设连续函数f(x)>0且单调递增,则积分I1=∫0π/2f(x)sinxdx,I2=∫0π/2f(x)cosxdx,I3=∫0π/2d(x)tanxdx的大小关系为( )

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问题 设连续函数f(x)>0且单调递增,则积分I1=∫0π/2f(x)sinxdx,I2=∫0π/2f(x)cosxdx,I3=∫0π/2d(x)tanxdx的大小关系为(     )

选项 A、I1﹥I2﹥I3
B、I1﹥I3﹥I2
C、I2﹥I3﹥I1
D、I3﹥I1﹥I2

答案D

解析 由于积分区间相同,比较被积函数的大小,
I1-I2=∫0π/2f(x)(sinx-cosx)dx
=∫0π/4f(x)(sinx-cosx)dx+∫π/4π/2f(x)(sinx-cosx)dx,
令x=-t则
π/4π/2f(x)(sinx-cosx)dx=∫0π/4f(-t)[sin(-t)-cos(-t)]dt
    =∫0π/4f (-t)(cosx-sinx)dx,
则有I1-I2=∫0π/4[f(x) -f(-x)] (sinx-cosx)dx,
已知函数d(x)﹥0且单调递增,则f(x)-﹤0,而当0﹤x﹤时,sinx﹤cosx,可以得到
被积函数
从而I1﹥I2
I3-I1=∫0π/2f(x)(tanx-sinx)dx,
当x﹥0时,有tanx-sinx﹥0,又因为f(x)﹥0,所以f(x)(tanx-sinx)﹥0,即I3-I1﹥0,综上可得I3﹥I1﹥I2
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