Question

高数题，求解！

这题怎么写？

Answer 1

积分值为√3－1+π/12

定积分求值。先把绝对值去了，也就是要把积分区间分解。然后在每个区间上去绝对值然后积分。详细过程如图

Answer 2

∫(0->π/2) | 1/2 -sinx| dx
=∫(0->π/6) ( 1/2 -sinx) dx - ∫(π/6->π/2) ( 1/2 -sinx) dx
=[(1/2)x +cosx]|(0->π/6) - [(1/2)x +cosx]|(π/6->π/2)
= (π/12 +√3/2) - 1 - [ π/4 -(π/12 +√3/2)]
= π/6 +√3 - 1 -π/4
=(√3 - 1) -π/12

Answer 3

解:把区间分为(0,π/6),(π/6,π/2)

∫(0,π/2)|(1/2)-sinx| dx

=∫(0,π/6)[(1/2)-sinx]dx+∫(π/6,π/2)[sinx-(1/2)]dx

=[(x/2)+cosx]|(0,π/6)+[-cosx-(x/2)]|(π/6,π/2)

=(√3)-1-π/12

Answer 4

变成两个积分之和。（0，π/4）绝对值里面是大于零，然后小于零

Answer 5

详情如图所示

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