第1个回答 2019-05-30
(3) y = √(2x-x^2), y' = (1-x)/√(2x-x^2)
I = ∫<L> xdx - ye^(2x-x^2)dy
= ∫<0, 1> [x - √(2x-x^2)e^(2x-x^2)(1-x)/√(2x-x^2)]dx
= ∫<0, 1> [x - e^(2x-x^2)(1-x)]dx
= [x^2/2]<0, 1> - (1/2)∫<0, 1> e^(2x-x^2)d(2x-x^2)
= 1/2 - (1/2)[e^(2x-x^2)]<0, 1> = 1 - e/2
(4) I = ∫<L> (y^2-x^2)dx + 2yzdy - x^2dz
= ∫<0, 1> [(t^4-t^2)dt + 2t^5dt^2 - t^2dt^3]
= ∫<0, 1> [(t^4-t^2) + 4t^6 - 3t^4]dt
= ∫<0, 1> (4t^6 - 2t^4 - t^2)dt
= 4/7 - 2/5 -1/3 = -17/105本回答被提问者和网友采纳
I = ∫<L> xdx - ye^(2x-x^2)dy
= ∫<0, 1> [x - √(2x-x^2)e^(2x-x^2)(1-x)/√(2x-x^2)]dx
= ∫<0, 1> [x - e^(2x-x^2)(1-x)]dx
= [x^2/2]<0, 1> - (1/2)∫<0, 1> e^(2x-x^2)d(2x-x^2)
= 1/2 - (1/2)[e^(2x-x^2)]<0, 1> = 1 - e/2
(4) I = ∫<L> (y^2-x^2)dx + 2yzdy - x^2dz
= ∫<0, 1> [(t^4-t^2)dt + 2t^5dt^2 - t^2dt^3]
= ∫<0, 1> [(t^4-t^2) + 4t^6 - 3t^4]dt
= ∫<0, 1> (4t^6 - 2t^4 - t^2)dt
= 4/7 - 2/5 -1/3 = -17/105本回答被提问者和网友采纳
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