求曲线y=sinx,y=cosx与直线x=0.x=π⼀2所围成平面图形的面积 (图中阴影部分)

2025-05-11 06:23:32
推荐回答(1个)
回答1:

所求面积=∫(cosx-sinx)dx+∫(sinx-cosx)dx
=(sinx+cosx)│+(-cosx-sinx)│
=(sin(π/4)+cos(π/4)-sin(0)-cos(0))+(-cos(π/2)-sin(π/2)+cos(π/4)+sin(π/4))
=(√2/2+√2/2-0-1)+(-0-1+√2/2+√2/2)
=2(√2-1).