求积分？过程谢谢！

显示全部楼层 · 2013-3-16 22:15:02

积分区间给错了吧，是[-1,1]曲线与x轴围成一个半圆，积分结果就是半圆面积π/2.或者直接积分∫√(1-x^2)dx=x√(1-x^2)-∫x·d[√(1-x^2)]=x√(1-x^2)+∫[x^2/(√(1-x^2))]dx(分部积分法)x^2/(√(1-x^2))=1/(√(1-x^2))-√(1-x^2) ∫x^2/(√(1-x^2))dx=∫1/(√(1-x^2))dx-∫√(1-x^2)dx∫√(1-x^2)dx=x√(1-x^2)+∫1/(√(1-x^2))dx-∫√(1-x^2)dx2∫√(1-x^2)dx=x√(1-x^2)+∫1/(√(1-x^2))dx=x√(1-x^2)+arcsinx∫√(1-x...

千问 · 2013-3-16 22:15:02

曲线与x轴围成一个半圆，积分结果就是半圆面积π/2....