求第二题
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(2)
(1)
f(x) = cos2x
f(-x) = f(x)
∫(-π/4->π/4) cos2x dx
=2∫(0->π/4) cos2x dx
=[sin2x]|(0->π/4)
=1
(2)
f(x) =x^4.sinx
f(-x) =-f(x)
∫(-π->π) x^4.sinx dx
=0
(1)
f(x) = cos2x
f(-x) = f(x)
∫(-π/4->π/4) cos2x dx
=2∫(0->π/4) cos2x dx
=[sin2x]|(0->π/4)
=1
(2)
f(x) =x^4.sinx
f(-x) =-f(x)
∫(-π->π) x^4.sinx dx
=0
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