求解定积分(2x-x^2)^(1
求解定积分 (2x-x^2)^(1/2)dx 积分上限是1 积分下限是0
∫(2x-x^2)^(1/2)dx
=∫√(-x^2+2x)dx
=∫√[-(x^2-2x+1)+1]dx
=∫√[1-(x-1)^2]dx
令(x-1)=sint,则x=sint+1
那么,dx=d(sint+1)=costdt
且,x=0时,sint=-1;x=1时,sint=0
则t∈[-π/2,0]
此时:√[1-(x-1)^2]=√(1-sin^2 t)=√(cos^2 t)=|cost|=cost
则原定积分=∫cost*costdt
=∫cos^2 tdt
=∫[(cos2t+1)/2]dt
=(1/2...全部
求解定积分 (2x-x^2)^(1/2)dx 积分上限是1 积分下限是0
∫(2x-x^2)^(1/2)dx
=∫√(-x^2+2x)dx
=∫√[-(x^2-2x+1)+1]dx
=∫√[1-(x-1)^2]dx
令(x-1)=sint,则x=sint+1
那么,dx=d(sint+1)=costdt
且,x=0时,sint=-1;x=1时,sint=0
则t∈[-π/2,0]
此时:√[1-(x-1)^2]=√(1-sin^2 t)=√(cos^2 t)=|cost|=cost
则原定积分=∫cost*costdt
=∫cos^2 tdt
=∫[(cos2t+1)/2]dt
=(1/2)∫(cos2t+1)dt
=(1/4)(sin2t)|+(1/2)(t)|
=(1/4)*(0-0)+(1/2)[0-(-π/2)]
=π/4。
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