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2(2) 联立解 y = x^2, y = x 得交点 O(0, 0), A(1, 1)
原式 = ∫<0, 1>dy∫<y, √y>xe^(x^2/y)dx
= ∫<0, 1>dy∫<y, √y>(y/2)e^(x^2/y)d(x^2/y)
= ∫<0, 1>(y/2)dy[e^(x^2/y)]<y, √y> = ∫<0, 1>(y/2)(e-e^y)dy
= (1/2)∫<0, 1>(ey - ye^y)dy
= (e/4)[y^2]<0, 1> - (1/2)∫<0, 1>yde^y
= e/4 - (1/2)[(y-1)e^y]<0, 1> = e/4 + 1/2
原式 = ∫<0, 1>dy∫<y, √y>xe^(x^2/y)dx
= ∫<0, 1>dy∫<y, √y>(y/2)e^(x^2/y)d(x^2/y)
= ∫<0, 1>(y/2)dy[e^(x^2/y)]<y, √y> = ∫<0, 1>(y/2)(e-e^y)dy
= (1/2)∫<0, 1>(ey - ye^y)dy
= (e/4)[y^2]<0, 1> - (1/2)∫<0, 1>yde^y
= e/4 - (1/2)[(y-1)e^y]<0, 1> = e/4 + 1/2
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