∫ [上限π,下限0] |cosx| dx
问题描述：
∫ [上限π,下限0] |cosx| dx
答
∫(0,π)|cosx|dx
= ∫(0,π/2)cosxdx + ∫(π/2,π)(-cosx)dx
= [sin(π/2)-sin0] -[sinπ-sin(π/2)]
= 2