设f(x)=∫0x(ecost—e—cost)dt,则( )

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问题 设f(x)=∫0x(ecost—e—cost)dt,则(     )

选项 A、f(x)=f(x+2π)
B、f(x)>f(x + 2π)
C、f(x)< f(x +2π)
D、当x>0时,f(x)>f(x +2π);当x<0时,f(x)<f(x+2π)

答案A

解析 考查f(x +2π)—f(x)=∫xx+2π(ecost—e—cost)dt,被积函数以2π为周期且为偶函数,由周期函数的积分性质得
f(x+2π)—f(x)=∫—ππ(ecost—e—cost)dt=2∫0π(ecost—e—cost)dt
2∫0π(ecost—e—cost)du,
因此,f(x+2π)— f(x)=0,故选A。
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