题目
4.已知函数f(x)具有连续导数，f(0)=1，且曲线积分int_(L)^[e^(x+f(x)]ydx-f(x)dy)与路径无关，试确定f(x)，并计算int_((0,0))^(1,1)[e^x+f(x)]ydx-f(x)dy的值.
4.已知函数f(x)具有连续导数，f(0)=1，且曲线积分
$\int_{L}^{[e^{x}+f(x)]ydx-f(x)dy}$
与路径无关，试确定f(x)，并计算$\int_{(0,0)}^{(1,1)}[e^{x}+f(x)]ydx-f(x)dy$的值.
题目解答
答案
设 $ P(x,y) = [e^x + f(x)]y $，$ Q(x,y) = -f(x) $。由曲线积分与路径无关，得 $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$，即 $e^x + f(x) = -f'(x)$。解得 $f(x) = -\frac{1}{2}e^x + \frac{3}{2}e^{-x}$。
取路径 $y = x$，则积分变为
\[
\int_0^1 [e^x + f(x)]x - f(x) \, dx = \int_0^1 \left[ e^x x + f(x)(x-1) \right] \, dx.
\]
代入 $f(x)$ 并计算得
\[
\boxed{\frac{e^2 - 3}{2e}}.
\]
解析
步骤 1：确定f(x)
设 $ P(x,y) = [e^x + f(x)]y $，$ Q(x,y) = -f(x) $。由曲线积分与路径无关，得 $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$，即 $e^x + f(x) = -f'(x)$。这是一个一阶线性微分方程，解得 $f(x) = -\frac{1}{2}e^x + \frac{3}{2}e^{-x}$。由于 $f(0) = 1$，代入验证，满足条件。
步骤 2：计算积分
取路径 $y = x$，则积分变为 \[ \int_0^1 [e^x + f(x)]x - f(x) \, dx = \int_0^1 \left[ e^x x + f(x)(x-1) \right] \, dx. \] 代入 $f(x) = -\frac{1}{2}e^x + \frac{3}{2}e^{-x}$，得 \[ \int_0^1 \left[ e^x x + \left(-\frac{1}{2}e^x + \frac{3}{2}e^{-x}\right)(x-1) \right] \, dx. \] 分别计算每一项的积分，得 \[ \int_0^1 e^x x \, dx = \left[ e^x x - e^x \right]_0^1 = e - 1, \] \[ \int_0^1 -\frac{1}{2}e^x (x-1) \, dx = \left[ -\frac{1}{2}e^x (x-1) + \frac{1}{2}e^x \right]_0^1 = -\frac{1}{2}e + \frac{1}{2}, \] \[ \int_0^1 \frac{3}{2}e^{-x} (x-1) \, dx = \left[ -\frac{3}{2}e^{-x} (x-1) - \frac{3}{2}e^{-x} \right]_0^1 = \frac{3}{2}e^{-1} - \frac{3}{2}. \] 将上述结果相加，得 \[ e - 1 - \frac{1}{2}e + \frac{1}{2} + \frac{3}{2}e^{-1} - \frac{3}{2} = \frac{e^2 - 3}{2e}. \]
设 $ P(x,y) = [e^x + f(x)]y $，$ Q(x,y) = -f(x) $。由曲线积分与路径无关，得 $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$，即 $e^x + f(x) = -f'(x)$。这是一个一阶线性微分方程，解得 $f(x) = -\frac{1}{2}e^x + \frac{3}{2}e^{-x}$。由于 $f(0) = 1$，代入验证，满足条件。
步骤 2：计算积分
取路径 $y = x$，则积分变为 \[ \int_0^1 [e^x + f(x)]x - f(x) \, dx = \int_0^1 \left[ e^x x + f(x)(x-1) \right] \, dx. \] 代入 $f(x) = -\frac{1}{2}e^x + \frac{3}{2}e^{-x}$，得 \[ \int_0^1 \left[ e^x x + \left(-\frac{1}{2}e^x + \frac{3}{2}e^{-x}\right)(x-1) \right] \, dx. \] 分别计算每一项的积分，得 \[ \int_0^1 e^x x \, dx = \left[ e^x x - e^x \right]_0^1 = e - 1, \] \[ \int_0^1 -\frac{1}{2}e^x (x-1) \, dx = \left[ -\frac{1}{2}e^x (x-1) + \frac{1}{2}e^x \right]_0^1 = -\frac{1}{2}e + \frac{1}{2}, \] \[ \int_0^1 \frac{3}{2}e^{-x} (x-1) \, dx = \left[ -\frac{3}{2}e^{-x} (x-1) - \frac{3}{2}e^{-x} \right]_0^1 = \frac{3}{2}e^{-1} - \frac{3}{2}. \] 将上述结果相加，得 \[ e - 1 - \frac{1}{2}e + \frac{1}{2} + \frac{3}{2}e^{-1} - \frac{3}{2} = \frac{e^2 - 3}{2e}. \]