求定积分√(2-x^2),上限为√2,下限为0设t=√(2-x^2),x=√(2-t^2),dx=[(2-x^2)]^(-1/2)dt;当x=0,t=√2,当x=√2,t=0;∫_0^√2[√(2-x^2)]dx=∫_√2^0{t[(2-x^2)]^(-1/2)}dt=(-1/2)∫_√2^0{[(2-x^2)]^(-1/2)}d(2-x^2)=(-1/2)[2-x^2](上为0,

来源:学生作业帮助网 编辑:作业帮 时间:2024/12/02 13:05:27
xSNQ~v ef 0tcI9L6hEhVi)>@VBsk6u3߿Xj|b';vI=${Iȑtng (0m#.DT=d(ن&4]fYТcbEKRM4b>'7~w&2{xq,Fck.r1b}B`Mp*1SO4VF۟ގ\ )WqW?=rPQp.KBUKn {s>&)a]w)c5{# \(L-ƙ J@F{yO<.Iػ>$._KY|vot Ԝ5PTܳң!a:hP7LB!xpA;CSip$_ OG.A:S19|* c-I{EʻtVBZB*v!^-Ae46z|7xH$f8 0{};Nj mFnjs'\t:,;A qm)ԧ$C(b |ږmX\0$KY iT2 Js4 hZ#? G|`q