(B) (int )_({x)_(1)}^(x_{3)}pi [ (f)^2(x)-(g)^2(x)] dx.-|||-(C)(int )_(x)^(x_{3)}pi |(f)^2(x)-(g)^2(x)|dx (D) |(int )_({x)_(1)}^(x_{3)}pi [ (f)^2(x)-(g)^2(x)] dx|

设(X_(1),X_(2),X_(3),X_(4))是总体X的简单随机样本，X~N(0,4)，F=C(X_(1)^2)/(X_(2)^2+X_{3)^2+X_(

(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}dfrac (|x|sin x)(1+{cos )^3x}dx=(int )_(-

二次型f(x_(1),x_(2),x_(3))=x_(1)^2+4x_(2)^2+4x_(3)^2+2lambda x_(1)x_(2)-2x_(1)x_(3)

已知实二次型f(x_(1)，x_(2)，x_(3))=x_(1)^2+4x_(2)^2+4x_(3)^2+2lambda x_(1)x_(2)-2x_(1)x_

设 M=int_(-(pi)/(2))^(pi)/(2) (sin x)/(1+x^2) cos^4 x dx，N=int_(-(pi)/(2))^(pi)/(

实 二 次 型 f(x_(1),x_(2),x_(3))=3x_(1)^2+4x_(1)x_(2)+4x_(2)^2-4x_(2)x_(3)+5x_(3)^2是

求-|||-(1) (int )_(-1)^1f(x)dx;-|||-(2)f(x)dx;-|||-(3) (int )_(3)^-1g(x)dx;-|||-(

设f(x)=arctan x，g(x)=sin (2x+pi )/(3)，则g[f(-1)]= .设$f\left(x\right)=arc\ta

2f=overline(x)_(1)x_(1)+3overline(x)_(1)x_(3)+4overline(x)_(2)x_(2)-3overline(x)

(int )_(-pi )^pi sqrt ({pi )^2-(x)^2}dx= )(int )_(-pi )^pi sqrt ({pi )^2-(x)^2}d