设S(x)=∫0x|cost|dt. (1)证明:当nπ≤x<(n+1)π时,2n≤S(x)<2(n+1); (2)求.

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问题 设S(x)=∫0x|cost|dt.
(1)证明:当nπ≤x<(n+1)π时,2n≤S(x)<2(n+1);
(2)求

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答案(1)当nπ≤x<(n+1)π时,∫0|cost|dt≤∫0x|cost|dt<∫0(n+1)π|cost|dt, ∫0|cost|dt=n∫0π|cost|dt=n[*]|cost|dt=2n[*]costdt=2n, ∫0(n+1)π|cost|dt=2(n+1),则2n≤S(x)<2(n+1) (2)由nπ≤x<(n+1)π,得[*], 从而[*],根据夹逼定理得[*].

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