求不定积分∫[x√(4-x²)]dx

来源:学生作业帮助网 编辑:作业帮 时间:2024/12/01 15:17:28
求不定积分∫[x√(4-x²)]dx
x){Ɏާf=_iGۣ:fiV)kƦT$RΆSPقO? Rg$HhŵqF aH;LF 16V\h/

求不定积分∫[x√(4-x²)]dx
求不定积分∫[x√(4-x²)]dx

求不定积分∫[x√(4-x²)]dx
原式=1/2∫根号下(4-x^2)dx^2=1/2∫根号下(4-t)dt=-1/2*2/3*(4-t)^(3/2)+C=-1/3*(4-x^2)^(3/2)+C