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第1个回答 2018-12-07
(4)
∫(-e->-1) 1/x dx
=[ln|x|]|(-e->-1)
=-1
(5)
∫(0->π/2) 2[sin(x/2)]^2 dx
=∫(0->π/2) (1- cosx) dx
= [ x - sinx]|(0->π/2)
=π/2 -1
(6)
∫(0->2π) | sinx| dx
=∫(0->π) sinx dx -∫(π->2π) sinx dx
=-[cosx]|(0->π) +[cosx]|(π->2π)
=4
∫(-e->-1) 1/x dx
=[ln|x|]|(-e->-1)
=-1
(5)
∫(0->π/2) 2[sin(x/2)]^2 dx
=∫(0->π/2) (1- cosx) dx
= [ x - sinx]|(0->π/2)
=π/2 -1
(6)
∫(0->2π) | sinx| dx
=∫(0->π) sinx dx -∫(π->2π) sinx dx
=-[cosx]|(0->π) +[cosx]|(π->2π)
=4
第2个回答 2018-12-07
如图
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