Let $$y=y(x)$$ be the solution of the differential equation
$$\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{2 x}{\left(1+x^2\right)^2} y=x \mathrm{e}^{\frac{1}{\left(1+x^2\right)}} ; y(0)=0.$$
Then the area enclosed by the curve $$f(x)=y(x) \mathrm{e}^{-\frac{1}{\left(1+x^2\right)}}$$ and the line $$y-x=4$$ is ________.
Let the mean and the standard deviation of the probability distribution
$$\mathrm{X}$$ | $$\alpha$$ | 1 | 0 | $$-$$3 |
---|---|---|---|---|
$$\mathrm{P(X)}$$ | $$\frac{1}{3}$$ | $$\mathrm{K}$$ | $$\frac{1}{6}$$ | $$\frac{1}{4}$$ |
be $$\mu$$ and $$\sigma$$, respectively. If $$\sigma-\mu=2$$, then $$\sigma+\mu$$ is equal to ________.
Let the maximum and minimum values of $$\left(\sqrt{8 x-x^2-12}-4\right)^2+(x-7)^2, x \in \mathbf{R}$$ be $$\mathrm{M}$$ and $$\mathrm{m}$$, respectively. Then $$\mathrm{M}^2-\mathrm{m}^2$$ is equal to _________.
Let the point $$(-1, \alpha, \beta)$$ lie on the line of the shortest distance between the lines $$\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$$ and $$\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$$. Then $$(\alpha-\beta)^2$$ is equal to _________.