根号下[1-(x-1)^2] - x 用定积分的性质和几何意义求出这个的值

√[1-(x-1)^2]-x楼主没有说积分上下限,这里x∈(0,1)∫√[1-(x-1)^2]-xdx=∫√[1-(x-1)^2]dx-x^2/2设x-1=sint则原式=∫(cost)^2dt-x^2/2=1/2*∫(1+cos2t)dt-x^2/2=t/2+(sin2t)/4-x^2/2+C这里t∈(-π/2,0),x∈(0,1)定...