4.证明：int_(0)^1x^m(1-x)^ndx=int_(0)^1x^n(1-x)^mdx(m,nin N).

2.[判断题] 判断：int_(0)^1x^m(1-x)^ndx=int_(0)^1x^n(1-x)^mdx,(m,nin N).（）A. 对B. 错

9.设a_(n)=int_(0)^1x^nsqrt(1-x^2)dx,b_(n)=int_(0)^(pi)/(2)sin^ntdt,则极限lim_(ntoinf

9.设a_(n)=int_(0)^1x^nsqrt(1-x^2)dx,b_(n)=int_(0)^(pi)/(2)sin^ntdt,则极限lim_(ntoinf

20.设a_(n)=int_(0)^1x^nsqrt(1-x^2)dx(n=0,1,2,...),则lim_(ntoinfty)((a_(n))/(a_(n-2

I=int_(-1)^1x^2(1+sin^3x)dx=A. -1B. 1C. $\frac{1}{3}$D. $\frac{2}{3}$

I=int_(-1)^1x^2(1+sin^3x)dx=A. -1B. 1C. $\frac{1}{3}$D. $\frac{2}{3}$

(6) int_((1)/(sqrt(2)))^1(sqrt(1-x^2))/(x^2)dx;(6) $\int_{\frac{1}{\sqrt{2}}}^{1

设 iint_(D) f(x, y)dx dy = int_(0)^1 dx int_(0)^1-x f(x, y)dy，则改变其积分次序后为A. $\int_

24.证明: (int )_(0)^1dx(int )_(0)^xf(x-y)dy=(int )_(0)^1f(x)(1-x)dx-|||-"f(x-y)dy=

5【判断题】 判断:反常积分int_(0)^(1)/(2)(1)/(sqrt(x(1-x)))dx是发散的.（）A. 对B. 错