求极限lim _(xarrow 0)dfrac ({int )_(0)^2xln (1+t)dt}(xsin x)求极限
(2) (int )_(1)^exln xdx;
(int )_(0)^+infty x(2)^-xdx=( )(int )_(0)^+infty x(2)^-xdx=( )(int )_(0)^+inft
5.设 _(1)=(int )_(0)^1xdx , _(2)=(int )_(0)^1sin xdx , _(3)=(int )_(0)^1tan xdx,
int 1(e)^-2tdt-|||-.(int )_(1)^2xdx.-|||-.fxsin xcos xdx.-|||-15. int (x)^2(cos
定积分(int )_(0)^3(2)^xdx=( ).A.(int )_(0)^3(2)^xdx=B.(int )_(0)^3(2)^xdx=C.(in
(1) int (x)^2arctan xdx;
(int )_(0)^1dfrac (1)(2)xdx= A.1B.0C.(int )_(0)^1dfrac (1)(2)xdx=D.2A.1B.0C.D.2
7.计算下列定积分:-|||-(1) (int )_(0)^1x(e)^-xdx ;-|||-(2)(int )_(1)^xxln xdx;-|||-(3) (
3.判断题int_(1)^2 xsin xdx=int_(1)^2 tsin tdtA. 对B. 错