求文档: 求曲线y=sinx,y=cosx与直线x=-π/4,x=π/4所围成图形的面积

网友回答

当x∈[-π/4,π/4]时,有cosx>sinx∴A=∫(cosx-sinx)dx 积分限为[-π/4,π/4]
=sinx+cosx
=[sin(π/4)+cos(π/4)]-[sin(-π/4)+cos(-π/4)]
=[√2/2+√2/2]-[-√2/2+√2/2]
=√2