# 2011年考研数二第21题解析：二重积分、分部积分

## 解析

$$\iint_{D} f(x,y) dxdy = a$$

$$\int_{0}^{1} dx \int_{0}^{1}f(x,y) dy = a.$$

$$I = \iint_{D} xy f_{xy}^{”}(x,y) dxdy =$$

$$\int_{0}^{1} x dx \int_{0}^{1} y f_{xy}^{”}(x,y) dy =$$

$$\int_{0}^{1} x dx \int_{0}^{1} y \cdot d[f_{x}^{‘}(x,y)].$$

$$\int_{0}^{1} y \cdot d[f_{x}^{‘}(x,y)] = ①$$

$$y f_{x}^{‘}(x,y) |_{0}^{1} – \int_{0}^{1} f_{x}^{‘}(x,y) dy = ②$$

$$1 \cdot f_{x}^{‘} (x,1) – \int_{0}^{1} f_{x}^{‘} (x,y) dy = ③$$

$$f_{x}^{‘} (x,1) – \int_{0}^{1} f_{x}^{‘} (x,y) dy.$$

$$\int_{0}^{1} x dx \int_{0}^{1} y \cdot d[f_{x}^{‘}(x,y)] =$$

$$\int_{0}^{1} x dx [f_{x}^{‘}(x,1) – \int_{0}^{1} f_{x}^{‘} (x,y) dy] =$$

$$\int_{0}^{1} x f_{x}^{‘} (x,1) dx – \int_{0}^{1} x dx \int_{0}^{1} f_{x}^{‘}(x,y) dy.$$

$$\int_{0}^{1} x f_{x}^{‘}(x,1) dx =$$

$$x f(x,1)|_{0}^{1} – \int_{0}^{1} f(x,1) dx =$$

$$x \cdot 0|_{0}^{1} – \int_{0}^{1} f(x,1) dx =$$

$$(-1) \cdot \int_{0}^{1} f(x,1) dx = 0.$$

$$\int_{0}^{1} x f_{x}^{‘} (x,1) dx – \int_{0}^{1} x dx \int_{0}^{1} f_{x}^{‘}(x,y) dy =$$

$$(-1) \cdot \int_{0}^{1} x dx \int_{0}^{1} f_{x}^{‘} (x,y) dy =$$

$$(-1) \cdot \int_{0}^{1} dy \int_{0}^{1} x f_{x}^{‘} (x,y) dx.$$

$$\int_{0}^{1} x f_{x}^{‘} (x,y) dx =$$

$$x f(x,y)|_{0}^{1} – \int_{0}^{1} f(x,y) dx =$$

$$1 \cdot f(1,y) – 0 – \int_{0}^{1} f(x,y) dx =$$

$$(-1) \cdot \int_{0}^{1} f(x,y) dx.$$

$$(-1) \cdot \int_{0}^{1} dy \int_{0}^{1} x f_{x}^{‘} (x,y) dx =$$

$$(-1) \cdot (-1) \cdot \int_{0}^{1} dy \int_{0}^{1} f(x,y) dx =$$

$$\int_{0}^{1} dy \int_{0}^{1} f(x,y) dx =$$

$$\int_{0}^{1} dx \int_{0}^{1} f(x,y) dy = a.$$

$$I = \iint_{D} xy f_{xy}^{”}(x,y) dxdy = a.$$