^{3-\frac{3}{2}y} \left(1-\frac{x+y}{3}\right) dx\,dy = \int_0^2 \left[x-\frac{1}{6}x^2-\frac{y}{3}x\right]_0^{3-\frac{3}{2}y} dy = \int_0^2 \frac{3}{2}-y+\frac{1}{8}y^2\,dy

=[32y−12y2+124y3]02=43= \left[\frac{3}{2}y-\frac{1}{2}y^2+\frac{1}{24}y^3\right]_0^2 = \frac{4}{3}

(4)

∫01∫x2x2x2y dy dx=∫01[x2y2]x2xdx=∫014x4−x4 dx=∫013x4 dx=[35x5]01=35\int_0^1 \int_x^{2x} 2x^2 y \, dy\, dx = \int_0^1 \left[x^2y^2\right]_x^{2x} dx = \int_0^1 4x^4 - x^4 \, dx = \int_0^1 3x^4 \, dx = \left[\frac{3}{5}x^5\right]_0^1 = \frac{3}{5}

順序入れ替え

∫01∫y2y2x2y dx dy+∫12∫y212x2y dx dy=∫01[23x3y]y2ydx+∫12[23x3y]y21dx\int_0^1 \int_{\frac{y}{2}}^{y} 2x^2 y\, dx\, dy + \int_1^2 \int_{\frac{y}{2}}^{1} 2x^2 y\, dx\, dy = \int_0^1 \left[\frac{2}{3}x^3y\right]_{\frac{y}{2}}^{y} dx + \int_1^2 \left[\frac{2}{3}x^3y\right]_{\frac{y}{2}}^{1} dx

=∫01712y4 dy+∫1223y−112y4 dy=[760y5]01+[13y2−160y5]12=35= \int_0^1 \frac{7}{12}y^4\, dy + \int_1^2 \frac{2}{3}y - \frac{1}{12}y^4\, dy = \left[\frac{7}{60}y^5\right]_0^1 + \left[\frac{1}{3}y^2 - \frac{1}{60}y^5\right]_1^2 = \frac{3}{5}

解説: r31bn1z