Question

1.求过由曲线y=sinX,y=cosX及直线x=0,x=π/2所围成的图形的面积

Answer 1

1.在区间[0,π/2]上,函数 sinx 与 cosx 交于 (π/4,根号2/2),而在[0,π/4)上 cosx>sinx;在[π/4,π/2]上,sinx>cosx,所以所求面积为
S
=∫(0->π/2) |sinx-cosx| dx
=∫(0->π/4) (cosx-sinx) dx + ∫(π/4->π/2) (sinx-cosx) dx
由于 ∫(cosx-sinx) dx = sinx+cosx+C,所以
∫(0->π/4) (cosx-sinx) dx = (sin π/4 + cos π/4)-(sin0 + cos0)=根号2-1;
同理,∫(sinx-cosx) dx = -sinx-cosx,所以
∫(π/4->π/2) (sinx-cosx) dx = (-sin π/2-cos π/2)-(-sin π/4-cos π/4)=根号2-1;
因此原式=∫(0->π/2) |sinx-cosx| = 2*(根号2-1) = 2根号2-2.
2.y'=27-3x^2.
令y'0 时 y''