1、设f(x)连续,且f(x)=x+2∫(下限为0,上限为pai/2)f(x)cosxdx ,求f(x);
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/24 10:03:53
1、设f(x)连续,且f(x)=x+2∫(下限为0,上限为pai/2)f(x)cosxdx ,求f(x);
2、设f(x)在R上可导且f(0)=0,f'(x)>=0,证明(∫(下限为0,上限为x)f(t)dt)^2
2、设f(x)在R上可导且f(0)=0,f'(x)>=0,证明(∫(下限为0,上限为x)f(t)dt)^2
1.
f(x)=x+2∫f(x)dsinx
=x+2sinxf(x)-2∫sinxf'(x)dx
f'(x)=[x+2∫f(x)dsinx ]'=1
f(0)=0
f(π/2)=π/2+2∫f(π/2)dx=π/2+2(0-π/2)f(π/2)
f(π/2)=π/2-πf(π/2)
f(π/2)=π/2(1-π)
f(x)=x+π/2(1-π)-2∫sinxdx
=x+π/2(1-π)+2cos0-2cosπ/2
=x+π/2(1-π)+2
f(x)=x+2∫f(x)dsinx
=x+2sinxf(x)-2∫sinxf'(x)dx
f'(x)=[x+2∫f(x)dsinx ]'=1
f(0)=0
f(π/2)=π/2+2∫f(π/2)dx=π/2+2(0-π/2)f(π/2)
f(π/2)=π/2-πf(π/2)
f(π/2)=π/2(1-π)
f(x)=x+π/2(1-π)-2∫sinxdx
=x+π/2(1-π)+2cos0-2cosπ/2
=x+π/2(1-π)+2
设f(x)=x+2∫f(t)dt,积分上限是1,下限是0 其中f(x)为连续函数,求f(x)
设函数F(X)具有二阶连续导数,且满足F(X)=[微分(上限X下限0)F(1-t)dt]+1,求F(X)
设函数f(x)可导,且满足f(x)-∫(上限为x,下限为0)f(t)dt=e^x,求f(x) 需要详解,
设f(2x)=xe^x,求∫f(x)dx 上限为6,下限为0
设f(x)为可导函数,且满足∫(上限为x下限为0)tf(t)dt=x^2+f(x),求f(x)
设函数f(x)在【0,1】上连续,在(0,1)内可导,且3∫f(x)dx=f(0),(上限为1,下限为2/3),证明:
3.设f(x)是连续函数,且:∫(0为下限,x为上限)(x-t)f(t)dt=ln(x+根号(1+x^2)),求f(x)
设函数f(x)可导,且满足f(x)=1+2x+∫(上限x下限0)tf(t)dt-x∫(上限x下限0)f(t)dt,试求函
f(x)=定积分上限为x^2,下限为0的cos2tdt,则f'(根号下pai/x)=
急.f(x)为连续的偶函数,求证∫(上限为a,下限为-a)f(x)dx=2∫(上限为a,下限为0)f(x)dx
设函数f(x)具有连续一阶导数,且满足f(x)=∫(上限是x下限是0)(x^2-t^2)f^,(t)dt+x^2求f(x
设f(x)为连续函数且F(x)=∫f(t)dt上限为lnx下限为1/x 则F'(x)=?