第8章 8.1

(1)

∫01∫12(x−y)dydx=∫01[xy−12y2]12dx=∫012x−2−x+12dx=∫01x−32dx=[12x2−32x]01=−1\int_0^1\int_1^2 (x-y)dydx = \int_0^1\left[xy-\frac{1}{2}y^2\right]_1^2 dx = \int_0^1 2x-2-x+\frac{1}{2}dx = \int_0^1 x-\frac{3}{2}dx = \left[\frac{1}{2}x^2-\frac{3}{2}x\right]_0^1 = -1

(2)

∫−11∫01xy2dxdy=∫−11[12x2y2]01dy=∫−1112y2dy=[16y3]−11=13\int_{-1}^1\int_0^1 xy^2 dxdy = \int_{-1}^1\left[\frac{1}{2}x^2y^2\right]_0^1 dy = \int_{-1}^1 \frac{1}{2}y^2 dy = \left[\frac{1}{6}y^3\right]_{-1}^1 = \frac{1}{3}

(3)

∫0a∫0bxy(x−y)dydx=∫0a[12x2y2−13xy3]0bdx=∫0a12b2x2−13b3xdx=[16b2x3−16b3x2]0a\int_0^a\int_0^b xy(x-y)dydx = \int_0^a\left[\frac{1}{2}x^2y^2-\frac{1}{3}xy^3\right]_0^b dx = \int_0^a \frac{1}{2}b^2x^2-\frac{1}{3}b^3xdx = \left[\frac{1}{6}b^2x^3-\frac{1}{6}b^3x^2\right]_0^a

=16(a3b2−a2b3)= \frac{1}{6}(a^3b^2-a^2b^3)

解説: r31bn1z