• 2022-06-14
    求(1-x)^2019的积分
    A: -1/2020(1-x)^2020+C
    B: 2020(1-x)^2020+C
    C: -1/2019(1-x)^2020+C
    D: 1/2019(1-x)^2020+C
  • A

    内容

    • 0

      17da42840675a6d.jpg,计算[img=19x34]17da4275482315f.jpg[/img]实验命令为(). A: syms x;f=diff(asinsqrt(x))f=1/2/x^(1/2)/(1-x)^(1/2) B: f=diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2)

    • 1

      若f′(cos2x)=sinx,则f(x)等于:() A: (1/3)(1-x)+c B: (2/3)(1-x)+c C: -(1/3)(1-x)+c D: (1-x)+c

    • 2

      y=ln(1-x²)的定义域为: -1≤x≤1|-1<x<1|0≤x≤1|0<x<1

    • 3

      已知x不等于1,计算(1+x)(1-x)=1-x^2,(1-x)(1+x+x^2)=1-x^3,(1-x)(1+x+x^2+x^3)=1-x^4

    • 4

      下面级数求和错误的是 A: $\sum_{n=0}^\infty q^n = \frac{1}{1-q} (0\lt q\lt1) $ B: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{x}{1-x} (|x|\lt 1) $ C: $\sum_{n=1}^\infty \frac{1}{{n!}} = e $ D: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{1}{1-x} (x>1) $