求解道定积分题,要详解,
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/10 17:14:15
求解道定积分题,要详解,
![](http://img.wesiedu.com/upload/6/21/621fa07aba1f88b69e2243f8eb10734c.jpg)
![](http://img.wesiedu.com/upload/6/21/621fa07aba1f88b69e2243f8eb10734c.jpg)
![求解道定积分题,要详解,](/uploads/image/z/16736187-3-7.jpg?t=%E6%B1%82%E8%A7%A3%E9%81%93%E5%AE%9A%E7%A7%AF%E5%88%86%E9%A2%98%2C%E8%A6%81%E8%AF%A6%E8%A7%A3%2C)
∫(0,1)(√(1-(1-x)^2)-x)dx
令1-x=cost,则√(1-(1-x)^2)=sint,x=1-cost,dx=sintdt,
原式=∫(0, π/2)(sint-1+cost)sintdt=-π/2+∫(0, π/2)sint^2dt+∫(0, π/2)sintcostdt
=-π/2+1/2*∫(0, π/2)(1-cos2t)dt+1/2*∫(0, π/2)sin2tdt
=-π/2+1/2*(t-1/2*sin2t)+1/2*(-1/2*cos2t) (0, π/2)
= -π/2+1/2*(π/2-1/2*(-1+1))
= -π/2+π/4
=-π/4
再问: 答案不是这么多,但给了我很大的提示,还是谢谢你了。
令1-x=cost,则√(1-(1-x)^2)=sint,x=1-cost,dx=sintdt,
原式=∫(0, π/2)(sint-1+cost)sintdt=-π/2+∫(0, π/2)sint^2dt+∫(0, π/2)sintcostdt
=-π/2+1/2*∫(0, π/2)(1-cos2t)dt+1/2*∫(0, π/2)sin2tdt
=-π/2+1/2*(t-1/2*sin2t)+1/2*(-1/2*cos2t) (0, π/2)
= -π/2+1/2*(π/2-1/2*(-1+1))
= -π/2+π/4
=-π/4
再问: 答案不是这么多,但给了我很大的提示,还是谢谢你了。