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

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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)

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
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

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
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

1、定积分是一个数,所以可假设f(x)=x+a,易求得a=2-π,所以f(x)=x+2-π
2、两者相减,求导