设函数f(x)在R上连续,且满足 f(π4+x)=f(π4−x) (1)证明: ∫0x2f(x)sin2xdx=∫0x2f(x)cos2xd
(2)计算
∫0π2(x−π4)2sin2xdx
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(1) 证明:根据函数f(x)的性质,可以得到f(π/4+x)=f(π/4-x)。将x=π/4+t,则dx=dt,有∫0x2f(x)sin2xdx=∫0t2f(π/4+t)sin2(π/4+t)dt=∫0t2f(π/4+t)(sin2π/4cos2t+cos2π/4sin2t)dt=∫0t2f(π/4+t)(cos2t)dt=∫0x2f(x)cos2xdx(2) 计算∫0π2(x−π4)2sin2xdx=∫0π4(x−π4)2sin2xdx+∫π4π2(x−π4)2sin2xdx=∫0π4(x−π4)2sin2xdx+∫0π4(π-x−π4)2sin2(π-x)dx根据函数f(x)的性质,可以得到f(π-x)=f(x),所以=∫0π4(x−π4)2sin2xdx+∫0π4(π-x−π4)2sin2x dx=∫0π4(x−π4)2sin2xdx+∫0π4(π4-x)2sin2xdx=2∫0π4(x−π4)2sin2xdx=2∫0π4(x2−π8x)sin2xdx=2∫0π4x2sin2xd
咨询记录 · 回答于2022-12-28
∫0π2(x−π4)2sin2xdx
设函数f(x)在R上连续,且满足
f(π4+x)=f(π4−x)
(1)证明:
∫0x2f(x)sin2xdx=∫0x2f(x)cos2xd
(2)计算
设函数f(x)在R上连续,且满足
∫0π2(x−π4)2sin2xdx
(2)计算
∫0x2f(x)sin2xdx=∫0x2f(x)cos2xd
(1)证明:
f(π4+x)=f(π4−x)
设函数f(x)在R上连续,且满足
∫0π2(x−π4)2sin2xdx
(2)计算
∫0x2f(x)sin2xdx=∫0x2f(x)cos2xd
(1)证明:
f(π4+x)=f(π4−x)
设函数f(x)在R上连续,且满足
∫0π2(x−π4)2sin2xdx
(2)计算
∫0x2f(x)sin2xdx=∫0x2f(x)cos2xd
(1)证明:
f(π4+x)=f(π4−x)
设函数f(x)在R上连续,且满足
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