设p为常数,若反常积分 (int )_(0)^1dfrac (ln x)({x)^p((1-x))^1-p}dx 收敛,则p的取值范围是 () .-|||-(A) (-1,1) (B) (-1,2) (C) (-infty ,1) (D) (-infty ,2)

[题目]-|||-2.讨论下列瑕积分是否收敛?若收敛则求其值:-|||-(1) (int )_(a)^bdfrac (dx)({(x-a))^p}-|||-(2

(int )_(0)^1dfrac (dx)({x)^2sqrt (1-x)} 发散-|||-D (int )_(0)^1dfrac (dx)(xsqrt {1

设 gt 0 ,知 lim _(xarrow +infty )(x)^p((a)^dfrac (1{x)}-(a)^dfrac (1{x+1)}) 存在,则P的

1.以下各积分不属于反常积分的是 () .-|||-(A) (int )_(0)^+infty ln (1+x)dx (B) (int )_(0)^1dfrac

17.求p的取值范围，使得int_(1)^+inftysin(pi)/(x)cdot(dx)/(ln^p)x收敛.17.求p的取值范围，使得$\int_{1}^

设总体X具有分布律:P(X=-1)=(1-p) (X=1)=(p)^2,P(X=-1)=(1-p) (X=1)=(p)^2,是来自X的一个样本观察值,求参数p的

不定积分int dfrac (1)(1+sqrt {1-x)}dx=( ) int dfrac (1)(1+sqrt {1-x)}dx=int dfrac (

(int )_(0)^1dfrac (x+2)(sqrt {x+1)}dx=( )(int )_(0)^1dfrac (x+2)(sqrt {x+1)}dx=

(int )_(dfrac {3)(4)}^1dfrac (dx)(sqrt {1-x)-1}..

(4)若 (int )_(0)^1f(x)dx=agt 0, 则 (int )_(0)^1dfrac (1)(sqrt {x)}f(sqrt (x))dx=()