第5题怎么做？？？概率论的题目
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∫(0~π/4)∫(0~π/4) sin(x+y) dxdy
= ∫(0~π/4) -cos(π/4+y)+cos(y) dy
= -sin(π/4+y)+sin(y) | (0~π/4)
= -sin(π/2)+sin(π/4)+sin(π/4)-sin(0)
=-1+2/根号2
=根号2-1
c(根号2-1)=1
c=根号2+1
= ∫(0~π/4) -cos(π/4+y)+cos(y) dy
= -sin(π/4+y)+sin(y) | (0~π/4)
= -sin(π/2)+sin(π/4)+sin(π/4)-sin(0)
=-1+2/根号2
=根号2-1
c(根号2-1)=1
c=根号2+1
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