2.计算定积分 ∫π/2到0 xcosxdx

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/29 00:08:09
2.计算定积分 ∫π/2到0 xcosxdx

2.计算定积分 ∫π/2到0 xcosxdx
2.计算定积分 ∫π/2到0 xcosxdx

2.计算定积分 ∫π/2到0 xcosxdx
xsinx的导数是多少?
(xsinx)'=xcos+sinx 那么就把题目中的积分构造一个xsinx吧!
∫xcosxdx
=∫(xcosx+sinx)dx-∫sinxdx
=xsinx+cosx
所以答案就是
(π/2*1+0)-(0+1)=π/2-1

∫π/2到0 xcosxdx
原式=∫π/2到0 xdsinx
=xsinx-∫π/2到0 sinxdx
=xsinx+∫π/2到0 dcosx
=(xsinx+cosx)|π/2到0
=[π/2*sin(π/2)+cos(π/2)]-[0*sin0+cos0]
=π/2-1

全部展开

∫π/2到0 xcosxdx
原式=∫π/2到0 xdsinx
=xsinx-∫π/2到0 sinxdx
=xsinx+∫π/2到0 dcosx
=(xsinx+cosx)|π/2到0
=[π/2*sin(π/2)+cos(π/2)]-[0*sin0+cos0]
=π/2-1

【中学生数理化】团队为您解答!祝您学习进步
不明白可以追问!
满意请点击下面的【选为满意回答】按钮,O(∩_∩)O谢谢

收起

原式= ∫π/2到0 xcosxdx
= ∫π/2到0 xdsinx
=x*sinx|(π/2,0)- ∫π/2到0 sinxdx
=π/2+cosx|(π/2,0)
=π/2-1

∫(0->π/2) xcosxdx
=∫(0->π/2) xdsinx
= [xsinx](0->π/2) -∫(0->π/2) sinxdx
=π/2 +[cosx](0->π/2)
=π/2 -1