已知 '(x)=cos ((1-x))^2, 且 (0)=0, 则 (int )_(0)^1y(x)dx= __

参考答案与解析:

相关试题

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

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

  • 查看答案
  • 已知 '(x)cdot (int )_(0)^2f(x)dx=8, 且 (0)=0, (x)geqslant 0, 则 f(x)= __

    已知 (x)cdot (int )_(0)^2f(x)dx=8, 且 (0)=0, (x)geqslant 0, 则 f(x)= __

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

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

  • 查看答案
  • 计算(int )_(0)^1dx(int )_(1-x)^sqrt (1-{x^2)}dfrac (x+y)({x)^2+(y)^2}dy=-|||-dv=__________.

    计算(int )_(0)^1dx(int )_(1-x)^sqrt (1-{x^2)}dfrac (x+y)({x)^2+(y)^2}dy=-|||-dv=__

  • 查看答案
  • 3.已知 '(x)=(e)^-x ,且 f(0)=0 ,则 int f(-x)dx= ( ).()-|||-

    3.已知 (x)=(e)^-x ,且 f(0)=0 ,则 int f(-x)dx= ( ).()-|||-

  • 查看答案
  • (1) (int )_(0)^2dfrac (dx)({(1-x))^2}

    (1) (int )_(0)^2dfrac (dx)({(1-x))^2}

  • 查看答案
  • (int )_(0)^2dfrac (dx)({(1-x))^2};

    (int )_(0)^2dfrac (dx)({(1-x))^2};;

  • 查看答案
  • (int )_(0)^2dfrac (dx)({(1-x))^2};

    (int )_(0)^2dfrac (dx)({(1-x))^2};

  • 查看答案
  • 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

    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_

  • 查看答案
  • (int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx=( ) .(int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx=(

    (int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx=( ) .(int )_(0)^dfrac (pi {4)}dfr

  • 查看答案