• 2022-06-19
    设f(x)在[0,π]上二阶连续可导,且f(π)=2满足∫π0(f(x)+f″(x))sinxdx=5,试计算f(0)的值.
  • 由于∫π0f(x)sinxdx=∫π0f(x)(-cosx)′dx=f(π)+f(0)+∫π0f′(x)cosθdx,∫π0f″(x)sinxdx=f′(x)sinx|π0-∫π0f′(x)cosxdx=-∫π0f′(x)cosxdx,∴5=f(π)+f(0)=2+f(0),∴f(0)=3.

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    • 0

      设f(x)二阶可导,且u=f(xy)满足∂2u∂x∂y=0,f(x)=(  )

    • 1

      设f(x)二阶连续可导,且f(0)=1,f(2)=3,f"(2)=5,则∫01xf"(2x)dx=_______.

    • 2

      设f(x)是连续函数,并满足∫f(x)sinxdx=cos2x+C;又F(x)是f(x)的原函数,且满足F(0)=0,则F(x)=___________.

    • 3

      设函数f(x)在[0,1]上连续,在(0,1)内可导,且f"(x)<0,则____ A: f(0)<0 B: f(1)>0 C: f(1)>f(0) D: f(1)<f(0)

    • 4

      设f(x)可导,且F(x)=f(x)(1+|sinx|)在x=0处可导,则______. A: f(0)=0 B: f"(0)=0 C: f(0)=f"(0) D: f(0)=-f"(0)