一维线性谐振子处在基态 psi (x)=sqrt (alpha /sqrt {pi )}(e)^-dfrac (1{2)(a)^2(x)^2} ,求-|||-(1)势能的期望值 overline (U)=dfrac (1)(2)m(omega )^2overline ({x)^2} ;-|||-(2)动能的期望值 overline (T)=dfrac (1)(2m)overline (p)^2 ;-|||-(3)动量的概率分布函数.

氢原子处在基态 psi(r,theta,varphi)=(1)/(sqrt(pi a_0^3))e^-(r)/(a_0)，求：(1) r 的期望值；(2) 势能

A)(overline(X) - u_(alpha) (2)/(sqrt(n)), overline(X) + u_(alpha) (2)/(sqrt(n))

A)(overline(X)-u_((alpha)/(2))(4)/(sqrt(n)),overline(X)+u_((alpha)/(2))(4)/(sqrt

(D) (overline (x)-(u)_(a12)dfrac (2)(sqrt {n)},overline (x)+(u)_(a12)dfrac (2)(s

设=dfrac (arcsin x)(sqrt {1-{x)^2}}（1）证明：=dfrac (arcsin x)(sqrt {1-{x)^2}}（2）求=df

设随机变量X的概率密度为(x)=dfrac (1)(2sqrt {pi )}(e)^-dfrac ({(x-3)^2)(4)}((x)=dfrac (1)(2s

求lim _(xarrow infty )dfrac (ln (x+sqrt {{x)^2+1)}-ln (x+sqrt ({x)^2-1})}({({e)^d

求下列定积分:-|||-(int )_(dfrac {sqrt {2)}(2)}^1dfrac (sqrt {1-{x)^2}}({x)^2}dx

(1)求 (int )_(0)^dfrac (pi {2)}sqrt (1-sin 2x)dx;

求下列函数的微分:-|||-(1) =dfrac (1)(x)+2sqrt (x);-|||-(2) =xsin 2x;-|||-(3) =dfrac (x)(