# 2018年考研数二第06题解析

## 题目

$\int_{-1}^{0} dx \int_{-x}^{2-x^{2}} (1-xy) dy +\int_{0}^{1} dx \int_{x}^{2-x^{2}} (1-xy) dy=?$

$$A. \frac{5}{3}$$

$$B. \frac{5}{6}$$

$$C. \frac{7}{3}$$

$$D. \frac{7}{6}$$

## 解析

$$\int_{-x}^{2-x^{2}} (1-xy) dy =$$

$$y-\frac{1}{2} y^{2}x|_{-x}^{2-x^{2}} =$$

$$[2-x^{2} – \frac{1}{2} (2-x^{2})^{2} x]-[-x-\frac{1}{2}(-x)^{2}x]=$$

$$[2-x^{2} – \frac{1}{2}x(4+x^{4}-4x^{2})]-[-x-\frac{1}{2}x^{3}] =$$

$$2-x^{2} – 2x – \frac{1}{2}x^{5} + 2x^{3} + x + \frac{1}{2}x^{3} =$$

$$2-x^{2}-x + \frac{5}{2}x^{3}-\frac{1}{2}x^{5}.$$

$$\int_{-1}^{0} dx \int_{-x}^{2-x^{2}} (1-xy) dy =$$

$$\int_{-1}^{0} (2-x^{2}-x + \frac{5}{2}x^{3}-\frac{1}{2}x^{5}) dx =$$

$$2x – \frac{1}{3}x^{3} – \frac{1}{2}x^{2} + \frac{5}{2} \cdot \frac{1}{4}x^{4} – \frac{1}{2} \cdot \frac{1}{6} x^{6}|_{-1}^{0} =$$

$$[0]-[-2 + \frac{1}{3} -\frac{1}{2} + \frac{5}{8} – \frac{1}{12}] =$$

$$2 – \frac{1}{3} + \frac{1}{2} – \frac{5}{8} + \frac{1}{12} = \frac{13}{8}.$$

$$\int_{x}^{2-x^{2}} (1-xy) dy =$$

$$y-\frac{1}{2}y^{2}x|_{x}^{2-x^{2}} =$$

$$[2-x^{2} – \frac{1}{2}x(2-x^{2})^{2}]-[x – \frac{1}{2}x^{3}] =$$

$$[2-x^{2}-\frac{1}{2}x(4+x^{4}-4x^{2})] – x + \frac{1}{2}x^{3} =$$

$$2 – x^{2} – 2x – \frac{1}{2} x^{5} + 2x^{3} – x + \frac{1}{2}x^{3} =$$

$$2-x^{2}-3x+\frac{5}{2}x^{3}-\frac{1}{2}x^{5} =$$

$$\int_{0}^{1} dx \int_{x}^{2-x^{2}} (1-xy) dy=$$

$$\int_{0}^{1} 2-x^{2}-3x+\frac{5}{2}x^{3}-\frac{1}{2}x^{5} dx =$$

$$2x – \frac{1}{3}x^{3} – 3 \cdot \frac{1}{2} x^{2} + \frac{5}{2} \cdot \frac{1}{4}x^{4} – \frac{1}{2} \cdot \frac{1}{6} x^{6} |_{0}^{1} =$$

$$[2-\frac{1}{3} – \frac{3}{2} + \frac{5}{8} – \frac{1}{12}] – [0] = \frac{17}{24}$$

$$原式 = \frac{13}{8} + \frac{17}{24} = \frac{7}{3}.$$

$$\{ (x, y) | -1 \leqslant x \leqslant 0, -x \leqslant y \leqslant 2-x^{2} \}$$

$$\cup \{ (x, y) | 0 \leqslant x \leqslant 1, x \leqslant y \leqslant 2-x^{2} \}$$

$$\int_{-1}^{0} dx \int_{-x}^{2-x^{2}} (1-xy) dy +\int_{0}^{1} dx \int_{x}^{2-x^{2}} (1-xy) dy=$$

$$\iint_{D} (1-xy) dxdy=$$

$$\iint_{D} 1 dxdy – \iint_{D} xy dxdy =$$

$$\iint_{D} 1 dxdy – 0 =$$

$$2 \int_{0}^{1} dx \int_{x}^{2-x^{2}} 1 dy =$$

$$2 \int_{0}^{1} (2-x^{2}-x)dx =$$

$$2 \cdot (2x – \frac{1}{3}x^{3} – \frac{1}{2}x^{2}) |_{0}^{1} =$$

$$2 \cdot (2 – \frac{1}{3} – \frac{1}{2} – 0)=$$

$$2 \cdot \frac{7}{6} = \frac{7}{3}.$$

EOF